Sequences and Series jee mains questions | Sequences and Series Previous Year Question

Q 11 :

For the two positive numbers $a, b,$ if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}$ , 10 and $\frac{1}{b}$ are in an arithmetic progression, then $16 a + 12 b$ is equal to ________ . [2023]

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(3)

$a, b, \frac{1}{18}$ are in G.P. $\Rightarrow b^{2} = \frac{a}{18}$

$\Rightarrow a = 18 b^{2}$ $...(i)$

$\frac{1}{a}, 10, \frac{1}{b} are in A.P. \Rightarrow 20 = \frac{1}{a} + \frac{1}{b}$

$\Rightarrow a + b = 20 a b \Rightarrow 18 b^{2} + b = 20 \cdot 18 b^{2} b [using (i)]$

$\Rightarrow 18 b + 1 = 360 b^{2} \Rightarrow 360 b^{2} - 18 b - 1 = 0$

$\Rightarrow 360 b^{2} - 30 b + 12 b - 1 = 0 \Rightarrow (30 b + 1) (12 b - 1) = 0$

$which gives b = \frac{1}{12} (Taking only positive value)$

$So, a = \frac{1}{8} .$

$Hence, 16 a + 12 b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 2 + 1 = 3 .$

Q 12 :

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 $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to________ . [2023]

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(60)

Given, $a_{4} \cdot a_{6} = 9$

$\Rightarrow {(a_{5})}^{2} = 9 \Rightarrow a_{5} = 3 and a_{5} + a_{7} = 24$

$\Rightarrow a_{5} + a_{5} r^{2} = 24 \Rightarrow a_{5} (1 + r^{2}) = 24$

$\Rightarrow (1 + r^{2}) = 8 \Rightarrow r = \sqrt{7} \Rightarrow a_{1} = \frac{3}{49}$

$So, a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7} = 9 + 27 + 3 + 21 = 60$

Q 13 :

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 $c_{3} = \frac{13}{4}$ then $\sum_{k = 1}^{\infty} c_{k} - (12 a_{6} + 8 b_{4})$ is equal to ________ . [2023]

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(9)

Given, $c_{k} = a_{k} + b_{k}$

$a_{1} = b_{1} = 4$

Now, $a_{2} = 4 r_{1}$ $a_{3} = 4 r_{1}^{2}$

$b_{2} = 4 r_{2}$ $b_{3} = 4 r_{2}^{2}$

Now, $c_{2} = a_{2} + b_{2} = 5$

$c_{3} = a_{3} + b_{3} = \frac{13}{4}$

$\Rightarrow r_{1} + r_{2} = \frac{5}{4} and r_{1}^{2} + r_{2}^{2} = \frac{13}{6}$

So, $r_{1} r_{2} = \frac{3}{8} which gives r_{1} = \frac{1}{2}, r_{2} = \frac{3}{4}$

$\sum_{k = 1}^{\infty} c_{k} - (12 a_{6} + 8 b_{4}) = \frac{4}{1 - r_{1}} + \frac{4}{1 - r_{2}} - (\frac{48}{32} + \frac{27}{2})$

$= 24 - 15 = 9$

Q 14 :

Let the first three terms $2, p$ and $q,$ with $q \neq 2,$ of a G.P. be respectively the 7th, 8th and 13th terms of an A.P. If the 5th term of the G.P. is the $n^{th}$ term of the A.P., then $n$ is equal to [2024]

177
151
169
163

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4

(4)

Since, $2, p$ and $q$ are in G.P.

$∴$ $p^{2} = 2 q$ ...(i)

Let first term of the A.P. be $a$ and common difference be $d .$

$∴$ $T_{7} = a + 6 d = 2$ ...(ii)

$T_{8} = a + 7 d = p$ ...(iii)

and $T_{13} = a + 12 d = q$ ...(iv)

$∴$ From (ii) and (iii), we get $d = p - 2$

From (ii) and (iv), we get $6 d = q - 2 \Rightarrow 6 (p - 2) = q - 2$

$\Rightarrow 6 p = q + 10$ ...(v)

From (i) and (v), we get $p = 2$ or $p = 10$ $∴$ $q = 2$ or 50

But $q \neq 2$

Hence, $p = 10, q = 50$ $∴$ $d = 8$ and $a = - 46$

Since, 5th term of G.P. $= n^{th}$ term of A.P.

