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jaoe38c1lseyfkoa | maths | sequences-and-series | geometric-progression-(g.p) | <p>If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <p>$$\begin{aligned}
& a+a r+a r^2+a r^3+\ldots .+a r^{63} \\
& =7\left(a+a r^2+a r^4 \ldots .+a r^{62}\right) \\
& \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\
& r=6
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift | 8,062 |
1lsg3p684 | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$a$$ and $$b$$ be be two distinct positive real numbers. Let $$11^{\text {th }}$$ term of a GP, whose first term is $$a$$ and third term is $$b$$, is equal to $$p^{\text {th }}$$ term of another GP, whose first term is $$a$$ and fifth term is $$b$$. Then $$p$$ is equal to</p> | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "25"}] | ["C"] | null | <p>The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.</p><p>F... | mcq | jee-main-2024-online-30th-january-evening-shift | 8,064 |
luxwd2ht | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$a, a r, a r^2$$, ............ be an infinite G.P. If $$\sum_\limits{n=0}^{\infty} a r^n=57$$ and $$\sum_\limits{n=0}^{\infty} a^3 r^{3 n}=9747$$, then $$a+18 r$$ is equal to</p> | [{"identifier": "A", "content": "27"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "31"}, {"identifier": "D", "content": "46"}] | ["C"] | null | <p>$$\begin{array}{ll}
\sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\
\sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)}
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\
& \Rightarrow ... | mcq | jee-main-2024-online-9th-april-evening-shift | 8,065 |
lv0vxcyd | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let the first three terms 2, p and q, with $$q \neq 2$$, of a G.P. be respectively the $$7^{\text {th }}, 8^{\text {th }}$$ and $$13^{\text {th }}$$ terms of an A.P. If the $$5^{\text {th }}$$ term of the G.P. is the $$n^{\text {th }}$$ term of the A.P., then $n$ is equal to:</p> | [{"identifier": "A", "content": "151"}, {"identifier": "B", "content": "177"}, {"identifier": "C", "content": "163"}, {"identifier": "D", "content": "169"}] | ["C"] | null | <p>$$\begin{aligned}
& \text { Let } p=2 r, q=2 r^2 \\
& T_7=2, T_8=2 r, T_{13}=2 r^2 \\
& d=2 r-2=2(r-1) \\
& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\
& \Rightarrow r^2-6 r+5=0 \\
& \Rightarrow(r-1)(r-5)=0 \\
& \therefore r=1,5 \\
& r=1 \text { (rejected) as } q \neq 2 \\
& \therefore r=5
\end{aligned}$$</p>
<p>$$5^{\text... | mcq | jee-main-2024-online-4th-april-morning-shift | 8,066 |
lv3veby6 | maths | sequences-and-series | geometric-progression-(g.p) | <p>In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $$\frac{70}{3}$$ and the product of the third and fifth terms is 49. Then the sum of the $$4^{\text {th }}, 6^{\text {th }}$$ and $$8^{\text {th }}$$ terms is equal to:</p> | [{"identifier": "A", "content": "78"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "91"}, {"identifier": "D", "content": "84"}] | ["C"] | null | <p>Let's denote the first term of the geometric progression by $$a$$ and the common ratio by $$r$$. The terms of the geometric progression can be written as follows:</p>
<p>First term: $$a$$</p>
<p>Second term: $$ar$$</p>
<p>Third term: $$(ar^2)$$</p>
<p>Fourth term: $$(ar^3)$$</p>
<p>Fifth term: $$(ar^4)$$</p>
<... | mcq | jee-main-2024-online-8th-april-evening-shift | 8,067 |
lvb294nj | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$A B C$$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $$A B C$$ and the same process is repeated infinitely many times. If $$\mathrm{P}$$ is the sum of perimeters and $$Q$$ is be the sum of areas of all the triangles formed in this process, then ... | [{"identifier": "A", "content": "$$\\mathrm{P}^2=72 \\sqrt{3} \\mathrm{Q}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{P}^2=36 \\sqrt{3} \\mathrm{Q}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{P}=36 \\sqrt{3} \\mathrm{Q}^2$$\n"}, {"identifier": "D", "content": "$$\\mathrm{P}^2=6 \\sqrt{3} \\mathrm{Q}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11ef-aad1-15919f5484e3/file-1lwac04hp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11ef-aad1-15919f5484e3... | mcq | jee-main-2024-online-6th-april-evening-shift | 8,068 |
VdoDcdBuFZIwxtlr | maths | sequences-and-series | harmonic-progression-(h.p) | If $${{a_1},{a_2},....{a_n}}$$ are in H.P., then the expression $${{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$$ is equal to | [{"identifier": "A", "content": "$$n({a_1}\\, - {a_n})$$ "}, {"identifier": "B", "content": "$$(n - 1)({a_1}\\, - {a_n})$$ "}, {"identifier": "C", "content": "$$n{a_1}{a_n}$$ "}, {"identifier": "D", "content": "$$(n - 1)\\,\\,{a_1}{a_n}$$ "}] | ["D"] | null | $${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)
<br><br>Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$
<br><br>$$\,\,\,\,\,\,\,\,\,\... | mcq | aieee-2006 | 8,070 |
1l57nygjf | maths | sequences-and-series | harmonic-progression-(h.p) | <p>$$x = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } } $$, where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc $$\ne$$ 0, then :</p> | [{"identifier": "A", "content": "x, y, z are in A.P."}, {"identifier": "B", "content": "x, y, z are in G.P."}, {"identifier": "C", "content": "$${1 \\over x}$$, $${1 \\over y}$$, $${1 \\over z}$$ are in A.P."}, {"identifier": "D", "content": "$${1 \\over x}$$ + $${1 \\over y}$$ + $${1 \\over z}$$ = 1 $$-$$ (a + b + c)"... | ["C"] | null | <p>$$x = \sum\limits_{n = 0}^\infty {{a^n} = {1 \over {1 - a}};\,y = \sum\limits_{n = 0}^\infty {{b^n} = {1 \over {1 - b}};\,z = \sum\limits_{n = 0}^\infty {{c^n} = {1 \over {1 - c}}} } } $$</p>
<p>Now,</p>
<p>a, b, c $$\to$$ AP</p>
<p>1 $$-$$ a, 1 $$-$$ b, 1 $$-$$ c $$\to$$ AP</p>
<p>$${1 \over {1 - a}},\,{1 \over ... | mcq | jee-main-2022-online-27th-june-morning-shift | 8,071 |
r5pz1x4onggAO3Gd | maths | sequences-and-series | summation-of-series | The value of $$\,{2^{1/4}}.\,\,{4^{1/8}}.\,{8^{1/16}}...\infty $$ is | [{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "3/2 "}, {"identifier": "D", "content": "4"}] | ["B"] | null | The product is $$p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$$
<br><br>$$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$
<br><br>Now let
<br><br>$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$
<br><br>$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,... | mcq | aieee-2002 | 8,073 |
lUgZFx9XzBXy4yRV | maths | sequences-and-series | summation-of-series | The sum of the serier $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............$$ up to $$\infty $$ is equal to | [{"identifier": "A", "content": "$$\\log {\\,_e}\\left( {{4 \\over e}} \\right)\\,\\,$$ "}, {"identifier": "B", "content": "$$2\\,\\log {\\,_e}2$$ "}, {"identifier": "C", "content": "$$\\log {\\,_e}2 - 1\\,$$ "}, {"identifier": "D", "content": "$$\\log {\\,_e}2$$ "}] | ["A"] | null | $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $$
<br><br>$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$
<br><br>$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$
<br><br>$$ = \left( {{1 \over 1} - {1 \over 2}} \rig... | mcq | aieee-2003 | 8,074 |
PL99UcmXmNOIsJPr | maths | sequences-and-series | summation-of-series | The sum of the first n terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}$$ when n is even. When n is odd the sum is | [{"identifier": "A", "content": "$${\\left[ {{{n(n + 1)} \\over 2}} \\right]^2}$$ "}, {"identifier": "B", "content": "$${{{n^2}(n + 1)} \\over 2}$$ "}, {"identifier": "C", "content": "$${{n{{(n + 1)}^2}} \\over 4}$$ "}, {"identifier": "D", "content": "$$\\,{{3n(n + 1)} \\over 2}$$ "}] | ["B"] | null | If $$n$$ is odd, the required sum is
<br><br>$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$
<br><br>$$ = {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$
<br><br>[ As $$\left( {n - 1} \right)$$ is even
<br... | mcq | aieee-2004 | 8,075 |
n9ha6c4bh8MQnUGo | maths | sequences-and-series | summation-of-series | The sum of series $${1 \over {2\,!}} + {1 \over {4\,!}} + {1 \over {6\,!}} + ........$$ is | [{"identifier": "A", "content": "$${{\\left( {{e^2} - 2} \\right)} \\over e}\\,$$ "}, {"identifier": "B", "content": "$${{{{\\left( {e - 1} \\right)}^2}} \\over {2e}}$$ "}, {"identifier": "C", "content": "$${{\\left( {{e^2} - 1} \\right)} \\over {2e}}\\,$$ "}, {"identifier": "D", "content": "$${{\\left( {{e^2} - 1} \\r... | ["B"] | null | We know that
<br><br>$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$
<br><br>and
<br><br>$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$
<br><br>$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$$
<br><br>$$\the... | mcq | aieee-2004 | 8,076 |
M5gWDJeqiWRmB7hX | maths | sequences-and-series | summation-of-series | The sum of the series $$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$ ad inf. is | [{"identifier": "A", "content": "$${{e - 1} \\over {\\sqrt e }}\\,$$ "}, {"identifier": "B", "content": "$${{e + 1} \\over {\\sqrt e }}$$ "}, {"identifier": "C", "content": "$${{e - 1} \\over {2\\sqrt e }}$$ "}, {"identifier": "D", "content": "$${{e + 1} \\over {2\\sqrt e }}$$ "}] | ["D"] | null | $${{{e^x} + {e^{ - x}}} \over 2}$$
<br><br>$$ = 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........$$
<br><br>Putting $$x = {1 \over 2}$$ we get
<br><br>$$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$
<br><br>$$\infty = {{{e^{{1 \over 2}}} + {e^{{{ - 1} \over 2}}}... | mcq | aieee-2005 | 8,077 |
4spZL4XKlmGNYvXy | maths | sequences-and-series | summation-of-series | The sum of series $${1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - .......$$ upto infinity is | [{"identifier": "A", "content": "$${e^{ - {1 \\over 2}}}$$ "}, {"identifier": "B", "content": "$${e^{ + {1 \\over 2}}}$$"}, {"identifier": "C", "content": "$${e^{ - 2}}$$ "}, {"identifier": "D", "content": "$${e^{ - 1}}$$"}] | ["D"] | null | We know that $${e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty $$
<br><br>Put $$x=-1$$
<br><br>$$\therefore$$ $${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty $$
<br><br>$$\therefore$$ $${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {... | mcq | aieee-2007 | 8,078 |
dEB3aGohwKvD17AB | maths | sequences-and-series | summation-of-series | The sum to infinite term of the series $$1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + .....$$ is | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["A"] | null | We have
<br><br>$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)$$
<br><br>Multiplying both sides by $${1 \over 3}$$ we get
<br><br>$${1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over {{3^4}}} + .......\... | mcq | aieee-2009 | 8,079 |
HGW0sDmHzjbMJRmV | maths | sequences-and-series | summation-of-series | <p> <b> Statement-1: </b> The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000.
