question_id stringlengths 8 35 | subject stringclasses 3
values | chapter stringclasses 90
values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
values | explanation stringlengths 1 10.6k | question_type stringclasses 3
values | paper_id stringclasses 154
values | __index_level_0__ int64 2 13.4k |
|---|---|---|---|---|---|---|---|---|---|---|---|
1l6gglr6d | maths | sequences-and-series | am,-gm-and-hm | <p>Consider two G.Ps. 2, 2<sup>2</sup>, 2<sup>3</sup>, ..... and 4, 4<sup>2</sup>, 4<sup>3</sup>, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $${(2)^{{{225} \over 8}}}$$, then $$\sum\limits_{k = 1}^n {k(n - k)} $$ is equal to :</p> | [{"identifier": "A", "content": "560"}, {"identifier": "B", "content": "1540"}, {"identifier": "C", "content": "1330"}, {"identifier": "D", "content": "2600"}] | ["C"] | null | <p>Given G.P's 2, 2<sup>2</sup>, 2<sup>3</sup>, .... 60 terms</p>
<p>4, 4<sup>2</sup>, .... n terms</p>
<p>Now, G.M $$ = {2^{{{225} \over 8}}}$$</p>
<p>$${\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}$$</p>
<p>$$\left( {{2^{{{{n^2} + n + 1830} \over {60 + n}}}}} \right) = {2... | mcq | jee-main-2022-online-26th-july-morning-shift | 7,937 |
lgnybkbk | maths | sequences-and-series | am,-gm-and-hm | Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}, G_{3}$ be three geometric<br/><br/> means of two distinct positive numbers. Then $G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}$ is equal to : | [{"identifier": "A", "content": "$\\left(A_{1}+A_{2}\\right)^{2} G_{1} G_{3}$"}, {"identifier": "B", "content": "$\\left(A_{1}+A_{2}\\right) G_{1}^{2} G_{3}^{2}$"}, {"identifier": "C", "content": "$2\\left(A_{1}+A_{2}\\right) G_{1}^{2} G_{3}^{2}$"}, {"identifier": "D", "content": "$2\\left(A_{1}+A_{2}\\right) G_{1} G_{... | ["A"] | null | <p>Now, we have the following relations :</p>
<p>Arithmetic progression :</p>
<p>Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the commo... | mcq | jee-main-2023-online-15th-april-morning-shift | 7,938 |
1lgsuajhg | maths | sequences-and-series | am,-gm-and-hm | <p>Let $$a, b, c$$ and $$d$$ be positive real numbers such that $$a+b+c+d=11$$. If the maximum value of $$a^{5} b^{3} c^{2} d$$ is $$3750 \beta$$, then the value of $$\beta$$ is</p> | [{"identifier": "A", "content": "110"}, {"identifier": "B", "content": "108"}, {"identifier": "C", "content": "90"}, {"identifier": "D", "content": "55"}] | ["C"] | null | Given that $$a+b+c+d=11$$ and the maximum value of $$a^5 b^3 c^2 d$$ is $$3750\beta$$, you assumed the numbers to be $$\frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d$$.
<br/><br/>Applying the AM-GM inequality:
<br/><br/>$$\frac{\frac... | mcq | jee-main-2023-online-11th-april-evening-shift | 7,939 |
lsappvw8 | maths | sequences-and-series | am,-gm-and-hm | Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is : | [{"identifier": "A", "content": "-4"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "11"}] | ["D"] | null | <p>Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:</p>
<p>$$ d = a - 3 $$</p>
<p>The nth term of an A.P. is given by the formula:</p>
<p>$$ T_n = a + (n-1)d $$</p>
<p>So, using this formula, we can express $b$ and $c$ ... | mcq | jee-main-2024-online-1st-february-morning-shift | 7,940 |
jaoe38c1lse508pa | maths | sequences-and-series | am,-gm-and-hm | <p>For $$0 < c < b < a$$, let $$(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$$ and $$\alpha \neq 1$$ be one of its root. Then, among the two statements</p>
<p>(I) If $$\alpha \in(-1,0)$$, then $$b$$ cannot be the geometric mean of $a$ and $$c$$</p>
<p>(II) If $$\alpha \in(0,1)$$, then $$b$$ may be the geometric mean ... | [{"identifier": "A", "content": "only (II) is true\n"}, {"identifier": "B", "content": "Both (I) and (II) are true\n"}, {"identifier": "C", "content": "only (I) is true\n"}, {"identifier": "D", "content": "Neither (I) nor (II) is true"}] | ["B"] | null | <p>$$\begin{aligned}
& f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\
& f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\
& f(1)=0 \\
& \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\
& \alpha=\frac{c+a-2 b}{a+b-2 c} \\
& \text { If, }-1<\alpha<0 \\
& -1<\frac{c+a-2 b}{a+b-2 c}<0 \\
& b+c<2 a \text { and } b>\frac{a+c}{2}
\end{align... | mcq | jee-main-2024-online-31st-january-morning-shift | 7,941 |
lv2erzp0 | maths | sequences-and-series | am,-gm-and-hm | <p>Let three real numbers $$a, b, c$$ be in arithmetic progression and $$a+1, b, c+3$$ be in geometric progression. If $$a>10$$ and the arithmetic mean of $$a, b$$ and $$c$$ is 8, then the cube of the geometric mean of $$a, b$$ and $$c$$ is</p> | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "316"}, {"identifier": "C", "content": "312"}, {"identifier": "D", "content": "128"}] | ["A"] | null | <p>$$\begin{aligned}
& 2 b=a+c \quad \text{.... (1)}\\
& b^2=(a+1)(c+3) \quad \text{.... (2)}\\
& \frac{a+b+c}{3}=8 \quad \text{.... (3)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow & \frac{3 b}{3}=8 \\
& b=8 \\
\Rightarrow \quad & a c+3 a+c+3=64
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 3 a+c+a c=61 \quad \t... | mcq | jee-main-2024-online-4th-april-evening-shift | 7,942 |
wcwhgSoE2D9djeuf | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${{T_r}}$$ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, $$m \ne n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,$$ then a - d equals | [{"identifier": "A", "content": "$${1 \\over m} + {1 \\over n}$$ "}, {"identifier": "B", "content": "1 "}, {"identifier": "C", "content": "$${1 \\over {m\\,n}}$$ "}, {"identifier": "D", "content": "0 "}] | ["D"] | null | $${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$$
<br><br>$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$
<br><br>$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$
<br><br>$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn... | mcq | aieee-2004 | 7,944 |
GjqY94gBeyZFKgR8 | maths | sequences-and-series | arithmetic-progression-(a.p) | A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after | [{"identifier": "A", "content": "19 months"}, {"identifier": "B", "content": "20 months"}, {"identifier": "C", "content": "21 months "}, {"identifier": "D", "content": "18 months "}] | ["C"] | null | Let required number of months $$=n$$
<br><br>$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$
<br><br>$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$
<br><br>$$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \ri... | mcq | aieee-2011 | 7,947 |
9Gv7MfBVcVWqAbtn3Bhao | maths | sequences-and-series | arithmetic-progression-(a.p) | If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4$${^{{1 \\over 3}}}$$"}, {"identifier": "C", "content": "4$${^{{2 \\over 3}}}$$"}, {"identifier": "D", "content": "4"}] | ["A"] | null | a, b and c are in AP.
<br><br>$$ \therefore $$ a + c = 2b
<br><br>As, abc = 8
<br><br> $$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8
<br><br>$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4
<br><br>$$ \therefore $$ ac = 4 and a + c = 4
<br><br>Then,
<br><br>b = $$\left( {{{a + c} \over 2}} \right)$$ =... | mcq | jee-main-2017-online-9th-april-morning-slot | 7,949 |
0pRQnw9nIKu0GgXK | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that
<br/><br/>$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.
<br/><br/>$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "66"}, {"identifier": "C", "content": "68"}, {"identifier": "D", "content": "34"}] | ["D"] | null | a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> . . . a<sub>43</sub> are in AP
<br><br>So, a<sub>2</sub> = a<sub>1</sub> + d
<br><br>a<sub>3</sub> = a<sub>1</sub> + 2d
<br><br>.
<br><br>.
<br><br>.
<br><br>a<sub>49</sub> =a<sub>1</sub> + 48d
<br><br>Now given, $${a_9} + {a_{43}} = 66$$
<br><br>$$ \Rightarrow \,\,\,\,$$ ... | mcq | jee-main-2018-offline | 7,950 |
woTqZt9YlsJk9DcM8ygkw | maths | sequences-and-series | arithmetic-progression-(a.p) | If x<sub>1</sub>, x<sub>2</sub>, . . ., x<sub>n</sub> and $${1 \over {{h_1}}}$$, $${1 \over {{h_2}}}$$, . . . , $${1 \over {{h_n}}}$$ are two A.P..s such that x<sub>3</sub> = h<sub>2</sub> = 8 and x<sub>8</sub> = h<sub>7</sub> = 20, then x<sub>5</sub>.h<sub>10</sub> equals : | [{"identifier": "A", "content": "2560"}, {"identifier": "B", "content": "2650"}, {"identifier": "C", "content": "3200"}, {"identifier": "D", "content": "1600"}] | ["A"] | null | Assume d<sub>1</sub> is the common difference of A.P x<sub>1</sub>,x<sub>2</sub> ..... x<sub>n</sub><br><br>
Given x<sub>3</sub> = 8 and x<sub>8</sub> = 20<br><br>
$$ \therefore $$ x<sub>1</sub> + 2d<sub>1</sub> = 8 ..... <b>(i)</b><br>
and x<sub>1</sub> + 7d<sub>1</sub> = 20 .....<b> (ii)</b><br><br>
Solving <b>(i)</b... | mcq | jee-main-2018-online-15th-april-morning-slot | 7,951 |
6FiYbaaf6MERQvv8ffrYX | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,$$ (x<sub>i</sub> $$ \ne $$ 0 for i = 1, 2, ..., n) be in A.P. such that x<sub>1</sub>=4 and x<sub>21</sub> = 20. If n is the least positive integer for which $${x_n} > 50,$$ then $$\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)} $$ is equ... | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$${{13} \\over 8}$$"}, {"identifier": "D", "content": "$${{13} \\over 4}$$"}] | ["D"] | null | $$ \because $$$$\,\,\,$$ $${1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}}$$ are in A.P.
<br><br>x<sub>1</sub> = 4 and x<sub>21</sub> = 20
<br><br>Let 'd' be the common difference of this A.P.
