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

Q 11 :

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

760
755
750
757

1

2

3

4

(4)

We have, $a r + a r^{3} + a r^{5} = 21, a r^{7} + a r^{9} + a r^{11} = 15309$

$\Rightarrow a r (1 + r^{2} + r^{4}) = 21$ ... (i)

and $a r^{7} (1 + r^{2} + r^{4}) = 15309$ ... (ii)

Divide equation (ii) by (i), we get

$\Rightarrow \frac{a . r^{7}}{a r} = \frac{15309}{21} \Rightarrow r^{6} = 729$

$\Rightarrow r = 3 and a = \frac{7}{91}$

Now, $S_{9} = \frac{a \cdot (r^{9} - 1)}{r - 1}$

$= \frac{\frac{7}{91} (19683 - 1)}{2}$

$= \frac{7 \times 19682}{91 \times 2}$

$= \frac{9841}{13} = 757$ .

Q 12 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

812
526
784
628

1

2

3

4

(3)

Here, r > 0 and $a_{1}, a_{2}, a_{3}, . . . . . > 0$ as G.P. has increasing positive terms.

$a_{1} a_{5} = 28$ (Given)

$\Rightarrow a (a r^{4}) = 28 = a^{2} r^{4} = 28$ ... (i)

Also, $a_{2} + a_{4} = 29$

$\Rightarrow a r + a r^{3} = 29$

$\Rightarrow a r (1 + r^{2}) = 29$ ... (ii)

From (i) and (ii), we get

$\frac{r^{2}}{{(1 + r^{2})}^{2}} = \frac{28}{{(29)}^{2}}$

$\Rightarrow 841 r^{2} = 28 r^{4} + 56 r^{2} + 28$

$\Rightarrow 28 r^{4} - 785 r^{2} + 28 = 0$

$\Rightarrow (r^{2} - 28) (28 r^{2} - 1) = 0$

$\Rightarrow r^{2} = 28 or \frac{1}{28}$

$\Rightarrow r = \sqrt{28}$ $[∵ r \neq \frac{1}{28} as G . P . is increasing]$

$\Rightarrow a = \frac{1}{\sqrt{28}} and a_{6} = a r^{5} = \frac{1}{\sqrt{28}} \times {(\sqrt{28})}^{5} = 784$ .

Q 13 :

Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

$3 a_{99} - 100$
$3 a_{100} - 100$
$3 a_{100} + 100$
$3 a_{99} + 100$

1

2

3

4

(2)

We have, $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$

Let $2 x^{2} = 5 x - 3$

$\Rightarrow 2 x^{2} - 5 x + 3 = 0 \Rightarrow (2 x - 3) (x - 1) = 0$

$\Rightarrow x = 1, \frac{3}{2} ∴ a_{n} = A {(1)}^{n} + B {(\frac{3}{2})}^{n}$

Put n = 0; 0 = A + B

$n = 1; \frac{1}{2} = A + \frac{3}{2} B \Rightarrow B = 1, A = - 1$

$\Rightarrow a_{n} = (- 1) + {(\frac{3}{2})}^{n}$

$\Rightarrow \sum_{k = 1}^{100} a_{k} = \sum_{k = 1}^{100} (- 1) + {(\frac{3}{2})}^{k}$

$= - 100 + \frac{(\frac{3}{2}) ({(\frac{3}{2})}^{100} - 1)}{(\frac{3}{2} - 1)}$

$\Rightarrow - 100 + [3 ({(\frac{3}{2})}^{100} - 1)]$

$= 3 a_{100} - 100$ .

Q 14 :

Let the coefficients of three consecutive terms, $T_{r}, T_{r + 1}$ and $T_{r + 2}$ in the binomial expansion of ${(a + b)}^{12}$ be in a G.P. and let p be the number of all possible values or r. Let q be the sum of all rational terms in the binomial expansion of ${(\sqrt[4]{3} + \sqrt[3]{4})}^{12}$ . Then p + q is equal to : [2025]

299
287
295
283

1

2

3

4

(4)

Since, $T_{r}, T_{r + 1}$ and $T_{r + 2}$ are in G.P.