</p><p> <b> Statement-2: </b> $$\sum\limits_{k = 1}^n {\left( {{k^3} - {{(k - 1)}^3}} \right)} = {n^3}$$, for any natural number n.</p>
| [{"identifier": "A", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explana... | ["B"] | null | $$n$$<sup>th </sup> term of the given series
<br><br>$$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$
<br><br>$$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$
<br><br>$$ = {n^3} - {\left( {n - 1} \right)^3}$$
<br><br>$$ \Rightarrow {S_n} = \s... | mcq | aieee-2012 | 8,080 |
4PDGIePDoIDBwloQ | maths | sequences-and-series | summation-of-series | If $${(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},$$, then k is equal to : | [{"identifier": "A", "content": "100 "}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "$${{121} \\over {10}}$$ "}, {"identifier": "D", "content": "$${{441} \\over {100}}$$ "}] | ["A"] | null | Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$
<br><br>Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\... | mcq | jee-main-2014-offline | 8,082 |
ZdSGnALvwlIDEdX3 | maths | sequences-and-series | summation-of-series | The sum of first 9 terms of the series.
<br/><br/>$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$ | [{"identifier": "A", "content": "142"}, {"identifier": "B", "content": "192"}, {"identifier": "C", "content": "71"}, {"identifier": "D", "content": "96"}] | ["D"] | null | $${n^{th}}$$ term of series
<br><br>$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$
<br><br>Sum of $$n$$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$
<br><br>$$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$$
<br><br>$$ = {... | mcq | jee-main-2015-offline | 8,083 |
FaNkL9CszrtfuIWo | maths | sequences-and-series | summation-of-series | If the sum of the first ten terms of the series $${\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + .......is\,{{16} \over 5}m,$$ then m is equal to : | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "99"}, {"identifier": "C", "content": "102"}, {"identifier": "D", "content": "101"}] | ["D"] | null | $${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$$
<br><br>$$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}... | mcq | jee-main-2016-offline | 8,084 |
xuwfSpE7OV4mQBpm | maths | sequences-and-series | summation-of-series | For any three positive real numbers a, b and c,
<br/><br/>9(25$${a^2}$$ + b<sup>2</sup>) + 25(c<sup>2</sup> - 3$$a$$c) = 15b(3$$a$$ + c).
<br/>Then | [{"identifier": "A", "content": "b, c and $$a$$ are in G.P."}, {"identifier": "B", "content": "b, c and $$a$$ are in A.P."}, {"identifier": "C", "content": "$$a$$, b and c are in A.P."}, {"identifier": "D", "content": "$$a$$, b and c are in G.P."}] | ["B"] | null | 9(25$${a^2}$$ + b<sup>2</sup>) + 25(c<sup>2</sup> - 3$$a$$c) = 15b(3$$a$$ + c)
<br><br> $$ \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc$$
<br><br>$$ \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc$$
<br><br>$$ \Rightarrow $$ $${1 \over 2}\lef... | mcq | jee-main-2017-offline | 8,086 |
cNdhbmgFjjDuKylw4LXXW | maths | sequences-and-series | summation-of-series | Let
<br/><br/>S<sub>n</sub> = $${1 \over {{1^3}}}$$$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$$
<br/><br/>If 100 S<sub>n</sub> = n, then n is equal to : | [{"identifier": "A", "content": "199"}, {"identifier": "B", "content": "99"}, {"identifier": "C", "content": "200"}, {"identifier": "D", "content": "19"}] | ["A"] | null | n<sup>th</sup> term, T<sub>n</sub> = $${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$$
<br><br>T<sub>n</sub> = $${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ T<sub>n</sub> = $${2 \over {n\left( {n + 1} \right)}}$$ = $$... | mcq | jee-main-2017-online-9th-april-morning-slot | 8,087 |
OME7QaClOHVLgwxtyldwp | maths | sequences-and-series | summation-of-series | If the sum of the first n terms of the series $$\,\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$$ is $$435\sqrt 3 ,$$ then n equals : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "29"}] | ["B"] | null | Given,
<br><br>$$\sqrt 3 $$ + $$\sqrt {75} $$ + $$\sqrt {243} $$ + $$\sqrt {507} $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ + $$\sqrt {25 \times 3} $$ + $$\sqrt {81 \times 3} $$ + $$\sqrt {169 \times 3} $$ + . . . . . .+ n terms
<br><br>= $$\sqrt 3 $$ + 5$$\sqrt 3 $$ + 9$$\sqrt 3 $$ + 13$$\sqrt 3 $$ + . . . . .... | mcq | jee-main-2017-online-8th-april-morning-slot | 8,088 |
OJSjRBTgKunyFAcA | maths | sequences-and-series | summation-of-series | Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
<br/>1<sup>2</sup> + 2.2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup>2</sup> + 5<sup>2</sup> + 2.6<sup>2</sup> ...........
<br/>If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to | [{"identifier": "A", "content": "496"}, {"identifier": "B", "content": "232"}, {"identifier": "C", "content": "248"}, {"identifier": "D", "content": "464"}] | ["C"] | null | <b><u>Note</u> : </b>
<br><br>Sum of square of first n odd terms
<br><br>1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> + . . . . .+ n<sup>2</sup> = $${{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}$$
<br><br>Given,
<br><br>1<sup>2</sup> + 2. 2<sup>2</sup> + 3<sup>2</sup> + 2.4<sup>2</sup> + 5<sup>2</su... | mcq | jee-main-2018-offline | 8,089 |
QtfYNrtuTFProPJ2lotH6 | maths | sequences-and-series | summation-of-series | The sum of the first 20 terms of the series
<br/><br/>$$1 + {3 \over 2} + {7 \over 4} + {{15} \over 8} + {{31} \over {16}} + ...,$$ is : | [{"identifier": "A", "content": "$$38 + {1 \\over {{2^{19}}}}$$"}, {"identifier": "B", "content": "$$38 + {1 \\over {{2^{20}}}}$$"}, {"identifier": "C", "content": "$$39 + {1 \\over {{2^{20}}}}$$"}, {"identifier": "D", "content": "$$39 + {1 \\over {{2^{19}}}}$$"}] | ["A"] | null | 1 + $${3 \over 2}$$ + $${7 \over 4}$$ + $${15 \over 8}$$ + $${31 \over 16}$$ + . . . .
<br><br>= (2 $$-$$ 1) + (2 $$-$$ $${1 \over 2}$$ ) + (2 $$-$$ $${1 \over 4}$$) + (2 $$-$$ $${1 \over 8}$$) + . . . . .+ 20 terms
<br><br>= (2 + 2 + . . . . . 20 terms) $$-$$ (1 + $${1 \over 2}$$ + $${1 \over 4}$$ + .... | mcq | jee-main-2018-online-16th-april-morning-slot | 8,090 |
BLiwgey2PSssP0DBRJxwz | maths | sequences-and-series | summation-of-series | Let A<sub>n</sub> = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$$ $$-$$. . . . . + ($$-$$1)<sup>n-1</sup> $${\left( {{3 \over 4}} \right)^n}$$ and B<sub>n</sub> = 1 $$-$$ A<sub>n</sub>.