$∴$ $2 {(\frac{10}{2})}^{4} = - 46 + (n - 1) 8$

$\Rightarrow 1250 = - 46 + 8 n - 8 \Rightarrow 8 n = 1304 \Rightarrow n = 163$

Q 15 :

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 4th, 6th, and 8th terms is equal to [2024]

78
96
84
91

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(4)

Let the G.P. be $\frac{a}{r^{3}}, \frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}, a r^{3}, a r^{4}$

Now, $\frac{a}{r^{2}} + a r^{2} = \frac{70}{3}$ ...(i)

and $\frac{a}{r} \times a r = 49 \Rightarrow a^{2} = 49 \Rightarrow a = 7$ (G.P. is increasing)

Now, $\frac{7}{r^{2}} + 7 r^{2} = \frac{70}{3}$ [Using (i)]

$\Rightarrow 3 r^{4} - 10 r^{2} + 3 = 0$

$\Rightarrow (3 r^{2} - 1) (r^{2} - 3) = 0 \Rightarrow r^{2} = \frac{1}{3} or r^{2} = 3$

As the G.P. is increasing, $∴$ $r^{2} = 3 \Rightarrow r = \sqrt{3}$

Now, $a + a r^{2} + a r^{4} = 7 + 7 (3) + 7 (9) = 91$

Q 16 :

Let $a, a r, a r^{2}, \dots$ be an infinite G.P. If $\sum_{n = 0}^{\infty} a r^{n} = 57$ and $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747,$ then $a + 18 r$ is equal to [2024]

38
46
31
27

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(3)

Given, $\sum_{n = 0}^{\infty} a r^{n} = 57$

$\Rightarrow \frac{a}{1 - r} = 57$ ...(i)

[Sum of infinite G.P.]

Also, $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747$

i.e., $a^{3} + a^{3} r^{3} + \dots \infty = 9747$

$\Rightarrow \frac{a^{3}}{1 - r^{3}} = 9747 \Rightarrow \frac{(57 {(1 - r))}^{3}}{1 - r^{3}} = 9747$

$\frac{(1 - r) (1 + r + r^{2})}{{(1 - r)}^{3}} = 19 \Rightarrow 19 {(1 - r)}^{2} = 1 + r + r^{2}$

$\Rightarrow 19 + 19 r^{2} - 38 r = 1 + r + r^{2}$

$\Rightarrow 18 r^{2} - 39 r + 18 = 0$

$\Rightarrow r = \frac{2}{3}, \frac{3}{2} \Rightarrow r = \frac{2}{3}$ $[∵ | r | < 1]$

$∴$ $a = 57 (1 - \frac{2}{3}) \Rightarrow a = 19$

So, $a + 18 r = 19 + 12 = 31$

Q 17 :

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 [2024]

$13$
$- 4$
$- 1$
$11$

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(4)

Given that 3, $a, b, c,$ are in A.P.

So, $a - 3 = b - a$

$\Rightarrow 2 a = b + 3$ ...(i)

and $b - a = c - b$

$\Rightarrow 2 b = a + c$ ...(ii)

Also, given that 3, $a - 1, b + 1, c + 9$ are in G.P.

So, $\frac{a - 1}{3} = \frac{b + 1}{a - 1}$

$\Rightarrow {(a - 1)}^{2} = 3 b + 3 \Rightarrow a^{2} - 2 a + 1 = 3 b + 3$

$\Rightarrow a^{2} - 2 a = 3 (2 a - 3) + 2$ (Using (i))

$\Rightarrow a^{2} - 8 a + 7 = 0 \Rightarrow a^{2} - 7 a - a + 7 = 0$

$\Rightarrow a (a - 7) - 1 (a - 7) = 0$

$\Rightarrow (a - 1) (a - 7) = 0 \Rightarrow a = 1, 7$

By (i), $b = - 1$ and $b = 11$

Since, $b$ cannot be negative.

By (ii), $c = 15$

$∴$ A.M. of $a, b, c = \frac{a + b + c}{3} = \frac{15 + 11 + 7}{3} = \frac{33}{3} = 11$

Q 18 :

Let $α$ and $β$ be the roots of the equation $p x^{2} + q x - r = 0,$ where $p \neq 0 .$ If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{α} + \frac{1}{β} = \frac{3}{4},$ then the value of ${(α - β)}^{2}$ is: [2024]

$8$
$9$
$\frac{20}{3}$
$\frac{80}{9}$

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(4)

We have,

$α + β = - \frac{q}{p}$ ...(i) and $α β = - \frac{r}{p}$ ...(ii)

Now, $\frac{1}{α} + \frac{1}{β} = \frac{3}{4} \Rightarrow \frac{α + β}{α β} = \frac{3}{4} \Rightarrow \frac{q}{r} = \frac{3}{4} \Rightarrow \frac{r}{q} = \frac{4}{3}$