<br><br>$$\therefore\,\,\,$$ its 21<sup>st</sup> term = $${1 \over {{x_{21}}}} = {1 \over {{x_1}}... | mcq | jee-main-2018-online-16th-april-morning-slot | 7,952 |
d4VJ7d65BnAyLfgZ7j3rsa0w2w9jwy1gult | maths | sequences-and-series | arithmetic-progression-(a.p) | If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ............... a<sub>n</sub> are in A.P. and a<sub>1</sub> + a<sub>4</sub> + a<sub>7</sub> + ........... + a<sub>16</sub> = 114, then a<sub>1</sub> + a<sub>6</sub> + a<sub>11</sub> + a<sub>16</sub> is equal to : | [{"identifier": "A", "content": "38"}, {"identifier": "B", "content": "98"}, {"identifier": "C", "content": "76"}, {"identifier": "D", "content": "64"}] | ["C"] | null | 3(a<sub>1</sub> + a<sub>16</sub>) = 114<br><br>
$${a_1} + {a_{16}} = 38$$<br><br>
Now a<sub>1</sub> + a<sub>6</sub> + a<sub>11</sub> + a<sub>16</sub> = 2(a<sub>1</sub> + a<sub>16</sub>)<br><br>
= 2 × 38 = 76 | mcq | jee-main-2019-online-10th-april-morning-slot | 7,953 |
kAbLnyXnMBfNTGXpwC3rsa0w2w9jxb4m3oi | maths | sequences-and-series | arithmetic-progression-(a.p) | If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... are in A.P. such that a<sub>1</sub> + a<sub>7</sub> + a<sub>16</sub> = 40, then the sum of the first 15 terms of this A.P. is : | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "150"}, {"identifier": "D", "content": "280"}] | ["B"] | null | a<sub>1</sub> + a<sub>7</sub> + a<sub>16</sub> = 40<br><br>
$${a_1} + \left( {{a_1} + 6d} \right) + ({a_1} + 15d) = 40$$<br><br>
$$ \Rightarrow 3{a_1} + 21d = 40$$<br><br>
$$ \Rightarrow {a_1} + 7d = {{40} \over 3}$$<br><br>
$$ \Rightarrow {a_1} + {a_2}....... + {a_{15}} = {{15} \over 2}[{a_1} + {a_{15}}]$$<br><br>
$$ ... | mcq | jee-main-2019-online-12th-april-evening-slot | 7,954 |
MAuiUu0xJFaSUmPVeZ3rsa0w2w9jx6g3geq | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>n</sub> denote the sum of the first n terms of an A.P. If S<sub>4</sub> = 16 and S<sub>6</sub>= – 48, then S<sub>10</sub> is equal to : | [{"identifier": "A", "content": "- 320"}, {"identifier": "B", "content": "- 380"}, {"identifier": "C", "content": "- 460"}, {"identifier": "D", "content": "- 210"}] | ["A"] | null | S<sub>4</sub> = $${4 \over 2}\left( {2a + 3d} \right) = 16$$<br><br>
$$ \Rightarrow 2a + 3d = 8$$<br><br>
S<sub>4</sub> = $${6 \over 2}\left( {2a + 5d} \right) = -48$$<br><br>
$$ \Rightarrow 2a + 5d = -16$$<br><br>
$$ \therefore $$ d = -12 and a = 22, Now S<sub>10</sub> = $${{10} \over 2}\left( {44 - 108} \right) = - ... | mcq | jee-main-2019-online-12th-april-morning-slot | 7,955 |
mkSTbBbv8X594mlprT3rsa0w2w9jx2g4463 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>,......be an A.P. with a<sub>6</sub> = 2. Then the common difference of this A.P., which maximises the
product a<sub>1</sub>a<sub>4</sub>a<sub>5</sub>, is :
| [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${6 \\over 5}$$"}, {"identifier": "C", "content": "$${8 \\over 5}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["C"] | null | first term = a, Common difference = d
<br><br>
$$ \therefore $$ a + 5d = 2<br><br>
a<sub>1</sub>. a<sub>4</sub>. a<sub>5</sub> = a(a + 3d) (a + 4d)<br><br>
f(d) = (2 – 5d) (2 – 2d) (2 – d)<br><br>
$$ \Rightarrow $$ $$f'(d) = 0 \Rightarrow d = {2 \over 3},{8 \over 5}$$<br><br>
$$ \Rightarrow $$ $$f''(d) < 0\,at\,d ... | mcq | jee-main-2019-online-10th-april-evening-slot | 7,956 |
dYLoLXwINfh5DdmFfC18hoxe66ijvwubvnl | maths | sequences-and-series | arithmetic-progression-(a.p) | If the sum and product of the first three term in
an A.P. are 33 and 1155, respectively, then a value
of its 11<sup>th</sup> term is :- | [{"identifier": "A", "content": "\u201325"}, {"identifier": "B", "content": "\u201336"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "\u201335"}] | ["A"] | null | Let the three terms are a - d, a, a + d
<br><br>Given a - d + a + a + d = 33
<br><br>$$ \Rightarrow $$ 3a = 33
<br><br>$$ \Rightarrow $$ a = 11
<br><br>Also given,
<br><br>(a - d)a(a + d) = 1155
<br><br>$$ \Rightarrow $$ (a<sup>2</sup> - d<sup>2</sup>)a = 1155
<br><br>$$ \Rightarrow $$ (11<sup>2</sup> - d<sup>2</sup>)1... | mcq | jee-main-2019-online-9th-april-evening-slot | 7,957 |
9YvLd5bxYwk2gXlduoBxc | maths | sequences-and-series | arithmetic-progression-(a.p) | Let the sum of the first n terms of a non-constant
A.P., a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... be $$50n + {{n(n - 7)} \over 2}A$$, where
A is a constant. If d is the common difference of
this A.P., then the ordered pair (d, a<sub>50</sub>) is equal to | [{"identifier": "A", "content": "(A, 50+45A)"}, {"identifier": "B", "content": "(50, 50+45A)"}, {"identifier": "C", "content": "(A, 50+46A)"}, {"identifier": "D", "content": "(50, 50+46A)"}] | ["C"] | null | S<sub>n</sub> = $$50n + {{n(n - 7)} \over 2}A$$
<br><br>We know, n<sup>th</sup> tem
<br><br>T<sub>n</sub> = S<sub>n</sub> - S<sub>n - 1</sub>
<br><br>= $$50n + {{n(n - 7)} \over 2}A$$ - $$50\left( {n - 1} \right) - {{\left( {n - 1} \right)\left( {n - 8} \right)} \over 2}A$$
<br><br>= 50 + $${A \over 2}\left[ {{n^2} - 7... | mcq | jee-main-2019-online-9th-april-morning-slot | 7,958 |
jMQGhZPXti3X7SerSWxl2 | maths | sequences-and-series | arithmetic-progression-(a.p) | If 19<sup>th</sup> term of a non-zero A.P. is zero, then its (49<sup>th</sup> term) : (29<sup>th</sup> term) is : | [{"identifier": "A", "content": "2 : 1"}, {"identifier": "B", "content": "4 : 1"}, {"identifier": "C", "content": "1 : 3"}, {"identifier": "D", "content": "3 : 1"}] | ["D"] | null | a + 18d = 0 . . . . .(1)
<br><br>$${{a + 48d} \over {a + 28d}} = {{ - 18d + 48d} \over { - 18d + 28d}} = {3 \over 1}$$ | mcq | jee-main-2019-online-11th-january-evening-slot | 7,960 |
XiEmOz8m70Df51MsaRno1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is - | [{"identifier": "A", "content": "1356"}, {"identifier": "B", "content": "1256"}, {"identifier": "C", "content": "1365"}, {"identifier": "D", "content": "1465"}] | ["A"] | null | $$\sum\limits_{r = 2}^{13} {(7r + 2) = 7.{{2 + 13} \over 2}} \times 6 + 2 \times 12$$
<br><br>= 7 $$ \times $$90 + 24 = 654
<br><br>$$\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702$$
<br><br>Total = 654 + 702 = 1356 | mcq | jee-main-2019-online-10th-january-morning-slot | 7,961 |
kQCDkroRzgx0A6RuISTkL | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $${a_1},{a_2},.......,{a_{30}}$$ be an A.P.,
<br/><br/>$$S = \sum\limits_{i = 1}^{30} {{a_i}} $$ and $$T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}} $$.
<br/><br/>If $$a_5$$ = 27 and S - 2T = 75, then $$a_{10}$$ is equal to : | [{"identifier": "A", "content": "47"}, {"identifier": "B", "content": "42"}, {"identifier": "C", "content": "52"}, {"identifier": "D", "content": "57"}] | ["C"] | null | Let the common difference = d
<br><br>S = $$\sum\limits_{i = 1}^{30} {{a_i}} $$
<br><br>= $$a$$<sub>1</sub> + $$a$$<sub>2</sub> + . . . . . + $$a$$<sub>30</sub>
<br><br>$$ \therefore $$ S = $${{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]$$
<br><br>= 15 [$$a$$<sub>1</sub> + $$a$$<sub>1</sub> + 29d]
<br><b... | mcq | jee-main-2019-online-9th-january-morning-slot | 7,962 |
BFKROwAabU9SIEAGMs1jw | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a, b and c be the 7<sup>th</sup>, 11<sup>th</sup> and 13<sup>th</sup> terms respectively of a non-constant A.P. If these are also three consecutive terms of a G.P., then $${a \over c}$$ equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 13}$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | T<sub>7</sub> = A + 6d = a; T<sub>11</sub> = A + 10d = b; T<sub>13</sub> = A + 12d = c
<br><br>Now a, b, c are in G.P.
<br><br>$$ \therefore $$ b<sup>2</sup> = ac
<br><br>$$ \Rightarrow $$ (A + 10d)<sup>2</sup> = (A + 6d) (A + 12d)
<br><br>$$ \Rightarrow $$ A<sup>2</sup> + 100d<sup>2</... | mcq | jee-main-2019-online-9th-january-evening-slot | 7,963 |
r4WC0aKydDZkqNCmrD5bM | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all natural numbers 'n' such that
100 < n < 200 and H.C.F. (91, n) > 1 is : | [{"identifier": "A", "content": "3221"}, {"identifier": "B", "content": "3121"}, {"identifier": "C", "content": "3203"}, {"identifier": "D", "content": "3303"}] | ["B"] | null | $$ \because $$ 91 = 13 $$ \times $$ 7
<br><br>So the required numbers are either divisible by
7 or 13.
<br><br>S<sub>A</sub>
= sum of numbers between 100 and 200 which are divisible by 7.