<br/>Then, the least dd natural numbr p, so that B<sub>n</sub> >... | [{"identifier": "A", "content": "9 "}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "5"}] | ["B"] | null | A<sub>n</sub> = $$\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3} - .... + {\left( { - 1} \right)^{n - 1}}{\left( {{3 \over 4}} \right)^n}$$
<br><br>Which in a G.P. with a = $${{3 \over 4}}$$, r = $${{{ - 3} \over 4}}$$ and number of terms = n
<br><br>$$\therefore\,\,... | mcq | jee-main-2018-online-15th-april-evening-slot | 8,091 |
VB4N8NG5lJI3jk4QxM3rsa0w2w9jx5edf2b | maths | sequences-and-series | summation-of-series | For x $$\varepsilon $$ R, let [x] denote the greatest integer $$ \le $$ x, then the sum of the series
$$\left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right]$$ is : | [{"identifier": "A", "content": "- 153"}, {"identifier": "B", "content": "- 135"}, {"identifier": "C", "content": "- 133"}, {"identifier": "D", "content": "- 131"}] | ["C"] | null | $$\left[ {{{ - 1} \over 3}} \right] + \left[ {{{ - 1} \over 3} - {1 \over {100}}} \right] + \left[ {{{ - 1} \over 3} - {2 \over {100}}} \right] +$$<br><br>
$$ ....... + \left[ {{{ - 1} \over 3} - {{99} \over {100}}} \right]$$<br><br>
$$ \Rightarrow ( - 1 - 1 - 1 - .....67\,times) + ( - 2 - 2 - 2 - .....33\,times)$$<br... | mcq | jee-main-2019-online-12th-april-morning-slot | 8,092 |
JcmjVvCtosFouYLOtS3rsa0w2w9jx1zojez | maths | sequences-and-series | summation-of-series | The sum
<br/>$$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$$ is equal to : | [{"identifier": "A", "content": "620"}, {"identifier": "B", "content": "1240"}, {"identifier": "C", "content": "1860"}, {"identifier": "D", "content": "660"}] | ["A"] | null | $$Sum = \sum\limits_{n = 1}^{15} {{{{1^3} + {2^3} + .... + {n^3}} \over {1 + 2 + .... + n}}} - {1 \over 2}{{15 \times 16} \over 2}$$<br><br>
$$ = \sum\limits_{n = 1}^{15} {{{n(n + 1)} \over 2}} - 60$$<br><br>
$$ = {1 \over 2}\sum\limits_{n = 1}^{15} {{n^2}} + {1 \over 2}\sum\limits_{n = 1}^{15} {n - 60} $$<br><br>
$... | mcq | jee-main-2019-online-10th-april-evening-slot | 8,093 |
HCoDSSB4LAVjlrMorg3rsa0w2w9jwxuogih | maths | sequences-and-series | summation-of-series | The sum
<br/>$${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$$ upto 10 terms is: | [{"identifier": "A", "content": "600"}, {"identifier": "B", "content": "660"}, {"identifier": "C", "content": "680"}, {"identifier": "D", "content": "620"}] | ["B"] | null | $${T_r} = {{(2r + 1)({1^3} + {2^3} + {3^3} + ...... + {r^3})} \over {{1^2} + {2^2} + {3^2} + ...... + {r^2}}}$$<br><br>
$${T_r} = (2r + 1){\left( {{{r(r + 1)} \over 2}} \right)^2} \times {6 \over {r(r + 1)(2r + 1)}}$$<br><br>
$${T_r} = {{3r(r + 1)} \over 2}$$<br><br>
Now, $$S = \sum\limits_{r = 1}^{10} {{T_r}} = {3 \o... | mcq | jee-main-2019-online-10th-april-morning-slot | 8,094 |
7PpCk7Hi0qXclaNmDM18hoxe66ijvww22ey | maths | sequences-and-series | summation-of-series | The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term is :- | [{"identifier": "A", "content": "945"}, {"identifier": "B", "content": "916"}, {"identifier": "C", "content": "915"}, {"identifier": "D", "content": "946"}] | ["D"] | null | S = 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term
<br><br>General term T<sub>n</sub> = n (2n - 1)
<br><br>$$ \therefore $$ S<sub>n</sub> = $$\sum {{T_n}} $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$\sum {\left( {2{n^2} - n} \right)} $$
<br><br>$$ \Rightarrow $$ S<sub>n</sub> = $$2\sum {{n^2} - \sum r } $$
<br><br>$... | mcq | jee-main-2019-online-9th-april-evening-slot | 8,095 |
50wym4bFg664eLPHoskkF | maths | sequences-and-series | summation-of-series | The sum
$$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $$ is equal to | [{"identifier": "A", "content": "$$2 - {11 \\over {{2^{19}}}}$$"}, {"identifier": "B", "content": "$$2 - {3 \\over {{2^{17}}}}$$"}, {"identifier": "C", "content": "$$1 - {11 \\over {{2^{20}}}}$$"}, {"identifier": "D", "content": "$$2 - {21 \\over {{2^{20}}}}$$"}] | ["A"] | null | Let S = $$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $$
<br><br>$$ \Rightarrow $$ S = $${1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ....... + {{20} \over {{2^{20}}}}$$ .......(1)
<br><br>This is an Arithmetic Geometric Sequence. Here numerator is in A.P and denominator is in G.P.
<br><b... | mcq | jee-main-2019-online-8th-april-evening-slot | 8,096 |
wUMtiArnVk8GEbYuJ01FI | maths | sequences-and-series | summation-of-series | Let S<sub>k</sub> = $${{1 + 2 + 3 + .... + k} \over k}.$$ If $$S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}$$A, then A is equal to : | [{"identifier": "A", "content": "283"}, {"identifier": "B", "content": "156"}, {"identifier": "C", "content": "301"}, {"identifier": "D", "content": "303"}] | ["D"] | null | S<sub>k</sub> = $${{K + 1} \over 2}$$
<br><br>$$\sum {S_k^2} = {5 \over {12}}$$ A
<br><br>$$\sum\limits_{K = 1}^{10} {{{\left( {{{K + 1} \over 2}} \right)}^2}} = {{{2^2} + {3^2} + - - + {{11}^2}} \over 4} = {5 \over {12}}$$ A
<br><br>$${{11 \times 12 \times 23} \over 6} - 1 = {5 \over 3}$$ A
<br><br>505 $$ = {5 \... | mcq | jee-main-2019-online-12th-january-morning-slot | 8,098 |
qOmtyXLtSaFKZemyMR1Mj | maths | sequences-and-series | summation-of-series | The sum of the following series
<br/><br/>$$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9}$$
<br/><br/> $$ + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....$$ up to 15 terms, is : | [{"identifier": "A", "content": "7520"}, {"identifier": "B", "content": "7510"}, {"identifier": "C", "content": "7830"}, {"identifier": "D", "content": "7820"}] | ["D"] | null | $$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9} + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....\,15$$
<br><br>$$ = {{3\left( {{1^2}} \right)} \over 3} + {{6\left( {{1^2} + {2^2}} \right)} \over 5} + {{9\left( {{1^2} + {... | mcq | jee-main-2019-online-9th-january-evening-slot | 8,099 |
QRFlNrjhUGdaY1JzcO18hoxe66ijvwq8ul6 | maths | sequences-and-series | summation-of-series | Some identical balls are arranged in rows to form
an equilateral triangle. The first row consists of one
ball, the second row consists of two balls and so
on. If 99 more identical balls are addded to the total
number of balls used in forming the equilaterial
triangle, then all these balls can be arranged in a
square wh... | [{"identifier": "A", "content": "262"}, {"identifier": "B", "content": "190"}, {"identifier": "C", "content": "157"}, {"identifier": "D", "content": "225"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264552/exam_images/ooaanpllsv7wqqhp4zns.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Sequences and Series Question 185 English Explanation 1">
<br... | mcq | jee-main-2019-online-9th-april-evening-slot | 8,100 |
U9xxSdQCH7SEff1HPajgy2xukfg6hk2u | maths | sequences-and-series | summation-of-series | If 2<sup>10</sup> + 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> = S - 2<sup>11</sup>, then S is equal to : | [{"identifier": "A", "content": "$${{{3^{11}}} \\over 2} + {2^{10}}$$"}, {"identifier": "B", "content": "3<sup>11</sup> \u2014 2<sup>12</sup>"}, {"identifier": "C", "content": "2.3<sup>11</sup>"}, {"identifier": "D", "content": "3<sup>11</sup>"}] | ["D"] | null | Let S<sub>1</sub> = 2<sup>10</sup> + 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> ....(1)
<br><br>Also $${{3{S_1}} \over 2}$$ = 2<sup>9</sup>.3<sup>1</sup> + 2<sup>8</sup>
.3<sup>2</sup> +.....+ 2.3<sup>9</sup> + 3<sup>10</sup> + $${{{3^{11}}} \over 2}$$ ......(2)... | mcq | jee-main-2020-online-5th-september-morning-slot | 8,101 |
q6m7BJB7vRxsb0PkbGjgy2xukfqd7bft | maths | sequences-and-series | summation-of-series | If the sum of the first 20 terms of the series
<br/>$${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$$ is 460,
<br/>then x is equal to : | [{"identifier": "A", "content": "e<sup>2</sup>"}, {"identifier": "B", "content": "7<sup>1/2</sup>"}, {"identifier": "C", "content": "7<sup>2</sup>"}, {"identifier": "D", "content": "7<sup>46/21</sup>"}] | ["C"] | null | 460 = log<sub>7</sub> <sup>x</sup>·(2 + 3 + 4 + ..... + 20 + 21)
<br><br>$$ \Rightarrow $$ 460 = log<sub>7</sub> <sup>x</sup>. $$\left( {{{21 \times 22} \over 2} - 1} \right)$$
<br><br>$$ \Rightarrow $$ 460 = 230. log<sub>7</sub> <sup>x</sup>
<br><br>$$ \Rightarrow $$ log<sub>7</sub> <sup>x</sup> = 2
<br><br>$$ \Righta... | mcq | jee-main-2020-online-5th-september-evening-slot | 8,102 |
6CDZ22WJaLTtjOakazjgy2xukf7gbwk0 | maths | sequences-and-series | summation-of-series | If 1+(1–2<sup>2</sup>.1)+(1–4<sup>2</sup>.3)+(1-6<sup>2</sup>.5)+......+(1-20<sup>2</sup>.19)= $$\alpha $$ - 220$$\beta $$, <br/>then an ordered pair $$\left( {\alpha ,\beta } \right)$$ is equal to: | [{"identifier": "A", "content": "(11, 103)"}, {"identifier": "B", "content": "(10, 103)"}, {"identifier": "C", "content": "(10, 97)"}, {"identifier": "D", "content": "(11, 97)"}] | ["A"] | null | $$1 + (1 - {2^2}.1) + (1 - {4^2}.3) + (1 - {6^2}.5) + ....(1 - {20^2}.19)$$<br><br>$$S = 1 + \sum\limits_{r = 1}^{10} {\left[ {1 - {{(2r)}^2}(2r - 1)} \right] = 1 + \sum\limits_{r = 1}^{10} {\left( {1 - 8{r^3} + 4{r^2}} \right)} = 1 + 10 - } \sum\limits_{r = 1}^{10} {\left( {8{r^3} - 4{r^2}} \right)} $$<br><br>$$= 11 ... | mcq | jee-main-2020-online-4th-september-morning-slot | 8,103 |
iitp3b1OAO5bXqqUlkjgy2xukf44u0jj | maths | sequences-and-series | summation-of-series | If the sum of the series
<br/><br/>20 + 19$${3 \over 5}$$ + 19$${1 \over 5}$$ + 18$${4 \over 5}$$ + ...