So, $\frac{T_{r + 1}}{T_{r}} = \frac{T_{r + 2}}{T_{r + 1}}$

$\Rightarrow \frac{{}^{12}C_{r}}{{}^{12}C_{r - 1}} = \frac{{}^{12}C_{r + 1}}{{}^{12}C_{r}} \Rightarrow \frac{13 - r}{r} = \frac{12 - r}{r + 1}$

$\Rightarrow 12 r - r^{2} = 13 r + 13 - r^{2} - r \Rightarrow 13 = 0$ (not possible)

So, p = 0

Now, ${(3^{1 / 4} + 4^{1 / 3})}^{12} = {}^{12}C_{0} {(3^{1 / 4})}^{12} {(4^{1 / 3})}^{0} + {}^{12}C_{12} {(3^{1 / 4})}^{0} {(4^{1 / 3})}^{12}$

$3^{3} + 4^{4} = 27 + 256 = 283 ∴ q = 283$

Thus, p + q = 0 + 283 = 283.

Q 15 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64 and a_{1} + a_{3} + a_{5} = \frac{813}{7} .$ Then $a_{3} + a_{5} + a_{7}$ is equal to: [2026]

3256
3244
3248
3252

1

2

3

4

(4)

$a r \cdot a r^{2} \cdot a r^{3} = 64$

$a^{3} r^{6} = 64 \Rightarrow a r^{2} = 4$

$a + a r^{2} + a r^{4} = \frac{813}{7}$

$r^{2} = 28$

$a r^{2} + a r^{4} + a r^{6} = ?$

$a r^{2} (1 + r^{2} + r^{4}) = 4 (1 + 28 + 784) = 3252$

Q 16 :

Let 729, 81, 9, 1, ... be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence.

If $2 \sum_{n = 1}^{40} {(P_{n})}^{\frac{1}{n}} = \frac{3^{α} - 1}{3^{β}}$ and $\gcd (α, β) = 1$ , then $α + β$ is equal to: [2026]

74
76
75
73

1

2

3

4

(4)

$P_{n} = 729 \cdot 81 \cdot 9 \dots (n terms)$

$= 3^{6} \cdot 3^{4} \cdot 3^{2} \dots 3^{- 2 n + 8}$

$P_{n} = 3^{6 + 4 + 2 + \dots + (- 2 n + 8)} = 3^{n (7 - n)}$

$P_{n}^{1 / n} = 3^{7 - n}$

$\Rightarrow \sum_{n = 1}^{40} {(P_{n})}^{1 / n} = 3^{6} + 3^{5} + \dots (40 terms)$

$= 3^{6} [\frac{1 - {(\frac{1}{3})}^{40}}{1 - \frac{1}{3}}]$

$= 3^{6} \frac{[3^{40} - 1] \times 3^{1}}{3^{40} \times 2}$

$\sum {(P_{n})}^{1 / n} = \frac{(3^{40} - 1)}{2 \times 3^{33}},$

$α = 40, β = 33$

$α + β = 73$

Q 17 :

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^{2}}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3}$ is equal to: [2026]

$2^{25}$
$3^{26}$
$2^{26}$
$3^{25}$

1

2

3

4

(3)

$S = \frac{6}{3^{26}} + \frac{10}{3^{25}} [\frac{6^{25} - 1}{6 - 1}]$

$S = \frac{6}{3^{26}} + \frac{10}{3^{25}} [\frac{6^{25} - 1}{5}]$

$S = \frac{2}{3^{25}} + 2 [\frac{2^{25}}{3^{25}} - \frac{1}{3^{25}}]$

$S = 2^{26}$

Q 18 :

$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}) + (\frac{1}{3^{3}} + \frac{1}{3^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^{2}}{7^{2}} + \frac{4^{3}}{7^{3}}) + \dots$ upto infinite terms, is equal to [2026]

$\frac{6}{5}$
$\frac{4}{3}$
$\frac{7}{4}$
$\frac{5}{2}$

1

2

3

4

(4)

$Let a = \frac{4}{7}, b = \frac{1}{3}$

${Multiply N}^{r} {and D}^{r} by (a - b) = \frac{4}{7} - \frac{1}{3} = \frac{5}{21}$

$\frac{1}{a - b} [(a^{2} - b^{2}) + (a^{3} - b^{3}) + (a^{4} - b^{4}) + \dots \infty]$

$= \frac{1}{a - b} [\frac{a^{2}}{1 - a} - \frac{b^{2}}{1 - b}]$ $= \frac{21}{5} [\frac{\frac{16}{49}}{1 - \frac{4}{7}} - \frac{\frac{1}{9}}{1 - \frac{1}{3}}]$