<br><br>$$ \Rightarrow $$ S<sub>A</sub> = 105 + 112 + ..... + 196
<br><br>S<sub>A</sub> = $${{14} \over 2}\left[ {105 + 196} \right]... | mcq | jee-main-2019-online-8th-april-morning-slot | 7,964 |
WbC8tVOvR5LmL8klSE7k9k2k5khzuf1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of terms common to the two A.P.'s
3, 7, 11, ....., 407 and 2, 9, 16, ....., 709 is ______. | [] | null | 14 | First A.P. is 3, 7, 11, 15, 19, 23, ..... 407
<br><br>d<sub>1</sub> = 4
<br><br>Second A.P. is 2, 9, 16, 23, ..... 709
<br><br>d<sub>2</sub> = 7
<br><br>First common term = 23
<br><br>Common difference of new A.P using the common terms of the two given A.P's is d = L.C.M. (4, 7) = 28
<br><br>Last term $$ \le $$ 407
<br... | integer | jee-main-2020-online-9th-january-evening-slot | 7,965 |
qMKL8WjZUIAD5hRjSpjgy2xukfg6jne3 | maths | sequences-and-series | arithmetic-progression-(a.p) | If $${3^{2\sin 2\alpha - 1}}$$, 14 and $${3^{4 - 2\sin 2\alpha }}$$ are the first three terms of an A.P. for some $$\alpha $$, then the sixth
terms of this A.P. is: | [{"identifier": "A", "content": "66"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "65"}, {"identifier": "D", "content": "78"}] | ["A"] | null | Given that<br><br>$${3^{4 - \sin 2\alpha }} + {3^{2\sin 2\alpha - 1}} = 28$$<br><br>Let $${3^{2\sin 2\alpha }}$$ = t<br><br>$$ \Rightarrow $$ $${{81} \over t} + {t \over 3} = 28$$<br><br>$$ \Rightarrow $$t = 81, 3<br><br>$$ \therefore $$ $${3^{2\sin 2\alpha }}$$ = 3<sup>1</sup>, 3<sup>4</sup><br><br>$$\sin 2\alpha = ... | mcq | jee-main-2020-online-5th-september-morning-slot | 7,967 |
cFq6eDhgpo0Oel3P33jgy2xukfakjhs2 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, ..., an be a given A.P. whose<br/> common difference is an integer and <br/>S<sub>n</sub> = a<sub>1</sub> + a<sub>2</sub> + .... + a<sub>n</sub>. If a<sub>1</sub> = 1, a<sub>n</sub> = 300 and 15 $$ \le $$ n $$ \le $$ 50, then <br/>the ordered pair (S<sub>n-4</sub>, a<sub>n–4</sub>) is ... | [{"identifier": "A", "content": "(2480, 249) "}, {"identifier": "B", "content": "(2480, 248)"}, {"identifier": "C", "content": "(2490, 248)"}, {"identifier": "D", "content": "(2490, 249)"}] | ["C"] | null | $${a_n} = {a_1} + (n - 1)d$$<br><br>$$ \Rightarrow 300 = 1 + (n - 1)d$$<br><br>$$ \Rightarrow (n - 1)d = 299 = 13 \times 23$$<br><br>since, n $$ \in $$[15, 50]<br><br>$$ \therefore $$ n = 24 and d = 13<br><br>$${a_{n - 4}} = {a_{20}} = 1 + 19 \times 13 = 248$$<br><br>$$ \Rightarrow {a_{n - 4}} = 248$$<br><br>$${S_{n - ... | mcq | jee-main-2020-online-4th-september-evening-slot | 7,968 |
p6030E7IzHd7wg2E0Jjgy2xukf0p91yb | maths | sequences-and-series | arithmetic-progression-(a.p) | If the first term of an A.P. is 3 and the sum of
its first 25 terms is equal to the sum of its next
15 terms, then the common difference of this
A.P. is : | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}] | ["D"] | null | First 25 terms = a, a + d, .......,a + 24d
<br><br>Next 15 terms = a + 25d, a + 26d, ......, a + 39d
<br><br>$$ \therefore $$ $${{25} \over 2}\left[ {2a + 24d} \right] = {{15} \over 2}\left[ {2\left( {a + 25d} \right) + 14d} \right]$$
<br><br>$$ \Rightarrow $$ 50a + 600d = 15 [2a + 50d + 14d]
<br><br>$$ \Rightarrow $$ ... | mcq | jee-main-2020-online-3rd-september-morning-slot | 7,969 |
7cFUZoeafMOtnxiXOLjgy2xukez7017n | maths | sequences-and-series | arithmetic-progression-(a.p) | If the sum of first 11 terms of an A.P.,
<br/>a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ....
is 0 (a $$ \ne $$ 0), then the sum of the A.P.,
<br/>a<sub>1</sub>
, a<sub>3</sub>
, a<sub>5</sub>
,....., a<sub>23</sub> is ka<sub>1</sub>
, where k is equal to :
| [{"identifier": "A", "content": "$${{121} \\over {10}}$$"}, {"identifier": "B", "content": "-$${{121} \\over {10}}$$"}, {"identifier": "C", "content": "$${{72} \\over 5}$$"}, {"identifier": "D", "content": "-$${{72} \\over 5}$$"}] | ["D"] | null | Let common difference be d.
<br><br>$$ \because $$ a<sub>1</sub>
+ a<sub>2</sub>
+ a<sub>3</sub>
+ ... + a<sub>11</sub> = 0
<br><br>$$ \therefore $$ $${{11} \over 2}\left[ {2{a_1} + 10d} \right]$$ = 0
<br><br>$$ \Rightarrow $$ a<sub>1</sub>
+ 5d = 0
<br><br>$$ \Rightarrow $$ d = $${ - {{{a_1}} \over 5}}$$ .....(1... | mcq | jee-main-2020-online-2nd-september-evening-slot | 7,970 |
m38rP9Tkzz02doHgLE7k9k2k5hj1bjo | maths | sequences-and-series | arithmetic-progression-(a.p) | If the 10<sup>th</sup> term of an A.P. is $${1 \over {20}}$$ and its 20<sup>th</sup> term
is $${1 \over {10}}$$, then the sum of its first 200 terms is | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "$$100{1 \\over 2}$$"}, {"identifier": "C", "content": "$$50{1 \\over 4}$$"}, {"identifier": "D", "content": "50"}] | ["B"] | null | T<sub>10</sub> = a + 9d = $${1 \over {20}}$$ ....(1)
<br><br>T<sub>20</sub> = a + 19d = $${1 \over {10}}$$ .....(2)
<br><br>Equation (2) – (1)
<br><br>10d = $${1 \over {10}}$$ - $${1 \over {20}}$$
<br><br>$$ \Rightarrow $$ d = $${1 \over {200}}$$
<br><br>a + $${9 \over {200}}$$ = $${1 \over {20}}$$
<br><br>$$ \Rightarr... | mcq | jee-main-2020-online-8th-january-evening-slot | 7,971 |
MeExSco81dydzJRFtt7k9k2k5gpayn9 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let ƒ : <b>R</b> $$ \to $$ <b>R</b> be such that for all
x $$ \in $$ R <br/>(2<sup>1+x</sup> + 2<sup>1–x</sup>), ƒ(x) and (3<sup>x</sup> + 3<sup>–x</sup>) are in
A.P., <br/>then the minimum value of ƒ(x) is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["C"] | null | f(x) = $${{2\left( {{2^x} + {2^{ - x}}} \right) + \left( {{3^x} + {3^{ - x}}} \right)} \over 2} \ge 3$$
<br><br>As we know, A.M > G.M | mcq | jee-main-2020-online-8th-january-morning-slot | 7,972 |
UXHgHEDPLWdzlEEwzB1kluy1yat | maths | sequences-and-series | arithmetic-progression-(a.p) | The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _________. | [] | null | 1000 | Let N be the four digit number<br><br>gcd(N, 18) = 3<br><br>Hence N is an odd integer which is divisible by 3 but not by 9.<br><br>4 digit odd multiples of 3<br><br>1005, 1011, ..........., 9999 $$ \to $$ 1500<br><br>4 digit odd multiples of 9<br><br>1017, 1035, ..........., 9999 $$ \to $$ 500<br><br>Hence number of su... | integer | jee-main-2021-online-26th-february-evening-slot | 7,974 |
s8bWRo7c7px8aS6V1D1kmm3p42s | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>1</sub> be the sum of first 2n terms of an arithmetic progression. Let S<sub>2</sub> be the sum of first 4n terms of the same arithmetic progression. If (S<sub>2</sub> $$-$$ S<sub>1</sub>) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to : | [{"identifier": "A", "content": "7000"}, {"identifier": "B", "content": "1000"}, {"identifier": "C", "content": "3000"}, {"identifier": "D", "content": "5000"}] | ["C"] | null | S<sub>1</sub> = $${{2n} \over 2}$$[2a + (2n $$-$$ 1)d]<br><br>S<sub>2</sub> = $${{4n} \over 2}$$[2a + (4n $$-$$ 1)d]<br><br>(where a = T<sub>1</sub> and d is common difference)<br><br>S<sub>2</sub> $$-$$ S<sub>1</sub>$$ \Rightarrow $$ 2n[2a + (4n $$-$$ 1)d] $$-$$ n[2a + (2n $$-$$ 1)d] = 1000<br><br>$$ \Rightarrow $$ n[... | mcq | jee-main-2021-online-18th-march-evening-shift | 7,975 |
1krub79cd | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all the elements in the set {n$$\in$$ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to _____________. | [] | null | 1251 | 2040 = 2<sup>3</sup> $$\times$$ 3 $$\times$$ 5 $$\times$$ 17<br><br>n should not be multiple of 2, 3, 5 and 17.<br><br>Sum of all n = (1 + 3 + 5 + ...... + 99) $$-$$ (3 + 9 + 15 + 21 + ...... + 99) $$-$$ (5 + 25 + 35 + 55 + 65 + 85 + 95) $$-$$ (17)<br><br>= 2500 $$-$$ $${{17} \over 2}$$(3 + 99) $$-$$ 365 $$-$$ 17<br><b... | integer | jee-main-2021-online-22th-july-evening-shift | 7,977 |
1krvs2cgo | maths | sequences-and-series | arithmetic-progression-(a.p) | Let S<sub>n</sub> be the sum of the first n terms of an arithmetic progression. If S<sub>3n</sub> = 3S<sub>2n</sub>, then the value of $${{{S_{4n}}} \over {{S_{2n}}}}$$ is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "8"}] | ["A"] | null | Let a be first term and d be common diff. of this A.P.<br><br>Given, S<sub>3n</sub> = 3S<sub>2n</sub><br><br>$$ \Rightarrow {{3n} \over 2}[2a + (3n - 1)d] = 3{{2n} \over 2}[2a + (2n - 1)d]$$<br><br>$$ \Rightarrow 2a + (3n - 1)d = 4a + (4n - 2)d$$<br><br>$$ \Rightarrow 2a + (n - 1)d = 0$$<br><br>Now, $${{{S_{4n}}} \over... | mcq | jee-main-2021-online-25th-july-morning-shift | 7,978 |