<br/><br/>upto nth term is 488
and the n<sup>th</sup> term is negative, then : | [{"identifier": "A", "content": "n = 41"}, {"identifier": "B", "content": "n = 60"}, {"identifier": "C", "content": "n<sup>th</sup> term is \u20134"}, {"identifier": "D", "content": "n<sup>th</sup> term is -4$${2 \\over 5}$$"}] | ["C"] | null | $$S = {{100} \over 5} + {{98} \over 5} + {{96} \over 5} + {{94} \over 5} + ...\,n$$<br><br>$$\,{S_n} = {n \over 2}\left( {2 \times {{100} \over 5} + (n - 1)\left( {{{ - 2} \over 5}} \right)} \right) = 188$$<br><br>$$ \Rightarrow $$ $$n(100 - n + 1) = 488 \times 5$$<br><br>$$ \Rightarrow $$ $${n^2} - 101n + 488 \times 5... | mcq | jee-main-2020-online-3rd-september-evening-slot | 8,104 |
grhqiUH1FX5lb2FWm3jgy2xukezf3clg | maths | sequences-and-series | summation-of-series | Let S be the sum of the first 9 terms of the
series :
<br/>{x + k$$a$$} + {x<sup>2</sup> + (k + 2)$$a$$} + {x<sup>3</sup> + (k + 4)$$a$$}
<br/>+ {x<sup>4</sup> + (k + 6)$$a$$} + .... where a $$ \ne $$ 0 and x $$ \ne $$ 1.
<br/><br/>If S = $${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$$, then k is equal... | [{"identifier": "A", "content": "-3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "-5"}, {"identifier": "D", "content": "3"}] | ["A"] | null | S = {x + k$$a$$} + {x<sup>2</sup> + (k + 2)$$a$$} + {x<sup>3</sup> + (k + 4)$$a$$}
<br>+ {x<sup>4</sup> + (k + 6)$$a$$} + ....<br><br>
$$S = \left( {x + {x^2} + {x^3} + ....\,9terms} \right) + $$<br> $$a\left( {k + \left( {k + 2} \right) + (k + 4) + (k + 6) + ....9terms} \right)$$<br><br>
$$S = {{x\lef... | mcq | jee-main-2020-online-2nd-september-evening-slot | 8,105 |
iYlXL3LrAQdpyM6ikm7k9k2k5is98y3 | maths | sequences-and-series | summation-of-series | The product $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ... to $$\infty $$ is equal
to : | [{"identifier": "A", "content": "$${2^{{1 \\over 4}}}$$"}, {"identifier": "B", "content": "$${2^{{1 \\over 2}}}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ...
<br><br>= $${2^{{1 \over 4} + {2 \over {16}} + {3 \over {48}} + ...\infty }}$$
<br><br>= $${2^{{1 \over 4} + {1 \over 8} + {1 \over {16}} + ...\infty }}$$
<br><br>= $${2^{\left( {{{{1 \over 4}} \over {1 - {1 \over 2}}}} \right)}}... | mcq | jee-main-2020-online-9th-january-morning-slot | 8,106 |
mYMcFJJVdAchWsiX9b7k9k2k5h0a9ty | maths | sequences-and-series | summation-of-series | The sum $$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $$ is : | [] | null | 1540 | $$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $$
<br><br>= $$\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $$
<br><br>= $$\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $$
<br><br>= $${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 ... | integer | jee-main-2020-online-8th-january-morning-slot | 8,108 |
tUvK0XGzjI8Hr4plFFjgy2xukewmjblr | maths | sequences-and-series | summation-of-series | If |x| < 1, |y| < 1 and x $$ \ne $$ y, then the sum to infinity
of the following series
<br/><br/>(x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + .... | [{"identifier": "A", "content": "$${{x + y - xy} \\over {\\left( {1 + x} \\right)\\left( {1 + y} \\right)}}$$"}, {"identifier": "B", "content": "$${{x + y - xy} \\over {\\left( {1 - x} \\right)\\left( {1 - y} \\right)}}$$"}, {"identifier": "C", "content": "$${{x + y + xy} \\over {\\left( {1 + x} \\right)\\left( {1 + y}... | ["B"] | null | (x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + ....
<br><br>By multiplying and dividing x – y :
<br><br>$${{\left( {{x^2} - {y^2}} \right) + \left( {{x^3} - {y^3}} \right) + \left( {{x^4} - {y^4}} \right) + ...} \over {x - y}}$$
<br><br>= $${{\left( {{x^2} +... | mcq | jee-main-2020-online-2nd-september-morning-slot | 8,110 |
yuKByka9gb3axlQDX41kls445x1 | maths | sequences-and-series | summation-of-series | If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $$ and $$z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi } $$ then : | [{"identifier": "A", "content": "xy $$-$$ z = (x + y)z"}, {"identifier": "B", "content": "xyz = 4"}, {"identifier": "C", "content": "xy + z = (x + y)z"}, {"identifier": "D", "content": "xy + yz + zx = z"}] | ["C"] | null | $$x = 1 + {\cos ^2}\theta + ..........\infty $$<br><br>$$x = {1 \over {1 - {{\cos }^2}\theta }} = {1 \over {{{\sin }^2}\theta }}$$ .......(1)<br><br>$$y = 1 + {\sin ^2}\phi + ........\infty $$<br><br>$$y = {1 \over {1 - {{\sin }^2}\phi }} = {1 \over {{{\cos }^2}\phi }}$$ ....... (2)<br><br>$$z = {1 \over {1 - {{\cos ... | mcq | jee-main-2021-online-25th-february-morning-slot | 8,111 |
KooM1QkVr1JhZisA3F1kluga4m4 | maths | sequences-and-series | summation-of-series | The sum of the infinite series <br/>$$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$$ is equal to : | [{"identifier": "A", "content": "$${9 \\over 4}$$"}, {"identifier": "B", "content": "$${13 \\over 4}$$"}, {"identifier": "C", "content": "$${15 \\over 4}$$"}, {"identifier": "D", "content": "$${11 \\over 4}$$"}] | ["B"] | null | $$S = 1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + ....$$<br><br>$${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {7 \over {{3^3}}} + {{12} \over {{3^4}}} + ....$$<br><br>$$2S = 1 + {1 \over 3} + {5 \over {{3^2}}} + {5 \over {{3^3}}} + {5 \over {{3^4}}} + ....$$ + up to infinite ... | mcq | jee-main-2021-online-26th-february-morning-slot | 8,112 |
ZBMT4NvKr5BSRHqBGT1kluxngbh | maths | sequences-and-series | summation-of-series | The sum of the series <br/><br/>$$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $$ is equal to : | [{"identifier": "A", "content": "$${{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} - 10$$"}, {"identifier": "B", "content": "$${{41} \\over 8}e - {{19} \\over 8}{e^{ - 1}} - 10$$"}, {"identifier": "C", "content": "$${{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} + 10$$"}, {"identifier": "D", "content": "$$ - {{41} \\over 8}... | ["B"] | null | $$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $$<br><br>Put 2n + 1 = r, where r = 3, 5, 7, .......<br><br>$$ \Rightarrow n = {{r - 1} \over 2}$$<br><br>$${{{n^2} + 6n + 10} \over {(2n + 1)!}} = {{{{\left( {{{r - 1} \over 2}} \right)}^2} + 3r - 3 + 10} \over {r!}} $$
<br><br>$$= {{{r^2} + 10r + 2... | mcq | jee-main-2021-online-26th-february-evening-slot | 8,113 |
zd4RTaPsITd0LFIGzU1kmiz8sd6 | maths | sequences-and-series | summation-of-series | S<sub>n</sub>(x) = log<sub>a<sup>1/2</sup></sub>x + log<sub>a<sup>1/3</sup></sub>x + log<sub>a<sup>1/6</sup></sub>x + log<sub>a<sup>1/11</sup></sub>x + log<sub>a<sup>1/18</sup></sub>x + log<sub>a<sup>1/27</sup></sub>x + ...... up to n-terms, where a > 1. If S<sub>24</sub>(x) = 1093 and S<sub>12</sub>(2x) = 265, then... | [] | null | 16 | $${S_n}(x) = {\log _a}{x^2} + {\log _a}{x^3} + {\log _a}{x^6} + {\log _a}{x^{11}}$$<br><br>$${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$$<br><br>$${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$$<br><br>$${S_r} = 2 + 3 + 6 + 11$$
<br><br>$$ \therefore $$ T<sub>n</sub> = 2 + (1 + 3 + 5 +..... | integer | jee-main-2021-online-16th-march-evening-shift | 8,114 |
qu0cXEoBuHthtOcLsa1kmlisv2k | maths | sequences-and-series | summation-of-series | If $$\alpha$$, $$\beta$$ are natural numbers such that <br/>100<sup>$$\alpha$$</sup> $$-$$ 199$$\beta$$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($$\alpha$$, $$\beta$$) and origin is : | [{"identifier": "A", "content": "540"}, {"identifier": "B", "content": "550"}, {"identifier": "C", "content": "530"}, {"identifier": "D", "content": "510"}] | ["B"] | null | RHS = $$\sum\limits_{r = 0}^{99} {(100 - r)(100 + r)} $$<br><br>$$ = {(100)^3} - {{99 \times 100 \times 199} \over 6} = {(100)^3} - (1650)199$$<br><br>LHS = (100)<sup>$$\alpha$$</sup> $$-$$ (199)<sup>$$\beta$$</sup><br><br>So, $$\alpha$$ = 3, $$\beta$$ = 1650<br><br>Slope = tan$$\theta$$ = $${\beta \over \alpha }$$<br... | mcq | jee-main-2021-online-18th-march-morning-shift | 8,115 |
XHol4qKySWK9HAFu5P1kmlix67q | maths | sequences-and-series | summation-of-series | $${1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$$ is equal to | [{"identifier": "A", "content": "$${{101} \\over {404}}$$"}, {"identifier": "B", "content": "$${{25} \\over {101}}$$"}, {"identifier": "C", "content": "$${{101} \\over {408}}$$"}, {"identifier": "D", "content": "$${{99} \\over {400}}$$"}] | ["B"] | null | $$S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}} $$<br><br>$$ = \sum\limits_{r = 1}^{100} {{1 \over {(2n + 1 + 1)(2n + 1 - 1)}}} $$<br><br>$$ = \sum\limits_{r = 1}^{100} {{1 \over {2n(2n + 2)}}} $$<br><br>$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {{1 \over {n(n + 1)}}} $$<br><br>$$ = {1 \over 4}\sum\lim... | mcq | jee-main-2021-online-18th-march-morning-shift | 8,116 |