$= \frac{21}{5} [\frac{16}{21} - \frac{1}{6}]$ $= \frac{21}{5} [\frac{96 - 21}{21.6}]$

$= \frac{75}{5 \cdot 6}$ $= \frac{15}{6} = \frac{5}{2}$

Q 19 :

Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \dots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to: [2026]

$2 (2 - \sqrt{2})$
$2 (\sqrt{2} - 1)$
$2 - \sqrt{2}$
$\sqrt{2} - 1$

1

2

3

4

(2)

$\frac{a_{2}}{2 a_{1}} = \frac{a_{3}}{2 a_{2}} = \frac{a_{4}}{2 a_{3}} = \dots = \frac{a_{10}}{2 a_{9}} = \frac{1}{\sqrt{2}}$

$∴$ $a_{1}, a_{2}, a_{3}, \dots, a_{10} are in G.P. with common ratio \sqrt{2} .$

$\sum_{i = 1}^{10} a_{i} = \frac{a_{1} ({(\sqrt{2})}^{10} - 1)}{\sqrt{2} - 1} = 62$

$\Rightarrow a_{1} = 2 (\sqrt{2} - 1)$

Q 20 :

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ then $9 (a^{2} + b^{2} + c^{2})$ is equal to______. [2026]

(9)

$a = b - d, c = b + d \Rightarrow b = \frac{1}{3}$

$\Rightarrow 4 b^{4} = a^{2} c^{2}$

$\Rightarrow 4 b^{4} = {[(b - d) (b + d)]}^{2}$

$\Rightarrow \frac{4}{81} = {(\frac{1}{9} - d^{2})}^{2}$

$\Rightarrow (\frac{1}{9} - d^{2}) = \pm \frac{2}{9}$

$d^{2} = \frac{1}{3} \Rightarrow d = \pm \frac{1}{\sqrt{3}} (as a < b < c)$

$∴$ $9 (a^{2} + b^{2} + c^{2})$

$= 9 [{(\frac{1}{3} - \frac{1}{\sqrt{3}})}^{2} + {(\frac{1}{3})}^{2} + {(\frac{1}{3} + \frac{1}{\sqrt{3}})}^{2}]$

$= 9 [\frac{1}{3} + \frac{2}{3}] = 3 + 6 = 9$

Q 21 :

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $ℝ - (a, b), then a^{2} + b^{2}$ is equal to_____. [2026]

(90)

$Let first three terms of G.P. are \frac{A}{r}, A, A r$

$\frac{A}{r} \cdot A \cdot A r = 27$

$A^{3} = 27 \Rightarrow A = 3$

$3 (\frac{1}{r} + 1 + r) = 3 + 3 (r + \frac{1}{r})$

$We know, r + \frac{1}{r} \geq 2 or r + \frac{1}{r} \leq - 2$

$S \in ℝ - (- 3, 9)$

$a^{2} + b^{2} = 9 + 81 = 90$

Q 22 :

If the third term of a G.P is 4 then the product of its first 5 terms is

$4^{3}$
$4^{4}$
$4^{5}$
$4^{2}$

1

2

3

4

(3)

Take the terms as $\frac{4}{r^{2}}, \frac{4}{r}, 4, 4 r, 4 r^{2}$

Q 23 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be an A.P. and $g_{1} = a_{1}, g_{2}, g_{3}, . . .$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ , then $a_{10} + g_{5}$ is equal to: [2026]

81
76
62
55

1

2

3

4

(4)

$a_{1}, a_{2}, a_{3}, . . .$ be an A.P.and $g_{1}, g_{2}, . . .$ be a G.P.

$g_{2} = g_{1} r = a_{1} r$ ( $∵ g_{1} = a_{1}$ )

$\Rightarrow g_{2} = r$ ( $∵ a_{1} = 1$ )

$g_{3} = a_{1} r^{2} = r^{2}$

Now, $a_{1} = a_{2} + g_{2} = 1$

$\Rightarrow (1 + d) + r = 1$ [ $∵ a_{2} = a_{1} + d = 1 + d and g_{2} = r$ ]

$\Rightarrow d + r = 0 \Rightarrow d = - r$

Also, $a_{3} + g_{3} = 4$

$(1 + 2 d) + (r^{2}) = 4$

$\Rightarrow 1 - 2 r + r^{2} = 4 \Rightarrow r^{2} - 2 r - 3 = 0$

$\Rightarrow (r - 3) (r + 1) = 0$

$\Rightarrow r = 3$ (Neglecting r = –1 as it is an increasing G.P.)