1ks0bytdw | maths | sequences-and-series | arithmetic-progression-(a.p) | If $${\log _3}2,{\log _3}({2^x} - 5),{\log _3}\left( {{2^x} - {7 \over 2}} \right)$$ are in an arithmetic progression, then the value of x is equal to _____________. | [] | null | 3 | $$2{\log _3}({2^x} - 5) = {\log _2} + {\log _3}\left( {{2^x} - {7 \over 2}} \right)$$<br><br>Let $${2^x} = t$$<br><br>$${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$$<br><br>$${(t - 5)^2} = 2t - 7$$<br><br>$${t^2} - 12t + 32 = 0$$<br><br>$$(t - 4)(t - 8) = 0$$<br><br>$$\Rightarrow$$ 2<sup>x</sup> =... | integer | jee-main-2021-online-27th-july-morning-shift | 7,979 |
1ktd2u4n1 | maths | sequences-and-series | arithmetic-progression-(a.p) | The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is _____________. | [] | null | 7744 | 209, 220, 231, ..........., 495<br><br>Sum = $${{27} \over 2}$$(209 + 495) = 9504<br><br>Number containing 1 at unit place $$\matrix{
{\underline 2 } & {\underline 3 } & {\underline 1 } \cr
{\underline 3 } & {\underline 4 } & {\underline 1 } \cr
{\underline 4 } & {\underline 5 } & {... | integer | jee-main-2021-online-26th-august-evening-shift | 7,980 |
1ktkdo86s | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________. | [] | null | 5143 | A = 4-digit numbers divisible by 3<br><br>A = 1002, 1005, ....., 9999.<br><br>9999 = 1002 + (n $$-$$ 1)3<br><br>$$\Rightarrow$$ (n $$-$$ 1)3 = 8997 $$\Rightarrow$$ n = 3000<br><br>B = 4-digit numbers divisible by 7<br><br>B = 1001, 1008, ......., 9996<br><br>$$\Rightarrow$$ 9996 = 1001 + (n $$-$$ 1)7<br><br>$$\Rightarr... | integer | jee-main-2021-online-31st-august-evening-shift | 7,982 |
1kto9lnec | maths | sequences-and-series | arithmetic-progression-(a.p) | Let a<sub>1</sub>, a<sub>2</sub>, ..........., a<sub>21</sub> be an AP such that $$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}} $$. If the sum of this AP is 189, then a<sub>6</sub>a<sub>16</sub> is equal to : | [{"identifier": "A", "content": "57"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "36"}] | ["B"] | null | $$\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} } $$<br><br>$$ = {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)} $$<br><br>$$ \Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \righ... | mcq | jee-main-2021-online-1st-september-evening-shift | 7,983 |
1l54tsnl8 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let 3, 6, 9, 12, ....... upto 78 terms and 5, 9, 13, 17, ...... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ________.</p> | [] | null | 2223 | <p>1st AP :</p>
<p>3, 6, 9, 12, ....... upto 78 terms</p>
<p>t<sub>78</sub> = 3 + (78 $$-$$ 1)3</p>
<p>= 3 + 77 $$\times$$ 3</p>
<p>= 234</p>
<p>2nd AP :</p>
<p>5, 9, 13, 17, ...... upto 59 terms</p>
<p>t<sub>59</sub> = 5 + (59 $$-$$ 1)4</p>
<p>= 5 + 58 $$\times$$ 4</p>
<p>= 237</p>
<p>Common term's AP :</p>
<p>First t... | integer | jee-main-2022-online-29th-june-evening-shift | 7,984 |
1l567rodu | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let A = {1, a<sub>1</sub>, a<sub>2</sub> ....... a<sub>18</sub>, 77} be a set of integers with 1 < a<sub>1</sub> < a<sub>2</sub> < ....... < a<sub>18</sub> < 77. <br/><br/>Let the set A + A = {x + y : x, y $$\in$$ A} contain exactly 39 elements. Then, the value of a<sub>1</sub> + a<sub>2</sub> + ........ | [] | null | 702 | If we write the elements of $A+A$, we can certainly find 39 distinct elements as $1+1,1+a_{1}, 1+a_{2}, \ldots .1$ $+a_{18}, 1+77, a_{1}+77, a_{2}+77, \ldots \ldots a_{18}+77,77+77$.<br/><br/> It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.
<br... | integer | jee-main-2022-online-28th-june-morning-shift | 7,985 |
1l56q8kl9 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> ...... and b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub> ....... are A.P., and a<sub>1</sub> = 2, a<sub>10</sub> = 3, a<sub>1</sub>b<sub>1</sub> = 1 = a<sub>10</sub>b<sub>10</sub>, then a<sub>4</sub> b<sub>4</sub> is equal to -</p> | [{"identifier": "A", "content": "$${{35} \\over {27}}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${{27} \\over {28}}$$"}, {"identifier": "D", "content": "$${{28} \\over {27}}$$"}] | ["D"] | null | <p>a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> .... are in A.P. (Let common difference is d<sub>1</sub>)</p>
<p>b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub> .... are in A.P. (Let common difference is d<sub>2</sub>)</p>
<p>and a<sub>1</sub> = 2, a<sub>10</sub> = 3, a<sub>1</sub>b<sub>1</sub> = 1 = a<sub>10</sub>b<sub>10<... | mcq | jee-main-2022-online-27th-june-evening-shift | 7,986 |
1l5c1fk8r | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>If $$\{ {a_i}\} _{i = 1}^n$$, where n is an even integer, is an arithmetic progression with common difference 1, and $$\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120} $$, then n is equal to :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "92"}, {"identifier": "D", "content": "104"}] | ["B"] | null | <p>$$\sum\limits_{i = 1}^n {{a_i} = 192} $$</p>
<p>$$\Rightarrow$$ a<sub>1</sub> + a<sub>2</sub> + a<sub>3</sub> + ...... + a<sub>n</sub> = 192</p>
<p>$$ \Rightarrow {n \over 2}[{a_1} + {a_n}] = 192$$</p>
<p>$$ \Rightarrow {a_1} + {a_n} = {{384} \over n}$$ ..... (1)</p>
<p>Now, $$\sum\limits_{i = 1}^{{n \over 2}} {{a_{... | mcq | jee-main-2022-online-24th-june-morning-shift | 7,987 |
1l6dx827f | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a, b$$ be two non-zero real numbers. If $$p$$ and $$r$$ are the roots of the equation $$x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$$ and $$\mathrm{q}$$ and s are the roots of the equation $$x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$$, such that $$\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}... | [] | null | 38 | $\because$ Roots of $2 a x^{2}-8 a x+1=0$ are $\frac{1}{p}$ and $\frac{1}{r}$ and roots of $6 b x^{2}+12 b x+1=0$ are $\frac{1}{q}$ and $\frac{1}{s}$.
<br/><br/>
Let $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$
<br/><br/>
So sum of roots $2 \alpha-... | integer | jee-main-2022-online-25th-july-morning-shift | 7,988 |
1l6i041q0 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ___________.</p> | [] | null | 53 | <p>$${d_1} = {{199 - 100} \over 2} \notin I$$</p>
<p>$${d_2} = {{199 - 100} \over 3} = 33$$</p>
<p>$${d_3} = {{199 - 100} \over 4} \notin I$$</p>
<p>$${d_n} = {{199 - 100} \over {i + 1}} \in I$$</p>
<p>$${d_i} = 33 + 11,\,9$$</p>
<p>Sum of CD's $$ = 33 + 11 + 9$$</p>
<p>$$ = 53$$</p> | integer | jee-main-2022-online-26th-july-evening-shift | 7,989 |
1l6p3bg7l | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a_{1}, a_{2}, a_{3}, \ldots$$ be an A.P. If $$\sum\limits_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$$, then $$4 a_{2}$$ is equal to _________.</p> | [] | null | 16 | <p>Given</p>
<p>$$S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\infty $$</p>
<p>$${{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + ... | integer | jee-main-2022-online-29th-july-morning-shift | 7,991 |
1ldo7ckhv | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>The sum of the common terms of the following three arithmetic progressions.</p>
<p>$$3,7,11,15, \ldots ., 399$$,</p>
<p>$$2,5,8,11, \ldots ., 359$$ and</p>
<p>$$2,7,12,17, \ldots ., 197$$,</p>
<p>is equal to _____________.</p> | [] | null | 321 | $$
\begin{array}{ll}
3,7,11,15, \ldots \ldots \ldots . .399 : & \mathrm{~d}_1=4 \\\\
2,5,8,11, \ldots \ldots \ldots \ldots, 359 : & \mathrm{~d}_2=3 \\\\
2,7,12,17, \ldots \ldots, 197 : & \mathrm{~d}_3=5 \\\\
\operatorname{LCM}\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=60 &
\end{array}
$$
<br/><br/>Common t... | integer | jee-main-2023-online-1st-february-evening-shift | 7,992 |
ldo8c1vp | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $a_1, a_2, a_3, \ldots$ be an A.P. If $a_7=3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n !-4 a_{n(n+2)}$ is equal to : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "$\\frac{381}{4}$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "$\\frac{33}{4}$"}] | ["A"] | null | $a_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d$
<br/><br/>$$
\begin{aligned}
& a_{1} \cdot a_{4}=a(a+3 d) \\\\
& \Rightarrow(3-6 d)(3-6 d+3 d) \\\\
& \Rightarrow 3(1-2 d) 3(1-d) \\\\
& \Rightarrow 9\left(2 d^{2}-3 d+1\right)
\end{aligned}
$$
<br/><br/>Let $f(d)=2 d^{2}-3 d+1$
<br/><br/>$f^{\prime}(d)=4 d-3 \Righta... | mcq | jee-main-2023-online-31st-january-evening-shift | 7,993 |
1ldptn1y5 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$a_{1}, a_{2}, \ldots, a_{n}$$ be in A.P. If $$a_{5}=2 a_{7}$$ and $$a_{11}=18$$, then
<br/><br/>$$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________.</p> | [] | null | 8 | $a_{11}=18$
<br/><br/>$$
\begin{aligned}
& a+10 d=18 \\\\
& a_{5}=2 a_{7} \\\\
& a+4 d=2(a+6 d) \\\\
& a=-8 d
\end{aligned}
$$
<br/><br/>(i) and (ii) $\Rightarrow a=-72, d=9$.