1krrx9mvg | maths | sequences-and-series | summation-of-series | Let $$\left\{ {{a_n}} \right\}_{n = 1}^\infty $$ be a sequence such that a<sub>1</sub> = 1, a<sub>2</sub> = 1 and $${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$$ for all n $$\ge$$ 1. Then the value of $$47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}} $$ is equal to ______________. | [] | null | 7 | $${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$$, let $$\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} = P$$<br><br>Divide by 8<sup>n</sup> we get<br><br>$${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$$<br><br>$$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1... | integer | jee-main-2021-online-20th-july-evening-shift | 8,119 |
1krw23dgt | maths | sequences-and-series | summation-of-series | If the value of<br/><br/> $${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$$<br/><br/> is $$l$$, then $$l$$<sup>2</sup> is equal to _______________. | [] | null | 3 | $$l = {\left( {\underbrace {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}}}_S + ....} \right)^{{{\log }_{0.25}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ...} \right)}}$$<br><br>$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$$<br><br>$${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + ... | integer | jee-main-2021-online-25th-july-morning-shift | 8,120 |
1ktbcep0a | maths | sequences-and-series | summation-of-series | The sum of the series <br/><br/>$${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$ when x = 2 is : | [{"identifier": "A", "content": "$$1 + {{{2^{101}}} \\over {{4^{101}} - 1}}$$"}, {"identifier": "B", "content": "$$1 + {{{2^{100}}} \\over {{4^{101}} - 1}}$$"}, {"identifier": "C", "content": "$$1 - {{{2^{100}}} \\over {{4^{100}} - 1}}$$"}, {"identifier": "D", "content": "$$1 - {{{2^{101}}} \\over {{2^{400}} - 1}}$$"}] | ["D"] | null | $$S = {1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$<br><br>$$S + {1 \over {1 - x}} = {1 \over {1 - x}} + {1 \over {x + 1}} + ...... = {2 \over {1 - {x^2}}} + {2 \over {1 + {x^2}}} + ....$$<br><br>$$S + {1 \over {1 - x}} = {{{2^{101}}} \ove... | mcq | jee-main-2021-online-26th-august-morning-shift | 8,121 |
1ktehkmgu | maths | sequences-and-series | summation-of-series | If 0 < x < 1, then $${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$$, is equal to : | [{"identifier": "A", "content": "$$x\\left( {{{1 + x} \\over {1 - x}}} \\right) + {\\log _e}(1 - x)$$"}, {"identifier": "B", "content": "$$x\\left( {{{1 - x} \\over {1 + x}}} \\right) + {\\log _e}(1 - x)$$"}, {"identifier": "C", "content": "$${{1 - x} \\over {1 + x}} + {\\log _e}(1 - x)$$"}, {"identifier": "D", "conten... | ["A"] | null | Let $$t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty $$<br><br>$$ = \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty $$<br><br>$$ = 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}... | mcq | jee-main-2021-online-27th-august-morning-shift | 8,122 |
1ktehsn1f | maths | sequences-and-series | summation-of-series | If for x, y $$\in$$ R, x > 0, y = log<sub>10</sub>x + log<sub>10</sub>x<sup>1/3</sup> + log<sub>10</sub>x<sup>1/9</sup> + ...... upto $$\infty$$ terms <br/><br/>and $${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$$, then the ordered pair (x, y) is equal to : | [{"identifier": "A", "content": "(10<sup>6</sup>, 6)"}, {"identifier": "B", "content": "(10<sup>4</sup>, 6)"}, {"identifier": "C", "content": "(10<sup>2</sup>, 3)"}, {"identifier": "D", "content": "(10<sup>6</sup>, 9)"}] | ["D"] | null | $${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$$<br><br>$$ \Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}$$<br><br>Now, <br><br>$$y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty $$<b... | mcq | jee-main-2021-online-27th-august-morning-shift | 8,123 |
1ktg3am8n | maths | sequences-and-series | summation-of-series | If 0 < x < 1 and $$y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$$, then the value of e<sup>1 + y</sup> at $$x = {1 \over 2}$$ is : | [{"identifier": "A", "content": "$${1 \\over 2}{e^2}$$"}, {"identifier": "B", "content": "2e"}, {"identifier": "C", "content": "$${1 \\over 2}\\sqrt e $$"}, {"identifier": "D", "content": "2e<sup>2</sup>"}] | ["A"] | null | $$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$<br><br>$$ = ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$<br><br>$$ = {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ..... | mcq | jee-main-2021-online-27th-august-evening-shift | 8,124 |
1ktioa9ro | maths | sequences-and-series | summation-of-series | The sum of 10 terms of the series
<br/><br/>$${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${{120} \\over {121}}$$"}, {"identifier": "C", "content": "$${{99} \\over {100}}$$"}, {"identifier": "D", "content": "$${{143} \\over {144}}$$"}] | ["B"] | null | $$S = {{{2^2} - {1^2}} \over {{1^2} \times {2^2}}} + {{{3^2} - {2^2}} \over {{2^2} \times {3^2}}} + {{{4^2} - {3^2}} \over {{3^2} \times {4^2}}} + ...$$<br><br>$$ = \left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right] + \left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] + \left[ {{1 \over {{3^2}}} - {1 \over {{4^... | mcq | jee-main-2021-online-31st-august-morning-shift | 8,125 |
1ktkeexuz | maths | sequences-and-series | summation-of-series | If $$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$$, then 160 S is equal to ________. | [] | null | 305 | $$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$$<br><br>$${1 \over 5}S = {7 \over 5} + {9 \over {{5^3}}} + {{13} \over {{5^4}}} + ....$$<br><br>On subtracting<br><br>$${4 \over 5}S = {7 \over 5} + {2 \over {{5^2}}} + {4 \over {{5^3}}} + {6 \over {{5^4}}} + ....$$<br><br>$$S =... | integer | jee-main-2021-online-31st-august-evening-shift | 8,126 |
1kto7yln2 | maths | sequences-and-series | summation-of-series | Let S<sub>n</sub> = 1 . (n $$-$$ 1) + 2 . (n $$-$$ 2) + 3 . (n $$-$$ 3) + ..... + (n $$-$$ 1) . 1, n $$\ge$$ 4.<br/><br/>The sum $$\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$ is equal to : | [{"identifier": "A", "content": "$${{e - 1} \\over 3}$$"}, {"identifier": "B", "content": "$${{e - 2} \\over 6}$$"}, {"identifier": "C", "content": "$${e \\over 3}$$"}, {"identifier": "D", "content": "$${e \\over 6}$$"}] | ["A"] | null | Let T<sub>r</sub> = r(n $$-$$ r)<br><br>T<sub>r</sub> = nr $$-$$ r<sup>2</sup><br><br>$$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{T_r} = \sum\limits_{r = 1}^n {(nr - {r^2})} } $$<br><br>$${S_n} = {{n\,.\,(n)(n + 1)} \over 2} - {{n(n + 1)(2n + 1)} \over 6}$$<br><br>$${S_n} = {{n(n - 1)(n + 1)} \over 6}$$<br><br>Now, ... | mcq | jee-main-2021-online-1st-september-evening-shift | 8,127 |
1l54apl39 | maths | sequences-and-series | summation-of-series | <p>The sum of the infinite series $$1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + {{35} \over {{6^4}}} + {{51} \over {{6^5}}} + {{70} \over {{6^6}}} + \,\,.....$$ is equal to :</p> | [{"identifier": "A", "content": "$${{425} \\over {216}}$$"}, {"identifier": "B", "content": "$${{429} \\over {216}}$$"}, {"identifier": "C", "content": "$${{288} \\over {125}}$$"}, {"identifier": "D", "content": "$${{280} \\over {125}}$$"}] | ["C"] | null | <p>$$S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + $$ ...... (1)</p>
<p>$${1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} + $$ ...... (2)</p>
<p>$$S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ........</p>
<p>$$ \Rightarrow {{5S} \over 6} ... | mcq | jee-main-2022-online-29th-june-evening-shift | 8,129 |
1l56q6r0z | maths | sequences-and-series | summation-of-series | <p>Let $$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + \,.....$$. Then 4S is equal to</p> | [{"identifier": "A", "content": "$${\\left( {{7 \\over 3}} \\right)^2}$$"}, {"identifier": "B", "content": "$${{{7^3}} \\over {{3^2}}}$$"}, {"identifier": "C", "content": "$${\\left( {{7 \\over 3}} \\right)^3}$$"}, {"identifier": "D", "content": "$${{{7^2}} \\over {{3^3}}}$$"}] | ["C"] | null | <p>$$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + $$ ..... ...... (i)</p>
<p>$${1 \over 7}S = {2 \over 7} + {6 \over {{7^2}}} + {{12} \over {{7^3}}} + {{20} \over {{7^4}}} + $$ .... ....... (ii)</p>
<p>(i) - (ii)</p>
<p>$${6 \over 7}S = 2 + {4 \over 7} + {6 \over {{7^2}}} +... | mcq | jee-main-2022-online-27th-june-evening-shift | 8,130 |
1l57p5mqd | maths | sequences-and-series | summation-of-series | <p>If the sum of the first ten terms of the series</p>
<p>$${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$$</p>
<p>is $${m \over n}$$, where m and n are co-prime numbers, then m + n is equal to ______________.</p> | [] | null | 276 | <p>$${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$$</p>
<p>$$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$$</p>
<p>$$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$$</p>
<p>$${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $$</... | integer | jee-main-2022-online-27th-june-morning-shift | 8,131 |