$∵$ $p, q$ and $r$ are in G.P $∴$ $\frac{q}{p} = \frac{r}{q} = \frac{4}{3}$

So, $α + β = - \frac{4}{3}$ and $α β = - {(\frac{r}{q})}^{2} = - \frac{16}{9}$

Now, ${(α - β)}^{2} = {(α + β)}^{2} - 4 α β = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$

Q 19 :

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 [2024]

7
4
5
6

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3

4

(4)

Sum of 64 terms = 7 $\times$ Sum of odd terms

$\Rightarrow a + a r + a r^{2} + \dots + a r^{63} = 7 (a + a r^{2} + a r^{4} + \dots + a r^{62})$

$\Rightarrow 1 + r + r^{2} + \dots + r^{63} = 7 (1 + r^{2} + r^{4} + \dots + r^{62})$

$\Rightarrow \frac{1 (1 - r^{64})}{1 - r} = \frac{7 \times (1 - {(r^{2})}^{32})}{1 - r^{2}} \Rightarrow \frac{1 - r^{64}}{(1 - r)} = \frac{7 \times (1 - r^{64})}{(1 - r) (1 + r)}$

$\Rightarrow 1 + r = 7 \Rightarrow r = 6$

Q 20 :

If each term of a geometric progression $a_{1}, a_{2}, a_{3}, \dots$ with $a_{1} = \frac{1}{8}$ and $a_{2} \neq a_{1},$ is the arithmetic mean of the next two terms and $S_{n} = a_{1} + a_{2} + \dots + a_{n},$ then $S_{20} - S_{18}$ is equal to [2024]

$- 2^{15}$
$2^{18}$
$2^{15}$
$- 2^{18}$

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(1)

Let $r$ be the common ratio.

Now, $a_{n} = A.M. of a_{n + 1} and a_{n + 2} = \frac{a_{n + 1} + a_{n + 2}}{2}$

$\Rightarrow a_{1} \cdot r^{n - 1} = \frac{1}{2} [a_{1} \cdot r^{n} + a_{1} \cdot r^{n + 1}] \Rightarrow 2 r^{n - 1} = r^{n} + r^{n + 1}$

$\Rightarrow 2 = r + r^{2} \Rightarrow r^{2} + r - 2 = 0 \Rightarrow (r + 2) (r - 1) = 0$

$\Rightarrow r = - 2 (∵ r \neq 1 as a_{1} \neq a_{2})$

Now, $S_{20} - S_{18} = a_{19} + a_{20} (∵ S_{n} = a_{1} + a_{2} + \dots + a_{n})$

$= \frac{1}{8} \cdot r^{18} + \frac{1}{8} \cdot r^{19} = \frac{1}{8} {(- 2)}^{18} [1 - 2] = \frac{- 2^{18}}{2^{3}} = - 2^{15}$

Topic Question Set

Q 11 :

For the two positive numbers $a, b,$ if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}$ , 10 and $\frac{1}{b}$ are in an arithmetic progression, then $16 a + 12 b$ is equal to ________ . [2023]

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Q 12 :

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 $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to________ . [2023]

0%

0%

Q 14 :

Let the first three terms $2, p$ and $q,$ with $q \neq 2,$ of a G.P. be respectively the 7th, 8th and 13th terms of an A.P. If the 5th term of the G.P. is the $n^{th}$ term of the A.P., then $n$ is equal to [2024]

Q 15 :

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 4th, 6th, and 8th terms is equal to [2024]

Q 16 :

Let $a, a r, a r^{2}, \dots$ be an infinite G.P. If $\sum_{n = 0}^{\infty} a r^{n} = 57$ and $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747,$ then $a + 18 r$ is equal to [2024]

Q 17 :

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 [2024]

Q 18 :

Let $α$ and $β$ be the roots of the equation $p x^{2} + q x - r = 0,$ where $p \neq 0 .$ If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{α} + \frac{1}{β} = \frac{3}{4},$ then the value of ${(α - β)}^{2}$ is: [2024]

Q 19 :

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 [2024]

Q 20 :

If each term of a geometric progression $a_{1}, a_{2}, a_{3}, \dots$ with $a_{1} = \frac{1}{8}$ and $a_{2} \neq a_{1},$ is the arithmetic mean of the next two terms and $S_{n} = a_{1} + a_{2} + \dots + a_{n},$ then $S_{20} - S_{18}$ is equal to [2024]

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