<br/><br/>On rationalising the denominator, given expression
<br/><br/>$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-... | integer | jee-main-2023-online-31st-january-morning-shift | 7,995 |
1ldybitxm | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>For three positive integers p, q, r, $${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$$ and r = pq + 1 such that 3, 3 log$$_yx$$, 3 log$$_zy$$, 7 log$$_xz$$ are in A.P. with common difference $$\frac{1}{2}$$. Then r-p-q is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $x^{p q^{2}}=y^{q r}=z^{p^{2} r}$
<br/><br/>
$$
3 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2}
$$
<br/><br/>
$$
\begin{aligned}
& \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\
& y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\
& \Rightarrow \frac{7}{6} p q^{... | mcq | jee-main-2023-online-24th-january-morning-shift | 7,996 |
1lgpy2jk4 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$s_{1}, s_{2}, s_{3}, \ldots, s_{10}$$ respectively be the sum to 12 terms of 10 A.P. s whose first terms are $$1,2,3, \ldots .10$$ and the common differences are $$1,3,5, \ldots \ldots, 19$$ respectively. Then $$\sum_\limits{i=1}^{10} s_{i}$$ is equal to :</p> | [{"identifier": "A", "content": "7360"}, {"identifier": "B", "content": "7220"}, {"identifier": "C", "content": "7260"}, {"identifier": "D", "content": "7380"}] | ["C"] | null | We have 10 arithmetic progressions (A.P.s) with the first terms $$a_i$$ and the common differences $$d_i$$, where $$i = 1, 2, \ldots, 10$$.
<br/><br/>The first terms are $$a_i = i$$ and the common differences are $$d_i = 2i - 1$$.
<br/><br/>Now, we need to find the sum of the first 12 terms for each A.P. The formula ... | mcq | jee-main-2023-online-13th-april-morning-shift | 7,997 |
1lgutxklh | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$x_{1}, x_{2}, \ldots, x_{100}$$ be in an arithmetic progression, with $$x_{1}=2$$ and their mean equal to 200 . If $$y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100$$, then the mean of $$y_{1}, y_{2}, \ldots, y_{100}$$ is :</p> | [{"identifier": "A", "content": "10051.50"}, {"identifier": "B", "content": "10049.50"}, {"identifier": "C", "content": "10100"}, {"identifier": "D", "content": "10101.50"}] | ["B"] | null | We have, mean of $x_1, x_2 \ldots \ldots x_{100}=200$
<br/><br/>Where, $x_1, x_2 \ldots x_{100}$ are in AP with first term as 2.
<br/><br/>$$
\begin{aligned}
\text { Mean } & =200 \\\\
& =\frac{\sum\limits_{i=1}^{100} x_i}{100}=200
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\frac{100}{2} \times[2 \times 2+99 d] =20... | mcq | jee-main-2023-online-11th-april-morning-shift | 7,998 |
1lgzztm49 | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>Let $$S_{K}=\frac{1+2+\ldots+K}{K}$$ and $$\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$$, where $$A, B, C, D \in \mathbb{N}$$ and $$A$$ has least value. Then</p> | [{"identifier": "A", "content": "$$A+B+C+D$$ is divisible by 5"}, {"identifier": "B", "content": "$$A+C+D$$ is not divisible by $$B$$"}, {"identifier": "C", "content": "$$A+B=5(D-C)$$"}, {"identifier": "D", "content": "$$A+B$$ is divisible by $$\\mathrm{D}$$"}] | ["D"] | null | $$
\begin{aligned}
& \because S_k=\frac{1+2+\ldots+k}{k} \\\\
& =\frac{k(k+1)}{2 k}=\frac{k+1}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\
& \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\rig... | mcq | jee-main-2023-online-8th-april-morning-shift | 7,999 |
lsamv2vn | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $\mathrm{S}_{15}-\mathrm{S}_5$ is equal to : | [{"identifier": "A", "content": "800"}, {"identifier": "B", "content": "890"}, {"identifier": "C", "content": "790"}, {"identifier": "D", "content": "690"}] | ["C"] | null | <p>To solve this problem, we will start by using the properties of an arithmetic progression (AP).</p>
<p>The sum of the first $n$ terms of an AP can be calculated using the formula:
$$ S_n = \frac{n}{2} (2a + (n-1)d) $$
where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common differ... | mcq | jee-main-2024-online-1st-february-evening-shift | 8,000 |
lsapy2m8 | maths | sequences-and-series | arithmetic-progression-(a.p) | Let $3,7,11,15, \ldots, 403$ and $2,5,8,11, \ldots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___________. | [] | null | 6699 | <p>To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).
<p>The first AP is:</p></p>
<p>$$3, 7, 11, 15, \ldots, 403$$</p... | integer | jee-main-2024-online-1st-february-morning-shift | 8,001 |
lsbl2vr8 | maths | sequences-and-series | arithmetic-progression-(a.p) | The number of common terms in the progressions <br/><br/>$4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and <br/><br/>$3,6,9,12, \ldots \ldots$, up to $37^{\text {th }}$ term is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>$$4,9,14,19, \ldots$$, up to $$25^{\text {th }}$$ term</p>
<p>$$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$$</p>
<p>$$3,6,9,12, \ldots$$, up to $$37^{\text {th }}$$ term</p>
<p>$$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$$</p>
<p>Common difference of $$\mathrm{I}^{\text {st }}$$ series $$\mathrm{d}_1=5$$</p>
<p>Common differen... | mcq | jee-main-2024-online-27th-january-morning-shift | 8,002 |
jaoe38c1lscn8k0e | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>$$\text { The } 20^{\text {th }} \text { term from the end of the progression } 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4} \text { is : }$$</p> | [{"identifier": "A", "content": "$$-115$$"}, {"identifier": "B", "content": "$$-100$$"}, {"identifier": "C", "content": "$$-110$$"}, {"identifier": "D", "content": "$$-118$$"}] | ["A"] | null | <p>$$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$$</p>
<p>This is A.P. with common difference</p>
<p>$$\begin{aligned}
& d_1=-1+\frac{1}{4}=-\frac{3}{4} \\
& -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20
\end{aligned}$$</p>
<p>This is also A.P. $$\mathrm{a}=... | mcq | jee-main-2024-online-27th-january-evening-shift | 8,003 |
lv3vegci | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>An arithmetic progression is written in the following way</p>
<p><img src="data:image/png;base64,UklGRjIHAABXRUJQVlA4ICYHAADwYgCdASoAA/MAP4G+2mW2L6wnILMJgsAwCWlu/HyZcetQ2f1w/1/rfMBOYO0eXvb59Dcv6Ip7TtadrTtadrTrnAmZpMwrqGmNsPadrTtadrTtTNrvzQeUko135oPKRhlK3bVgTkiVPHpm7CNG42w9p2tNko135oPKSUa8DEsz2N5RXJ3p1I6IU7Wna07Wna02... | [] | null | 1505 | <p>First term is each row form pattern</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4ntqyk/94850946-5efa-47dd-aab7-199c5589db35/d9f12fc0-10fc-11ef-aaa0-17ca36a32505/file-1lw4ntqyl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4ntqyk/94850946-5efa-47dd-aab7-199c55... | integer | jee-main-2024-online-8th-april-evening-shift | 8,007 |
lv9s20ci | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>For $$x \geqslant 0$$, the least value of $$\mathrm{K}$$, for which $$4^{1+x}+4^{1-x}, \frac{\mathrm{K}}{2}, 16^x+16^{-x}$$ are three consecutive terms of an A.P., is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "16"}] | ["A"] | null | <p>To determine the least value of $$\mathrm{K}$$ for which the terms $$4^{1+x} + 4^{1-x}, \frac{\mathrm{K}}{2}, 16^x + 16^{-x}$$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.</p>
<p>For three numbers to be in an arithmetic progression, the middle term must b... | mcq | jee-main-2024-online-5th-april-evening-shift | 8,009 |
lvb294px | maths | sequences-and-series | arithmetic-progression-(a.p) | <p>A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $$\mathrm{m}$$ is equal to... | [{"identifier": "A", "content": "125"}, {"identifier": "B", "content": "160"}, {"identifier": "C", "content": "150"}, {"identifier": "D", "content": "180"}] | ["C"] | null | <p>To determine the value of $$\mathrm{m}$$, we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:</p>
<p>Initially, there are $$\mathrm{m}$$ computers, and it is estimated that with these $$\mathrm{m}$$ computers, the assignmen... | mcq | jee-main-2024-online-6th-april-evening-shift | 8,010 |
1l5w0vbmu | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let for $$f(x) = {a_0}{x^2} + {a_1}x + {a_2},\,f'(0) = 1$$ and $$f'(1) = 0$$. If a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub> are in an arithmatico-geometric progression, whose corresponding A.P. has common difference 1 and corresponding G.P. has common ratio 2, then f(4) is equal to _____________.</p> | [] | null | 2 | <p>Given,</p>
<p>$$f(x) = {a_0}{x^2} + {a_1}x + {a_2}$$</p>
<p>$$f'(0) = 1$$</p>
<p>$$f'(1) = 0$$</p>
<p>a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub> are in A. G. P</p>
<p>Common difference of $$AP = 1$$</p>
<p>Common ratio of $$GP = 2$$</p>
<p>A.P terms = a, a + 1, a + 2</p>
<p>G.P terms = y, ry, r<sup>2</sup>y</p>
<p>... | integer | jee-main-2022-online-30th-june-morning-shift | 8,011 |
1lgypzf6s | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let $$0 < z < y < x$$ be three real numbers such that $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in an arithmetic progression and $$x, \sqrt{2} y, z$$ are in a geometric progression. If $$x y+y z+z x=\frac{3}{\sqrt{2}} x y z$$ , then $$3(x+y+z)^{2}$$ is equal to ____________.</p> | [] | null | 150 | $\because \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
<br/><br/>$$
\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y}
$$ ........... (i)
<br/><br/>and $x, \sqrt{2} y, z$ are in G.P.