1l59jzej6 | maths | sequences-and-series | summation-of-series | <p>The sum 1 + 2 . 3 + 3 . 3<sup>2</sup> + ......... + 10 . 3<sup>9</sup> is equal to :</p> | [{"identifier": "A", "content": "$${{2\\,.\\,{3^{12}} + 10} \\over 4}$$"}, {"identifier": "B", "content": "$${{19\\,.\\,{3^{10}} + 1} \\over 4}$$"}, {"identifier": "C", "content": "$$5\\,.\\,{3^{10}} - 2$$"}, {"identifier": "D", "content": "$${{9\\,.\\,{3^{10}} + 1} \\over 2}$$"}] | ["B"] | null | <p>Let $$S = 1\,.\,{3^0} + 2\,.\,{3^1} + 3\,.\,{3^2} + \,\,......\,\, + \,\,10\,.\,{3^9}$$</p>
<p>$$3S = 1\,.\,{3^1} + 2\,.\,{3^2} + \,\,..........\,\, + \,\,10\,.\,{3^{10}}$$</p>
<p>___________________________________________________________</p>
<p>$$ - 2S = (1\,.\,{3^0} + 1\,.\,{3^1} + 1\,.\,{3^2} + \,\,........\,\, ... | mcq | jee-main-2022-online-25th-june-evening-shift | 8,133 |
1l5ajy4fx | maths | sequences-and-series | summation-of-series | <p>The greatest integer less than or equal to the sum of first 100 terms of the sequence $${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$$ ...... is equal to ___________.</p> | [] | null | 98 | <p>$$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $$ ....</p>
<p>$$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $$</p>
<p>$$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$$</p>
<p>$$ = 98 + 2{\left( {{2 \ove... | integer | jee-main-2022-online-25th-june-morning-shift | 8,135 |
1l5vzhxr8 | maths | sequences-and-series | summation-of-series | <p>The value of $$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \,\,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,.....\,\, + \,\,11}}$$ is equal to:</p> | [{"identifier": "A", "content": "$${{20} \\over {11}}$$"}, {"identifier": "B", "content": "$${{11} \\over {6}}$$"}, {"identifier": "C", "content": "$${{241} \\over {132}}$$"}, {"identifier": "D", "content": "$${{21} \\over {11}}$$"}] | ["B"] | null | <p>Given,</p>
<p>$$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}}\, + \,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,....\,\, + \,11}}$$</p>
<p>General term,</p>
<p>$${T_n} = {1 \over {1 + 2 + 3\, + \,\,....\,\, + \,\,n}}$$</p>
<p>$$ = {1 \over {{{n(n + 1)} \over 2}}}$$</p>
<p>$$ = {2 \over {n(n + 1)}}$$</p>
<p>$$ = 2\left[... | mcq | jee-main-2022-online-30th-june-morning-shift | 8,136 |
1l6dx9fyy | maths | sequences-and-series | summation-of-series | <p>Let $$a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$$ and $$b_{n}=a_{n}+b_{n-1}$$ for every <br/><br/>natural number $$n \geqslant 2$$. Then $$\sum\limits_{n = 1}^{15} {{a_n}.{b_n}} $$ is equal to ___________.</p> | [] | null | 27560 | <p>Given,</p>
<p>$${a_n} = {a_{n - 1}} + 2$$</p>
<p>$$ \Rightarrow {a_n} - {a_{n - 1}} = 2$$</p>
<p>$$\therefore$$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.</p>
<p>Also given $${a_1} = 1$$</p>
<p>$$\therefore$$ Series is = 1, 3, 5, 7 ......</p>
<p>$$... | integer | jee-main-2022-online-25th-july-morning-shift | 8,137 |
1l6f0wdoi | maths | sequences-and-series | summation-of-series | <p>The sum $$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}} $$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{7}{87}$$"}, {"identifier": "B", "content": "$$\\frac{7}{29}$$"}, {"identifier": "C", "content": "$$\\frac{14}{87}$$"}, {"identifier": "D", "content": "$$\\frac{21}{29}$$"}] | ["B"] | null | <p>$$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}} = {3 \over 4}\sum\limits_{n = 1}^{21} {{1 \over {4n - 1}} - {1 \over {4n + 3}}} } $$</p>
<p>$$ = {3 \over 4}\left[ {\left( {{1 \over 3} - {1 \over 7}} \right) + \left( {{1 \over 7} - {1 \over {11}}} \right) + \,\,....\,\, + \,\,\left( {{1 \over {83}} - {1 \ove... | mcq | jee-main-2022-online-25th-july-evening-shift | 8,138 |
1l6gjc5oa | maths | sequences-and-series | summation-of-series | <p>The series of positive multiples of 3 is divided into sets : $$\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots$$ Then the sum of the elements in the $$11^{\text {th }}$$ set is equal to ____________.</p> | [] | null | 6993 | <p>Given series</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbx3j3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-1... | integer | jee-main-2022-online-26th-july-morning-shift | 8,139 |
1l6hzs0vf | maths | sequences-and-series | summation-of-series | <p>If $$\sum\limits_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$$, where m and n are co-prime, then $$m+n$$ is equal to _____________.</p> | [] | null | 166 | <p>$$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $$</p>
<p>$$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$$</p>
<p>$$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,..... | integer | jee-main-2022-online-26th-july-evening-shift | 8,140 |
1l6klic2q | maths | sequences-and-series | summation-of-series | <p>$$
\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+
\frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$$ is equal to _____________.</p> | [] | null | 120 | <p>$${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{2n(2{n^2} + 3n + 1) -... | integer | jee-main-2022-online-27th-july-evening-shift | 8,141 |
1l6m6bitt | maths | sequences-and-series | summation-of-series | <p>Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\ri... | [{"identifier": "A", "content": "$$-$$30"}, {"identifier": "B", "content": "$$-$$31"}, {"identifier": "C", "content": "$$-$$60"}, {"identifier": "D", "content": "$$-$$61"}] | ["C"] | null | <p>$${a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}$$</p>
<p>$$ \Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1$$</p>
<p>$$ \Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2$$</p>
<p>For</p>
<p>$$\matrix{
{n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr
{n = 2} & {{a_4}{a_3} - {a_3}{a_2} = ... | mcq | jee-main-2022-online-28th-july-morning-shift | 8,142 |
1l6npcsi3 | maths | sequences-and-series | summation-of-series | $${6 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3} = {2^n}\,.\,m$$, where m is odd, then m . n is equal to ____________. | [] | null | 12 | <p>$${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right... | integer | jee-main-2022-online-28th-july-evening-shift | 8,143 |
1l6p3pptf | maths | sequences-and-series | summation-of-series | <p>If $$\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}=\frac{\mathrm{k}}{101}$$, then 34 k is equal to _________.</p> | [] | null | 286 | <p>$$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$$</p>
<p>$$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$$</p>
<p>$$ = {1 \over 2}\left( {{1 \over 6} - {... | integer | jee-main-2022-online-29th-july-morning-shift | 8,144 |
1ldo67p4f | maths | sequences-and-series | summation-of-series | <p>The sum $$\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}} $$ is equal to :</p> | [{"identifier": "A", "content": "$${{11e} \\over 2} + {7 \\over {2e}}$$"}, {"identifier": "B", "content": "$${{13e} \\over 4} + {5 \\over {4e}} - 4$$"}, {"identifier": "C", "content": "$${{11e} \\over 2} + {7 \\over {2e}} - 4$$"}, {"identifier": "D", "content": "$${{13e} \\over 4} + {5 \\over {4e}}$$"}] | ["B"] | null | $\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} \\\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{2 n(2 n-1)+8 n+8}{(2 n) !} \\\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\\\ & e=1+1+\frac{1}{2 !}+\frac{1}{3 ... | mcq | jee-main-2023-online-1st-february-evening-shift | 8,146 |
ldoawens | maths | sequences-and-series | summation-of-series | The sum $1^{2}-2 \cdot 3^{2}+3 \cdot 5^{2}-4 \cdot 7^{2}+5 \cdot 9^{2}-\ldots+15 \cdot 29^{2}$ is _________. | [] | null | 6952 | $S=1^{2}-2.3^{2}+3.5^{2}-4.7^{2}+\ldots \ldots+15.29^{2}$
<br/><br/>Separating odd placed and even placed terms we get
<br/><br/>$$
\begin{aligned}
& \mathrm{S}=\left(1.1^2+3.5^2+\ldots .15 .(29)^2\right)-\left(2.3^2+4.7^2\right. \\
& +\ldots .+14 .(27)^2
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\sum_{r=1}^{8}(... | integer | jee-main-2023-online-31st-january-evening-shift | 8,147 |
1ldom86ia | maths | sequences-and-series | summation-of-series | <p>The sum of 10 terms of the series</p>
<p>$${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$$ is</p> | [{"identifier": "A", "content": "$${{58} \\over {111}}$$"}, {"identifier": "B", "content": "$${{56} \\over {111}}$$"}, {"identifier": "C", "content": "$${{55} \\over {111}}$$"}, {"identifier": "D", "content": "$${{59} \\over {111}}$$"}] | ["C"] | null | $$
\begin{aligned}
& T_n=\frac{n}{1+n^2+n^4} \\\\
& =\frac{n}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)} \\\\
& =\frac{1}{2}\left[\frac{\left(\mathrm{n}^2+\mathrm{n}+1\right)-\left(\mathrm{n}^2-\mathrm{n}+1\right)}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\... | mcq | jee-main-2023-online-1st-february-morning-shift | 8,148 |
ldqzs9si | maths | sequences-and-series | summation-of-series | The $8^{\text {th }}$ common term of the series
<br/><br/>$$
\begin{aligned}
& S_1=3+7+11+15+19+\ldots . . \\\\
& S_2=1+6+11+16+21+\ldots . .
\end{aligned}
$$
<br/><br/>is : | [] | null | 151 | <p>First common term is 11</p>
<p>Common difference of series of common terms is LCM (4, 5) = 20</p>
<p>$$a_8=a+7d$$</p>
<p>$$=11+7\times20=151$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 8,149 |
1ldr5km4h | maths | sequences-and-series | summation-of-series | <p>If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :</p> | [{"identifier": "A", "content": "$${{51} \\over {144}}$$"}, {"identifier": "B", "content": "$${{49} \\over {138}}$$"}, {"identifier": "C", "content": "$${{50} \\over {141}}$$"}, {"identifier": "D", "content": "$${{52} \\over {147}}$$"}] | ["C"] | null | <p>$$\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } } $$</p>
<p>$$ = \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)} $$</p>
<p>$$ = \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over... | mcq | jee-main-2023-online-30th-january-morning-shift | 8,150 |
1ldr7tnf5 | maths | sequences-and-series | summation-of-series | <p>Let $$\sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$$, where $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$$ and $$e=\sum_\limits{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n}... | [] | null | 26 | <p>$$\sum\limits_{n = 0}^\infty {{{{n^3}(2n!) + (2n - 1)(n!)} \over {n!\,.\,(2n)!}}} $$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{{{n^3}} \over {n!}} + {{2n - 1} \over {2n!}}} $$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{3 \over {(n - 2)!}} + {1 \over {(n - 3)!}} + {1 \over {(n - 1)!}} + {1 \over {(2n - 1)!}} - {1 \ov... | integer | jee-main-2023-online-30th-january-morning-shift | 8,151 |
1ldsg00ct | maths | sequences-and-series | summation-of-series | <p>Let $$a_1=b_1=1$$ and $${a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2$$. If $$S = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}} $$ and $$T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}} $$, then $${2^7}(2S - T)$$ is equal to ____________.</p> | [] | null | 461 | <p>$$\because$$ $${a_n} = {a_{n - 1}} + (n - 1)$$ and $${a_1} = {b_1} = 1$$</p>
<p>$${b_n} = {b_{n - 1}} + {a_{n - 1}}$$</p>
<p>$$\therefore$$ $${b_{n + 1}} = 2{b_n} - {b_{n - 1}} + n - 1$$</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;bord... | integer | jee-main-2023-online-29th-january-evening-shift | 8,152 |
1ldwxkndi | maths | sequences-and-series | summation-of-series | <p>If $${{{1^3} + {2^3} + {3^3}\, + \,...\,up\,to\,n\,terms} \over {1\,.\,3 + 2\,.\,5 + 3\,.\,7\, + \,...\,up\,to\,n\,terms}} = {9 \over 5}$$, then the value of $$n$$ is</p> | [] | null | 5 | Given $\frac{1^{3}+2^{3}+3^{3}+\ldots \text { up to } n \text { terms }}{1.3+2.5+3.7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$
<br/><br/>
Now
<br/><br/>
Let $S=1.3+2.5+3.7+\ldots$
<br/><br/>
$$
\begin{aligned}
& T_{n}=n \cdot(2 n+1) \\\\
& \therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\\\
& \Rig... | integer | jee-main-2023-online-24th-january-evening-shift | 8,153 |
lgnyx1yy | maths | sequences-and-series | summation-of-series | If the sum of the series
<br/><br/>$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+$
<br/><br/>$\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-... | [] | null | 7 | We can rewrite the given series as follows :
<br/><br/>$$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$$
<br/><br/>Th... | integer | jee-main-2023-online-15th-april-morning-shift | 8,154 |
1lgq0vrkc | maths | sequences-and-series | summation-of-series | <p>The sum to $$20$$ terms of the series $$2 \cdot 2^{2}-3^{2}+2 \cdot 4^{2}-5^{2}+2 \cdot 6^{2}-\ldots \ldots$$ is equal to __________.</p> | [] | null | 1310 | $$
\begin{aligned}
& \sum_{r=1}^{10}\left(2 \cdot(2 r)^2-(2 r+1)^2\right) \\\\
& =\sum_{r=1}^{10}\left(8 r^2-4 r^2-4 r-1\right) \\\\
& =\sum_{r=1}^{10}\left(4 r^2-4 r-1\right) \\\\
& =\frac{4 \cdot 10 \cdot 11 \cdot 21}{6}-4 \frac{10 \cdot 11}{2}-10 \\\\
& =44 \cdot 35-220-10 \\\\
& =1540-230=1310
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-morning-shift | 8,155 |
1lgreevjx | maths | sequences-and-series | summation-of-series | <p>Let $$< a_{\mathrm{n}} > $$ be a sequence such that $$a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$$. If $$28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$$, where $$\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$$ are the first $$\mathrm{m}$$ prime numbers,... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["C"] | null | Given the sum of the first n terms, $S_n = \frac{n^2+3n}{(n+1)(n+2)}$, we can find the n<sup>th</sup> term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms :
<br/><br/>So,
$$a_n = S_n - S_{n-1}$$
<br/><br/>Solving, we get :
<br/><br/>$$a_n = \frac{n^2+3n}{(n+1)(n+2)} ... | mcq | jee-main-2023-online-12th-april-morning-shift | 8,156 |
1lgsw5qty | maths | sequences-and-series | summation-of-series | <p>For $$k \in \mathbb{N}$$, if the sum of the series $$1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$$ is 10 , then the value of $$k$$ is _________.</p> | [] | null | 2 | From the given series :
<br/><br/>$$10 = 1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$$
<br/><br/>We isolate the 1 to get :
<br/><br/>$$9 = \frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$$ .......(1)
<br/><br/>Divide each term in the equation by $k$ :
<br/><br/>$$\frac{9}{k} = \fra... | integer | jee-main-2023-online-11th-april-evening-shift | 8,157 |
1lguwu480 | maths | sequences-and-series | summation-of-series | <p>Let $$S=109+\frac{108}{5}+\frac{107}{5^{2}}+\ldots .+\frac{2}{5^{107}}+\frac{1}{5^{108}}$$. Then the value of $$\left(16 S-(25)^{-54}\right)$$ is equal to ___________.</p> | [] | null | 2175 | We have, $S=109+\frac{108}{5}+\frac{107}{5^2}+\ldots+\frac{2}{5^{107}}+\frac{1}{5^{108}}$ ...........(i) <br/><br/>$\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2}+\frac{107}{5^3}+\ldots .+\frac{2}{5^{108}}+\frac{1}{5^{109}}$ .............(ii)
<br/><br/>On subtracting Eq. (ii) from Eq. (i), we get
<br/><br/>$$
\begin{aligned... | integer | jee-main-2023-online-11th-april-morning-shift | 8,158 |
1lgvpd2gn | maths | sequences-and-series | summation-of-series | <p>If $$\mathrm{S}_{n}=4+11+21+34+50+\ldots$$ to $$n$$ terms, then $$\frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "227"}, {"identifier": "B", "content": "226"}, {"identifier": "C", "content": "220"}, {"identifier": "D", "content": "223"}] | ["D"] | null | Given that
<br/><br/>$$
\begin{aligned}
& \mathrm{S}_n=4+11+21+24+50+\ldots+\mathrm{T}_n \\\\
& \mathrm{~S}_n=~~~~~~~~4+11+21+34++\mathrm{T}_{n-1}+\mathrm{T}_n \\
& -\quad-\quad-\quad-\quad-\quad-\quad- \\
& \hline 0=4+7+10+13+16+\ldots\left(\mathrm{T}_n-\mathrm{T}_{n-2}\right)-\mathrm{T}_n
\end{aligned}
$$
<br/><br/>$... | mcq | jee-main-2023-online-10th-april-evening-shift | 8,159 |
1lgxwg676 | maths | sequences-and-series | summation-of-series | <p>The sum of all those terms, of the arithmetic progression 3, 8, 13, ...., 373, which are not divisible by 3, is equal to ____________.</p> | [] | null | 9525 | The sum of all those terms, of the arithmetic progression $3,8,13 \ldots, 373$.