<br/><br/>$$
\Rightarrow 2 y^2=x z
$$ .......... (ii)
<br/><br/>from (i), $\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$
<br/><br/>$$... | integer | jee-main-2023-online-8th-april-evening-shift | 8,013 |
lsblc3ma | maths | sequences-and-series | arithmetico-geometric-progression | If $8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2 p)+\frac{1}{4^3}(3+3 p)+\cdots \cdots \infty$, then the value of $p$ is ____________. | [] | null | 9 | <p>$$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$$</p>
<p>$$\text { (sum of infinite terms of A.G.P }=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \text { ) }$$</p>
<p>$$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$$</p> | integer | jee-main-2024-online-27th-january-morning-shift | 8,014 |
lv9s20lt | maths | sequences-and-series | arithmetico-geometric-progression | <p>If $$1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots$$ upto $$\infty=2+\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$$, where a and b are integers with $$\operatorname{gcd}(a, b)=1$$, then $$\mathrm{11 a+1... | [] | null | 76 | <p>$$\begin{aligned}
& S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\\\
& =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty
\end{aligned}$$</p>
<p>$$\text { let } 1-\fr... | integer | jee-main-2024-online-5th-april-evening-shift | 8,015 |
lvc57uf2 | maths | sequences-and-series | arithmetico-geometric-progression | <p>Let the first term of a series be $$T_1=6$$ and its $$r^{\text {th }}$$ term $$T_r=3 T_{r-1}+6^r, r=2,3$$,
............ $$n$$. If the sum of the first $$n$$ terms of this series is $$\frac{1}{5}\left(n^2-12 n+39\right)\left(4 \cdot 6^n-5 \cdot 3^n+1\right)$$, then $$n$$ is equal to ___________.</p> | [] | null | 6 | <p>$$\begin{aligned}
& T_r=3 T_{r-1}+6^r \\
& \Rightarrow \text { solving homogenous part } \\
& T_r=3 T_{r-1} \\
& \Rightarrow x=3 \text { is the root }
\end{aligned}$$</p>
<p>$$\therefore T_r=a .3^r$$</p>
<p>Solving for particular part</p>
<p>$$\begin{aligned}
& T_r=b .6^r \\
& b .6^r=3 b 6^{r-1}+6^r \\
& \Rightarrow... | integer | jee-main-2024-online-6th-april-morning-shift | 8,016 |
jk3IWUpURQ7vO8z9 | maths | sequences-and-series | geometric-progression-(g.p) | Fifth term of a GP is 2, then the product of its 9 terms is | [{"identifier": "A", "content": "256"}, {"identifier": "B", "content": "512 "}, {"identifier": "C", "content": "1024"}, {"identifier": "D", "content": "none of these"}] | ["B"] | null | $$a{r^4} = 2$$
<br><br>$$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$$
<br><br>$$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$$ | mcq | aieee-2002 | 8,017 |
y3zqSDYlgLFRtylu | maths | sequences-and-series | geometric-progression-(g.p) | l, m, n are the $${p^{th}}$$, $${q^{th}}$$ and $${r^{th}}$$ term of a G.P all positive, $$then\,\left| {\matrix{
{\log \,l} & p & 1 \cr
{\log \,m} & q & 1 \cr
{\log \,n} & r & 1 \cr
} } \right|\,equals$$ | [{"identifier": "A", "content": "- 1"}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "1 "}, {"identifier": "D", "content": "0 "}] | ["D"] | null | $$l = A{R^{p - 1}}$$
<br><br>$$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$
<br><br>$$m = A{R^{q - 1}}$$
<br><br>$$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$
<br><br>$$n = A{R^{r - 1}}$$
<br><br>$$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$
<br><br>Now, $$\left| {\m... | mcq | aieee-2002 | 8,018 |
WsEUUUVc0kVsIZYL | maths | sequences-and-series | geometric-progression-(g.p) | Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is | [{"identifier": "A", "content": "5 "}, {"identifier": "B", "content": "3/5 "}, {"identifier": "C", "content": "8/5 "}, {"identifier": "D", "content": "1/5 "}] | ["B"] | null | Let $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$
<br><br>Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$
<br><br>Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$
<br><br>$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$
<br><br>Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$... | mcq | aieee-2002 | 8,019 |
dsXDAGYKTZx0KfLz | maths | sequences-and-series | geometric-progression-(g.p) | The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is | [{"identifier": "A", "content": "- 4"}, {"identifier": "B", "content": "- 12"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "4"}] | ["B"] | null | As per question,
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$$... | mcq | aieee-2008 | 8,021 |
cbq6hH68znDtn5vz | maths | sequences-and-series | geometric-progression-(g.p) | Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is : | [{"identifier": "A", "content": "$$2 - \\sqrt 3 $$ "}, {"identifier": "B", "content": "$$2 + \\sqrt 3 $$ "}, {"identifier": "C", "content": "$$\\sqrt 2 + \\sqrt 3 $$ "}, {"identifier": "D", "content": "$$3 + \\sqrt 2 $$ "}] | ["B"] | null | Let $$a,ar,a{r^2}$$ are in $$G.P.$$
<br><br>According to the question
<br><br>$$a,2ar,a{r^2}$$ are in $$A.P.$$
<br><br>$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$
<br><br>$$ \Rightarrow 4r = 1 + {r^2}$$
<br><br>$$ \Rightarrow {r^2} - 4r + 1 = 0$$
<br><br>$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$
<br><b... | mcq | jee-main-2014-offline | 8,022 |
t8Hle6z9XMRBLBrD | maths | sequences-and-series | geometric-progression-(g.p) | If the $${2^{nd}},{5^{th}}\,and\,{9^{th}}$$ terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is : | [{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "$${7 \\over 4}$$ "}, {"identifier": "C", "content": "$${8 \\over 5}$$ "}, {"identifier": "D", "content": "$${4 \\over 3}$$"}] | ["D"] | null | <p>The terms of an Arithmetic Progression (A.P.) are given by $a$, $a + d$, $a + 2d$, ..., where $a$ is the first term and $d$ is the common difference.</p>
<p>Given that the 2nd, 5th and 9th terms of an A.P. are in Geometric Progression (G.P.), we can denote them as follows :</p>
<p>2nd term = $a + d$</p>
<p>5th term ... | mcq | jee-main-2016-offline | 8,023 |
4jkMKIeMvDgkctyNVmMvG | maths | sequences-and-series | geometric-progression-(g.p) | If a, b, c are in A.P. and a<sup>2</sup>, b<sup>2</sup>, c<sup>2</sup> are in G.P. such that
<br/>a < b < c and a + b + c = $${3 \over 4},$$ then the value of a is : | [{"identifier": "A", "content": "$${1 \\over 4} - {1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over 4} - {1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over 4} - {1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 4} - {1 \\over {\\sqrt 2 }}$$"}] | ["C"] | null | $$ \because $$$$\,\,\,$$a, b, c are in A.P. then
<br><br>a + c = 2b
<br><br>also it is given that,
<br><br>a + b + c = $${{3 \over 4}}$$ . . . .(1)
<br><br>$$ \Rightarro... | mcq | jee-main-2018-online-15th-april-evening-slot | 8,025 |
jOGbNFYRMg8wfAaE013rsa0w2w9jx2b2rv6 | maths | sequences-and-series | geometric-progression-(g.p) | Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \ne $$ 0 and 0 < r $$ \le $$ $${1 \over 2}$$
. If 3$$a$$, 7b and 15c are the first three
terms of an A.P., then the 4<sup>th</sup> term of this A.P. is : | [{"identifier": "A", "content": "$$a$$"}, {"identifier": "B", "content": "$${7 \\over 3}a$$"}, {"identifier": "C", "content": "5$$a$$"}, {"identifier": "D", "content": "$${2 \\over 3}a$$"}] | ["A"] | null | a = a, b = ar and c = ar<sup>2</sup><br><br>
3a, 7b, 15c $$ \to $$ A.P.<br><br>
14b = 3a + 15c<br><br>
14(ar) = 3a + 15(ar<sup>2</sup>)<br><br>
15r<sup>2</sup> – 14r + 3 = 0<br><br>
$$ \Rightarrow r = {1 \over 3},{3 \over 5}(rejected)$$<br><br>
Common difference = 7b – 3a<br><br>
= 7ar – 3a<br><br>
$$ \Rightarrow $$ $$... | mcq | jee-main-2019-online-10th-april-evening-slot | 8,027 |
JASdaZ2MclBZMbjgwCTi1 | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>1</sub>, a<sub>2</sub>, . . . . . ., a<sub>10</sub> be a G.P. If $${{{a_3}} \over {{a_1}}} = 25,$$ then $${{{a_9}} \over {{a_5}}}$$ equals | [{"identifier": "A", "content": "5<sup>3</sup>"}, {"identifier": "B", "content": "2(5<sup>2</sup>)"}, {"identifier": "C", "content": "4(5<sup>2</sup>)"}, {"identifier": "D", "content": "5<sup>4</sup>"}] | ["D"] | null | a<sub>1</sub>, a<sub>2</sub>, . . . . ., a<sub>10</sub> are in G.P.,
<br><br>Let the common ratio be r
<br><br>$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$
<br><br>$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$ | mcq | jee-main-2019-online-11th-january-morning-slot | 8,029 |
sIssaJfB1HmzkLOvnfDcl | maths | sequences-and-series | geometric-progression-(g.p) | If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x <b>cannot</b> be | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "-2"}] | ["A"] | null | a, b, c are in G.P.