<br/><br/>Which are not divisible by 3 is
<br/><br/>$$
\begin{aligned}
& =(3+8+13+18+\ldots+373) -(3+18+33+\ldots+363) \\\\
& =\frac{75}{2}(3+373)-\frac{25}{2}(3+363) \\\\
& =\frac{75}{2} \times 376-\frac{25}{2} \times 366 \\\\
& =75 \times... | integer | jee-main-2023-online-10th-april-morning-shift | 8,160 |
1lgyl2od3 | maths | sequences-and-series | summation-of-series | <p>Let $$\mathrm{a}_{\mathrm{n}}$$ be the $$\mathrm{n}^{\text {th }}$$ term of the series $$5+8+14+23+35+50+\ldots$$ and $$\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$$. Then $$\mathrm{S}_{30}-a_{40}$$ is equal to :</p> | [{"identifier": "A", "content": "11280"}, {"identifier": "B", "content": "11290"}, {"identifier": "C", "content": "11310"}, {"identifier": "D", "content": "11260"}] | ["B"] | null | Let $\mathrm{S}_n=5+8+14+23+\ldots .+a_n$
<br/><br/>and $\mathrm{S}_n=0+5+8+14+\ldots .+a_n$
<br/><br/>On subtracting, we get
<br/><br/>$$
\begin{aligned}
& 0=5+3+6 \ldots-a_n \\\\
& \Rightarrow a_n=5+3+6+9+\ldots(n-1) \text { terms } \\\\
& =5+\left[\frac{(n-1)}{2}(6+(n-2) 3)\right]
\end{aligned}
$$
<br/><br/>$$
\be... | mcq | jee-main-2023-online-8th-april-evening-shift | 8,161 |
jaoe38c1lse57f85 | maths | sequences-and-series | summation-of-series | <p>The sum of the series $$\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$$ up to 10 -terms is</p> | [{"identifier": "A", "content": "$$\\frac{45}{109}$$\n"}, {"identifier": "B", "content": "$$-\\frac{55}{109}$$\n"}, {"identifier": "C", "content": "$$\\frac{55}{109}$$\n"}, {"identifier": "D", "content": "$$-\\frac{45}{109}$$"}] | ["B"] | null | <p>General term of the sequence,</p>
<p>$$\begin{aligned}
& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\
... | mcq | jee-main-2024-online-31st-january-morning-shift | 8,164 |
1lsg5cobe | maths | sequences-and-series | summation-of-series | <p>Let $$S_n$$ be the sum to $$n$$-terms of an arithmetic progression $$3,7,11$$,
If $$40<\left(\frac{6}{n(n+1)} \sum_\limits{k=1}^n S_k\right)<42$$, then $$n$$ equals ________.</p> | [] | null | 9 | <p>$$\begin{aligned}
& \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\
&=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\
&=2 \mathrm{n}^2+\mathrm{n} \\
& \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K... | integer | jee-main-2024-online-30th-january-evening-shift | 8,165 |
luxwdj8i | maths | sequences-and-series | summation-of-series | <p>If $$\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots . .+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$$, then $$\alpha$$ is equal to ___________.</p> | [] | null | 1011 | <p>$$\begin{aligned}
& \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\
& \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\
& \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) ... | integer | jee-main-2024-online-9th-april-evening-shift | 8,167 |
luy6z4j2 | maths | sequences-and-series | summation-of-series | <p>If the sum of the series $$\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$$ is equal to 5, then $$50 \mathrm{~d}$$ is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "20"}] | ["A"] | null | <p>$$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$$</p>
<p>Multiply and divide by $$d$$</p>
<p>$$\begin{aligned}
& \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\
& \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{... | mcq | jee-main-2024-online-9th-april-morning-shift | 8,168 |
lv2erzk5 | maths | sequences-and-series | summation-of-series | <p>The value of $$\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{305}{301}$$\n"}, {"identifier": "B", "content": "$$\\frac{306}{305}$$\n"}, {"identifier": "C", "content": "$$\\frac{32}{31}$$\n"}, {"identifier": "D", "content": "$$\\frac{31}{30}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n(n+1)^2}{\sum_\limits{n=1}^{100} n^2(n+1)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n^3+2 n^2+n}... | mcq | jee-main-2024-online-4th-april-evening-shift | 8,169 |
lv5gt51f | maths | sequences-and-series | summation-of-series | <p>Let the positive integers be written in the form :</p>
<p><img src="data:image/png;base64,UklGRvQEAABXRUJQVlA4IOgEAABwYwCdASoAA6UBP4HA3mc2MK4noHToSsAwCWlu4XdhG/P58iPxBcuiJs2bNmzZs2bNj6YykDSTJkyZMmTJkyZMmTJkyZMmTJkyZDQzmJcnECBAgQIECBAgQIECBAgQIECBAgJSFLZI8ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx4... | [] | null | 103 | <p>To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.</p>
<h3><strong>Understanding the Pattern</strong></h3>
<p><p><strong>First row ($k = 1$)</strong>: Contains 1 number.</p></p>
... | integer | jee-main-2024-online-8th-april-morning-shift | 8,170 |
lv7v4g04 | maths | sequences-and-series | summation-of-series | <p>If $$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$$ and $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=\mathrm{n}$$, then the point $$(\mathrm{m}, \mathrm{n})$$ lies on the line</p> | [{"identifier": "A", "content": "$$11(x-1)-100 y=0$$\n"}, {"identifier": "B", "content": "$$11 x-100 y=0$$\n"}, {"identifier": "C", "content": "$$11(x-1)-100(y-2)=0$$\n"}, {"identifier": "D", "content": "$$11(x-2)-100(y-1)=0$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\
& \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\
& =\frac... | mcq | jee-main-2024-online-5th-april-morning-shift | 8,172 |
lvb2953r | maths | sequences-and-series | summation-of-series | <p>If $$\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0$$, and $$(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$$, where $$a, b \in N$$, then $$(a+b)$$ equal to _________.</p> | [] | null | 3660 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4/file-1lwawri94.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4... | integer | jee-main-2024-online-6th-april-evening-shift | 8,173 |
WSabZw5gL3VAFWgcOz1qpqahkk8e9ppyl | maths | sets-and-relations | number-of-sets-and-relations | Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y $$ \subseteq $$ X, Z $$ \subseteq $$ X and Y $$ \cap $$ Z is empty, is : | [{"identifier": "A", "content": "3<sup>5</sup>"}, {"identifier": "B", "content": "2<sup>5</sup>"}, {"identifier": "C", "content": "5<sup>3</sup>"}, {"identifier": "D", "content": "5<sup>2</sup>"}] | ["A"] | null | For any element x<sub>i</sub> present in X, 4 cases arises while making subsets Y and Z.
<br><br><b>Case- 1 :</b> x<sub>i</sub> $$ \in $$ Y, x<sub>i</sub> $$ \in $$ Z $$ \Rightarrow $$ Y $$ \cap $$ Z $$ \ne $$ $$\phi $$
<br><br><b>Case- 2 :</b> x<sub>i</sub> $$ \in $$ Y, x<sub>i</sub> $$ \notin $$ Z $$ \Rightarrow $$ Y... | mcq | aieee-2012 | 8,174 |
l91h0jo8 | maths | sets-and-relations | number-of-sets-and-relations | Let A and B be two sets containing four and
two elements respectively. Then, the number
of subsets of the set A $\times$ B , each having atleast
three elements are | [{"identifier": "A", "content": "219"}, {"identifier": "B", "content": "256"}, {"identifier": "C", "content": "275"}, {"identifier": "D", "content": "510"}] | ["A"] | null | Given,<br/><br/>
$$
\begin{aligned}
&n(A)=4, n(B) =2 \\\\
&\Rightarrow n(A \times B) =8
\end{aligned}
$$<br/><br/>
Total number of subsets of set $(A \times B)=2^8$<br/><br/>
Number of subsets of set $A \times B$ having no element (i.e. $\phi)=1$<br/><br/>
Number of subsets of set $A \times B$ having one element $={ }^... | mcq | jee-main-2015-offline | 8,175 |
ITWzRryGcqebgm3ACxWNT | maths | sets-and-relations | number-of-sets-and-relations | Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$} <br/><br/>and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then | [{"identifier": "A", "content": "P $$ \\subset $$ Q and Q $$-$$ P $$ \\ne $$ $$\\phi $$"}, {"identifier": "B", "content": "Q $$ \\not\\subset $$ P "}, {"identifier": "C", "content": "P $$ \\not\\subset $$ Q"}, {"identifier": "D", "content": "P = Q"}] | ["D"] | null | Given,
<br><br>sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$
<br><br>$$ \Rightarrow... | mcq | jee-main-2016-online-10th-april-morning-slot | 8,176 |
DFFepIdax5IrxuOX | maths | sets-and-relations | number-of-sets-and-relations | Two sets A and B are as under :
<br/><br/>A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};
<br/><br/>B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)<sup>2</sup> + 9(b - 5)<sup>2</sup> $$ \le $$ 36 };
<br/><br/>Then | [{"identifier": "A", "content": "neither A $$ \\subset $$ B nor B $$ \\subset $$ A"}, {"identifier": "B", "content": "B $$ \\subset $$ A"}, {"identifier": "C", "content": "A $$ \\subset $$ B"}, {"identifier": "D", "content": "A $$ \\cap $$ B = $$\\phi $$ ( an empty set )"}] | ["C"] | null | Given,
<br><br>$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$
<br><br>Let $$a - 6 = x$$ and $$b - 5 = y$$
<br><br>$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$
<br><br>$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$
<br><br>This is a equation of ellipse.
<br><br>This el... | mcq | jee-main-2018-offline | 8,177 |
qjItDZ6yIHBa8CspKAyI4 | maths | sets-and-relations | number-of-sets-and-relations | Let Z be the set of integers.
<br/>If A = {x $$ \in $$ Z : 2<sup>(x + 2) (x<sup>2</sup> $$-$$ 5x + 6)</sup> = 1} and
<br/>B = {x $$ \in $$ Z : $$-$$ 3 < 2x $$-$$ 1 < 9},
<br/>then the number of subsets of the set A $$ \times $$ B, is | [{"identifier": "A", "content": "2<sup>12</sup>"}, {"identifier": "B", "content": "2<sup>18</sup>"}, {"identifier": "C", "content": "2<sup>10</sup>"}, {"identifier": "D", "content": "2<sup>15</sup>"}] | ["D"] | null | A ={x $$ \in $$ z : 2<sup>(x+2)(x<sup>2</sup> $$-$$ 5x + 6) </sup> = 1}
<br><br>2<sup>(x+2)(x<sup>2</sup> $$-$$ 5x + 6)</sup> = 2<sup>0</sup> $$ \Rightarrow $$ x = $$-$$ 2, 2, 3
<br><br>A = {$$-$$2, 2, 3}
<br><br>B = {x $$\varepsilon $$ Z : $$-$$ < 2x $$-$$ 1 < 9}
<br><br>B = {0, 1, 2, 3, 4}
<br><br>A $$ \times $... | mcq | jee-main-2019-online-12th-january-evening-slot | 8,179 |
uh9zBvgdVXVFTfGA1mjgy2xukewrzn2p | maths | sets-and-relations | number-of-sets-and-relations | If R = {(x, y) : x, y
$$ \in $$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$$ \le $$ 8} is a relation
on the set of integers Z, then the domain of R<sup>–1</sup> is : | [{"identifier": "A", "content": "{0, 1} "}, {"identifier": "B", "content": "{\u20132, \u20131, 1, 2}"}, {"identifier": "C", "content": "{\u20131, 0, 1}"}, {"identifier": "D", "content": "{\u20132, \u20131, 0, 1, 2}"}] | ["C"] | null | Given R = {(x, y) : x, y
$$ \in $$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$$ \le $$ 8}
<br><br>So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1)
<br>(-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}
<br><br>$$ \Rightarrow $$ R : { -2, -1, 0, 1, 2} $$ \to $$ {-1, 0, 1}
<br><br>$$ \therefore $$ R<sup>... | mcq | jee-main-2020-online-2nd-september-morning-slot | 8,181 |
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