<br><br>So, b = ar
<br><br>and c = ar<sup>2</sup>
<br><br>given a + b + c = xb
<br><br>$$ \Rightarrow $$ a + br + ar<sup>2</sup> = x(ar)
<br><br>$$ \Rightarrow $$ 1 + r + r<sup>2</sup> = xr
<br><br>$$ \Rightarrow $$ x = 1 + r + $${1 \over r}$$
<br><br>let... | mcq | jee-main-2019-online-9th-january-morning-slot | 8,032 |
4x6CObgJQXDZB4of137k9k2k5fj4i71 | maths | sequences-and-series | geometric-progression-(g.p) | Let $${a_1}$$
, $${a_2}$$
, $${a_3}$$
,....... be a G.P. such that <br/>$${a_1}$$
< 0, $${a_1}$$
+ $${a_2}$$
= 4 and $${a_3}$$
+ $${a_4}$$
= 16. <br/>If $$\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $$, then $$\lambda $$ is
equal to: | [{"identifier": "A", "content": "171"}, {"identifier": "B", "content": "-171"}, {"identifier": "C", "content": "-513"}, {"identifier": "D", "content": "$${{511} \\over 3}$$"}] | ["B"] | null | $${a_1}$$
+ $${a_2}$$
= 4
<br><br>$$ \Rightarrow $$ $${a_1}$$
+ $${a_1}$$r
= 4 ...(1)
<br><br>$${a_3}$$
+ $${a_4}$$
= 16
<br><br>$$ \Rightarrow $$ $${a_1}$$r<sup>2</sup>
+ $${a_1}$$r<sup>3</sup>
= 16 ...(2)
<br><br>Doing (1) $$ \div $$ (2), we get
<br><br>r = $$ \pm $$ 2
<br><br>If r = 2, then a<sub>1</sub> = $... | mcq | jee-main-2020-online-7th-january-evening-slot | 8,033 |
94MqkDoVjiepQoAoLy7k9k2k5khy3qy | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>n</sub> be the n<sup>th</sup> term of a G.P. of positive terms.<br/><br/>
$$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$ and $$\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $$,
<br/><br/>
then $$\sum\limits_{n = 1}^{200} {{a_n}} $$ is equal to : | [{"identifier": "A", "content": "150"}, {"identifier": "B", "content": "175"}, {"identifier": "C", "content": "225"}, {"identifier": "D", "content": "300"}] | ["A"] | null | $$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $$
<br><br>$$ \Rightarrow $$ a<sub>3</sub> + a<sub>5</sub> + a<sub>7</sub> + .... + a<sub>201</sub> = 200
<br><br>$$ \Rightarrow $$ $$a{r^2}{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}$$ = 200 ....(1)
<br><br>$$\sum\limits_{n = 1}^{100} {{a_{2n}}... | mcq | jee-main-2020-online-9th-january-evening-slot | 8,034 |
jvyQgYboQpU3UgCLRYjgy2xukf0znhpx | maths | sequences-and-series | geometric-progression-(g.p) | The value of $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$ is equal to ______. | [] | null | 4 | Given, $${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$$
<br><br>As sum of GP upto infinity = $${a \over {1 - r}}$$
<br><br>$$ \therefore $$ $${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $$ = $${{{1 \over 3}} \over {1 - {1 \over 3}}}$$ =... | integer | jee-main-2020-online-3rd-september-morning-slot | 8,036 |
uGscaepOHl8oit0YBgjgy2xukfqat3me | maths | sequences-and-series | geometric-progression-(g.p) | If the sum of the second, third and fourth terms
of a positive term G.P. is 3 and the sum of its
sixth, seventh and eighth terms is 243, then the
sum of the first 50 terms of this G.P. is : | [{"identifier": "A", "content": "$${2 \\over {13}}\\left( {{3^{50}} - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over {13}}\\left( {{3^{50}} - 1} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over {26}}\\left( {{3^{49}} - 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over {26}}\\left( {{3^... | ["D"] | null | Let first term = a > 0
<br><br>Common ratio = r > 0
<br><br>ar + ar<sup>2</sup>
+ ar<sup>3</sup>
= 3 ....(i)
<br><br>ar<sup>5</sup>
+ ar<sup>6</sup>
+ ar<sup>7</sup>
= 243 ....(ii)
<br><br>$$ \Rightarrow $$ r<sup>4</sup>(ar + ar<sup>2</sup> + ar<sup>3</sup>) = 243
<br><br>$$ \Rightarrow $$ r<sup>4</sup>(3) =... | mcq | jee-main-2020-online-5th-september-evening-slot | 8,037 |
C38geiuYcGLB2P3ZAY1klrmzqlk | maths | sequences-and-series | geometric-progression-(g.p) | The sum of first four terms of a geometric progression (G. P.) is $${{65} \over {12}}$$ and the sum of their respective reciprocals is $${{65} \over {18}}$$. If the product of first three terms of the G.P. is 1, and the third term is $$\alpha$$, then 2$$\alpha$$ is _________. | [] | null | 3 | Let the terms are $$a,ar,a{r^2},a{r^3}$$<br><br>$$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$$ ..........(1)<br><br>$${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$$<br><br>$${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$$ ........... | integer | jee-main-2021-online-24th-february-evening-slot | 8,039 |
qJBoOi84bbZpOZlOUr1kls5n6t7 | maths | sequences-and-series | geometric-progression-(g.p) | Let A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ....... be squares such that for each n $$ \ge $$ 1, the length of the side of A<sub>n</sub> equals the length of diagonal of A<sub>n+1</sub>. If the length of A<sub>1</sub> is 12 cm, then the smallest value of n for which area of A<sub>n</sub> is less than one, is _____... | [] | null | 9 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266517/exam_images/rfwj0y3zp4ddisjyhq5w.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - Sequences and Series Question 146 English Explanation">
... | integer | jee-main-2021-online-25th-february-morning-slot | 8,040 |
9AKsFRmmitCwyDhceD1kluhfa4b | maths | sequences-and-series | geometric-progression-(g.p) | In an increasing geometric series, the sum of the second and the sixth term is $${{25} \over 2}$$ and the product of the third and fifth term is 25. Then, the sum of 4<sup>th</sup>, 6<sup>th</sup> and 8<sup>th</sup> terms is equal to : | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "26"}, {"identifier": "D", "content": "35"}] | ["D"] | null | a, ar, ar<sup>2</sup>, .....<br><br>$${T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}$$<br><br>$${a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}$$ .... (1)<br><br>$${T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25$$<br><br>$${a^2}{r^6} = 25$$ .....(2)<br><br>On dividing (1) by (2)<br><br>$${{{{(... | mcq | jee-main-2021-online-26th-february-morning-slot | 8,041 |
y5OVsMxprZsXF3XuT31kmhzkrrq | maths | sequences-and-series | geometric-progression-(g.p) | Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ___________. | [] | null | 3 | A.P. from the set will be 11, 16, 21, 26 .....
<br><br>G.P. from the set will be 4, 8, 16, 32, 64, 128, 256,
512, 1024, 2048, 4096, 8192 .....
<br><br>So common terms are 16, 256, 4096. | integer | jee-main-2021-online-16th-march-morning-shift | 8,042 |
SVaD9jBaTrJAspxagr1kmizc846 | maths | sequences-and-series | geometric-progression-(g.p) | Let $${1 \over {16}}$$, a and b be in G.P. and $${1 \over a}$$, $${1 \over b}$$, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________. | [] | null | 14 | $${a^2} = {b \over {16}}$$ and $${2 \over b} = {1 \over a} + 6$$<br><br>Solving, we get $$a = {1 \over {12}}$$ or $$a = - {1 \over 4}$$ [rejected]<br><br>if $$a = {1 \over {12}} \Rightarrow b = {1 \over 9}$$<br><br>$$ \therefore $$ $$72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$$ | integer | jee-main-2021-online-16th-march-evening-shift | 8,043 |
1ktbfqwpt | maths | sequences-and-series | geometric-progression-(g.p) | If the sum of an infinite GP a, ar, ar<sup>2</sup>, ar<sup>3</sup>, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar<sup>2</sup>, ar<sup>4</sup>, ar<sup>6</sup>, ....... is : | [{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${25 \\over 2}$$"}, {"identifier": "D", "content": "$${9 \\over 2}$$"}] | ["B"] | null | Sum of infinite terms :<br><br>$${a \over {1 - r}} = 15$$ ..... (i)<br><br>Series formed by square of terms :<br><br>a<sup>2</sup>, a<sup>2</sup>r<sup>2</sup>, a<sup>2</sup>r<sup>4</sup>, a<sup>2</sup>r<sup>6</sup> .......<br><br>Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$<br><br>$$ \Rightarrow {a \over {1 - r}}.{a \ov... | mcq | jee-main-2021-online-26th-august-morning-shift | 8,044 |
1ktd3ojty | maths | sequences-and-series | geometric-progression-(g.p) | Let a<sub>1</sub>, a<sub>2</sub>, ......., a<sub>10</sub> be an AP with common difference $$-$$ 3 and b<sub>1</sub>, b<sub>2</sub>, ........., b<sub>10</sub> be a GP with common ratio 2. Let c<sub>k</sub> = a<sub>k</sub> + b<sub>k</sub>, k = 1, 2, ......, 10. If c<sub>2</sub> = 12 and c<sub>3</sub> = 13, then $$\sum\li... | [] | null | 2021 | $$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$$ are in AP common difference $$=-3$$<br/><br/> $$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$$ are in GP common ratio $$=2$$<br/><br/> Since, $$c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$$<br/><br/>
$$\therefore c_{2} =a_{2}+b_{2}=12$$<br/><br/>
$$ c_{3} =a_{3}+b_{3}=13$$<br/><br/>
Now,... | integer | jee-main-2021-online-26th-august-evening-shift | 8,045 |
1ktipg9jo | maths | sequences-and-series | geometric-progression-(g.p) | Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r<sup>2</sup>, then r<sup>2</sup> $$-$$ d is equal to : | [{"identifier": "A", "content": "7 $$-$$ 7$$\\sqrt 3 $$"}, {"identifier": "B", "content": "7 + $$\\sqrt 3 $$"}, {"identifier": "C", "content": "7 $$-$$ $$\\sqrt 3 $$"}, {"identifier": "D", "content": "7 + 3$$\\sqrt 3 $$"}] | ["B"] | null | Let numbers be $${a \over r}$$, a, ar $$\to$$ G.P.<br><br>$${a \over r}$$, 2a, ar $$\to$$ A.P. $$\Rightarrow$$ 4a = $${a \over r}$$ + ar $$\Rightarrow$$ r + $${1 \over r}$$ = 4<br><br>r = 2 $$\pm$$ $$\sqrt 3 $$<br><br>4<sup>th</sup> form of G.P. = 3r<sup>2</sup> $$\Rightarrow$$ ar<sup>2</sup> = 3r<sup>2</sup> $$\Righta... | mcq | jee-main-2021-online-31st-august-morning-shift | 8,046 |
1l55j37xx | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let for n = 1, 2, ......, 50, S<sub>n</sub> be the sum of the infinite geometric progression whose first term is n<sup>2</sup> and whose common ratio is $${1 \over {{{(n + 1)}^2}}}$$. Then the value of <br/><br/>$${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$ is eq... | [] | null | 41651 | <p>$${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$$</p>
<p>Now $${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$</p>
<p>$$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) +... | integer | jee-main-2022-online-28th-june-evening-shift | 8,047 |
1l566am8y | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ....... be an increasing geometric progression of positive real numbers. If A<sub>1</sub>A<sub>3</sub>A<sub>5</sub>A<sub>7</sub> = $${1 \over {1296}}$$ and A<sub>2</sub> + A<sub>4</sub> = $${7 \over {36}}$$, then the value of A<sub>6</sub> + A<sub>8</sub> + A<sub>10</... | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "37"}, {"identifier": "C", "content": "43"}, {"identifier": "D", "content": "47"}] | ["C"] | null | <p>$${{{A_4}} \over {{r^3}}}.\,{{{A_4}} \over r}.\,{A_4}r\,.\,{A_4}{r^3} = {1 \over {1296}}$$</p>
<p>$${A_4} = {1 \over 6}$$</p>
<p>$${A_2} = {7 \over {36}} - {1 \over 6} = {1 \over {36}}$$</p>
<p>So $${A_6} + {A_8} + {A_{10}} = 1 + 6 + 36 = 43$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 8,048 |
1l58h3pwb | maths | sequences-and-series | geometric-progression-(g.p) | <p>If a<sub>1</sub> (> 0), a<sub>2</sub>, a<sub>3</sub>, a<sub>4</sub>, a<sub>5</sub> are in a G.P., a<sub>2</sub> + a<sub>4</sub> = 2a<sub>3</sub> + 1 and 3a<sub>2</sub> + a<sub>3</sub> = 2a<sub>4</sub>, then a<sub>2</sub> + a<sub>4</sub> + 2a<sub>5</sub> is equal to ___________.</p> | [] | null | 40 | <p>Let G.P. be a<sub>1</sub> = a, a<sub>2</sub> = ar, a<sub>3</sub> = ar<sup>2</sup>, .........</p>
<p>$$\because$$ 3a<sub>2</sub> + a<sub>3</sub> = 2a<sub>4</sub></p>
<p>$$\Rightarrow$$ 3ar + ar<sup>2</sup> = 2ar<sup>3</sup></p>
<p>$$\Rightarrow$$ 2ar<sup>2</sup> $$-$$ r $$-$$ 3 = 0</p>
<p>$$\therefore$$ r = $$-$$1 or... | integer | jee-main-2022-online-26th-june-evening-shift | 8,049 |
1l6kjt9ea | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $$\frac{98}{25}$$. Then the sum of the first 21 terms of an AP, whose first term is $$10\mathrm{a r}, \mathrm{n}^{\text {th }}$$ term is $$\mathrm{a}_{\mathrm{n}}$$ and the common differenc... | [{"identifier": "A", "content": "$$21 \\,\\mathrm{a}_{11}$$"}, {"identifier": "B", "content": "$$22 \\,\\mathrm{a}_{11}$$"}, {"identifier": "C", "content": "$$15 \\,\\mathrm{a}_{16}$$"}, {"identifier": "D", "content": "$$14 \\,\\mathrm{a}_{16}$$"}] | ["A"] | null | <p>Let first term of G.P. be a and common ratio is r</p>
<p>Then, $${a \over {1 - r}} = 5$$ ...... (i)</p>
<p>$$a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}$$</p>
<p>$$\therefore$$ $${r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}$$</p>
<p... | mcq | jee-main-2022-online-27th-july-evening-shift | 8,050 |
1ldprmen7 | maths | sequences-and-series | geometric-progression-(g.p) | <p>If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["A"] | null | $\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \mathrm{ar}^{3}(\mathrm{a}, \mathrm{r}>0)$
<br/><br/>$a^{4} r^{6}=1296$
<br/><br/>$a^{2} r^{3}=36$
<br/><br/>$a=\frac{6}{r^{3 / 2}}$
<br/><br/>$a+a r+a r^{2}+a r^{3}=126$
<br/><br/>$\frac{1}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{2}}{\... | mcq | jee-main-2023-online-31st-january-morning-shift | 8,051 |
ldqw0prh | maths | sequences-and-series | geometric-progression-(g.p) | Let $a, b, c>1, a^3, b^3$ and $c^3$ be in A.P., and $\log _a b, \log _c a$ and $\log _b c$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is $-444$, then $a b c$ is equal to : | [{"identifier": "A", "content": "343"}, {"identifier": "B", "content": "216"}, {"identifier": "C", "content": "$\\frac{343}{8}$"}, {"identifier": "D", "content": "$\\frac{125}{8}$"}] | ["B"] | null | <p>$$2{b^3} = {a^3} + {c^3}$$</p>
<p>$${\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)$$</p>
<p>$$ \Rightarrow {(\log a)^3} = {(\log c)^3}$$</p>
<p>$$ \Rightarrow \log a = \log c$$</p>
<p>$$ \Rightarrow a = c$$</p>
<p>$$ \Rightarrow a... | mcq | jee-main-2023-online-30th-january-evening-shift | 8,052 |
1ldsgdvt5 | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$\{ {a_k}\} $$ and $$\{ {b_k}\} ,k \in N$$, be two G.P.s with common ratios $${r_1}$$ and $${r_2}$$ respectively such that $${a_1} = {b_1} = 4$$ and $${r_1} < {r_2}$$. Let $${c_k} = {a_k} + {b_k},k \in N$$. If $${c_2} = 5$$ and $${c_3} = {{13} \over 4}$$ then $$\sum\limits_{k = 1}^\infty {{c_k} - (12{a_6} +... | [] | null | 9 | <p>$$\{ {a_k}\} $$ be a G.P. with $${a_1} = 4,r = {r_1}$$</p>
<p>And</p>
<p>$$\{ {b_k}\} $$ be G.P. with $${b_1} = 4,r = {r_2}$$ $$({r_1} < {r_2})$$</p>
<p>Now</p>
<p>$${C_k} = {a_k} + {b_k}$$</p>
<p>$${c_1} = 4 + 4 = 8$$ and $${c_2} = 5$$<p>
<p>$${a_2} + {b_2} = 5$$</p>
<p>$$\therefore$$ $${r_1} + {r_2} = {5 \over 4}$... | integer | jee-main-2023-online-29th-january-evening-shift | 8,053 |
1ldswm61u | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$a_1,a_2,a_3,...$$ be a $$GP$$ of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $$a_1a_9+a_2a_4a_9+a_5+a_7$$ is equal to __________.</p> | [] | null | 60 | Let $r$ be the common ratio of the G.P
<br/><br/>
$\therefore a_{1} r^{3} \times a_{1} r^{5}=9$
<br/><br/>
$a_{1}^{2} r^{8}=9 \Rightarrow a_{1} r^{4}=3$
<br/><br/>
And
<br/><br/>
$$
\begin{aligned}
& a_{1}\left(r^{4}+r^{6}\right)=24 \\\\
\Rightarrow & 3\left(1+r^{2}\right)=24 \\\\
\therefore & r^{2}=7 \text { and } a_{... | integer | jee-main-2023-online-29th-january-morning-shift | 8,054 |
1ldyc0ddl | maths | sequences-and-series | geometric-progression-(g.p) | <p>The 4$$^\mathrm{th}$$ term of GP is 500 and its common ratio is $$\frac{1}{m},m\in\mathbb{N}$$. Let $$\mathrm{S_n}$$ denote the sum of the first n terms of this GP. If $$\mathrm{S_6 > S_5 + 1}$$ and $$\mathrm{S_7 < S_6 + \frac{1}{2}}$$, then the number of possible values of m is ___________</p> | [] | null | 12 | $T_{4}=500$
<br/><br/>
$$
a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}}
$$
<br/><br/>
Now,
<br/><br/>
$$
\begin{aligned}
& S_{6} > S_{5}+1 \\\\
& \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\
& a r^{5} > 1 \\\\
& \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\
& \Rightar... | integer | jee-main-2023-online-24th-january-morning-shift | 8,056 |
1lgow1w1l | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let a$$_1$$, a$$_2$$, a$$_3$$, .... be a G.P. of increasing positive numbers. Let the sum of its 6<sup>th</sup> and 8<sup>th</sup> terms be 2 and the product of its 3<sup>rd</sup> and 5<sup>th</sup> terms be $$\frac{1}{9}$$. Then $$6(a_2+a_4)(a_4+a_6)$$ is equal to</p> | [{"identifier": "A", "content": "2$$\\sqrt2$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3$$\\sqrt3$$"}, {"identifier": "D", "content": "3"}] | ["D"] | null |
<p>Given the conditions :</p>
<ol>
<li>$a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$</li>
<li>$a_3 \cdot a_5 = \frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \frac{1}{9} \Rightarrow a r^3 = \frac{1}{3}$</li>
</ol>
<p>From this, we can form the equation $\frac{r^2}{3} + \frac{r^4}{3} = 2$, which simplifies to $r^4 +... | mcq | jee-main-2023-online-13th-april-evening-shift | 8,057 |
1lgxw30j8 | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let the first term $$\alpha$$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to</p> | [{"identifier": "A", "content": "241"}, {"identifier": "B", "content": "231"}, {"identifier": "C", "content": "220"}, {"identifier": "D", "content": "210"}] | ["B"] | null | Given that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a, a r, a r^2$
<br/><br/>According to the question, $a^2+a^2 r^2+a^2 r^4=33033$
<br/><br/>$$
\begin{aligned}
\Rightarrow a^2\left(1+r^2+r^4\right) & =3 \times 7 \times 11 \times 11 \times 13... | mcq | jee-main-2023-online-10th-april-morning-shift | 8,058 |
1lh2z2v5w | maths | sequences-and-series | geometric-progression-(g.p) | <p>If
<br/><br/>$$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$$,
<br/><br/>then $$k$$ is equal to ___________.</p> | [] | null | 400 | $\begin{aligned} &(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17} \\ & \quad+\ldots \ldots+20(21)^{19}=k(20)^{19} \\\\ & \Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19} \\\\ & \Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\f... | integer | jee-main-2023-online-6th-april-evening-shift | 8,059 |
lsan9avi | maths | sequences-and-series | geometric-progression-(g.p) | If three successive terms of a G.P. with common ratio $\mathrm{r}(\mathrm{r}>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to _____________. | [] | null | 1 | <p>To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.</p>
<p>According to the triangle inequality theore... | integer | jee-main-2024-online-1st-february-evening-shift | 8,060 |
jaoe38c1lsd4o6jt | maths | sequences-and-series | geometric-progression-(g.p) | <p>Let $$2^{\text {nd }}, 8^{\text {th }}$$ and $$44^{\text {th }}$$ terms of a non-constant A. P. be respectively the $$1^{\text {st }}, 2^{\text {nd }}$$ and $$3^{\text {rd }}$$ terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -</p> | [{"identifier": "A", "content": "990"}, {"identifier": "B", "content": "980"}, {"identifier": "C", "content": "960"}, {"identifier": "D", "content": "970"}] | ["D"] | null | <p>$$\begin{aligned}
& 1+d, \quad 1+7 d, 1+43 d \text { are in GP } \\
& (1+7 d)^2=(1+d)(1+43 d) \\
& 1+49 d^2+14 d=1+44 d+43 d^2 \\
& 6 d^2-30 d=0 \\
& d=5 \\
& S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\
& \quad=10[2+95] \\
& \quad=970
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift | 8,061 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.