$\Rightarrow d = - 3$

Now, $a_{10} + g_{5} = (1 + 9 d) + (r^{4})$

$= 1 + 9 (- 3) + (3^{4}) = 8 - 27 = 55$

Q 24 :

Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ and $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty$ . Then the value of ${(0.2)}^{\log_{\sqrt{5}} (α)} + {(0.04)}^{\log_{5} (β)}$ is equal to: [2026]

4
5
8
25

1

2

3

4

(3)

We have, $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$

$α = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{2} = \frac{1}{2}$

and $β = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}$

Now, ${(0.2)}^{\log_{\sqrt{5}} (α)} {(0.04)}^{\log_{5} (β)}$

$= {(\frac{1}{5})}^{\log_{\sqrt{5}} (2^{- 1})} {+ (\frac{4}{100})}^{\log_{5} (\frac{1}{2})}$

$= 5^{2 \log_{5} (2)} {+ (\frac{1}{5^{2}})}^{\log_{5} (\frac{1}{2})}$

$= {4 + (\frac{1}{2})}^{\log_{5} (5^{- 2})} {= 4 + (\frac{1}{2})}^{- 2}$

$= 4 + 2^{2} = 4 + 4 = 8$

Q 25 :

Let $α, β$ be the roots of the equation $x^{2} - x + p = 0$ and $γ, δ$ be the roots of the equation $x^{2} - 4 x + q = 0$ ; $p, q \in Z$ . If $α, β, γ, δ$ are in G.P., then |p + q| equals [2026]

16
32
34
38

1

2

3

4

(3)

Given, $x^{2} - x + p = 0$

$α + β = 1 and α β = p$

Given, $x^{2} - 4 x + q = 0$

$γ + δ = 4 and γ δ = q$

Given, $α, β, γ, δ$ are in G.P.

Let $a, a r, a r^{2}, a r^{3}$ are in G.P.

Now, $α + β = a + a r = a (1 + r)$

But $α + β = 1 \Rightarrow a (1 + r) = 1$ ... (i)

Also, $γ + δ = a r^{2} + a r^{3} = a r^{2} (1 + r)$

But $γ + δ = 4 a r^{2} (1 + r) = 4$ ... (ii)

From(i) and (ii), $r^{2} = 4 \Rightarrow r = \pm 2$

If r = 2 then $a = \frac{1}{2} \notin I$ $∴ r \neq 2$

If r = –2 then a = –1, $a \in I$

$∴$ Roots are –1, 2, –4, 8

$∴ p = α β = - 2 and q = γ δ = - 32$

$∴$ |p + q| = |–2 –32| = 34

Q 26 :

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the GP and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is: [2026]

$\frac{34}{9}$
$\frac{34}{13}$
$\frac{32}{9}$
$\frac{32}{13}$

1

2

3

4

(1)

Sum of first 10 terms of A.P. $S_{10} = 160$

$\Rightarrow 5 [2 a + 9 d] = 160 \Rightarrow 2 a + 9 d = 32$

First term of G.P. = Common difference of A.P. = d

and ratio of G.P. = First term of A.P.

So, G.P. is d, da, $d a^{2}$ ...

Now, d + da = 8

$\Rightarrow d + d \frac{(32 - 9 d)}{2} = 9$

$\Rightarrow 2 d + 32 d - 9 d^{2} = 16$

$\Rightarrow 9 d^{2} = 34 d + 16 = 0$

Required sum $= d_{1} + d_{2} = \frac{34}{9}$

Q 27 :

Consider the quadratic equation $(n^{2} - 2 n + 2) x^{2} - 3 x + {(n^{2} - 2 n + 2)}^{2} = 0, n \in R$ . Let $α$ be the minimum value of the product of its roots and $β$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $α$ and the common ratio is $\frac{α}{β}$ , is: [2026]

$\frac{61}{37}$
$\frac{121}{81}$
$\frac{364}{243}$
$\frac{1093}{729}$

1

2

3

4

(3)

Given, $(n^{2} - 2 n + 2) x^{2} - 3 x + {(n^{2} - 2 n + 2)}^{2} = 0$

Let $k = n^{2} - 2 n + 2$

$\Rightarrow k = {(n - 1)}^{2} + 1$

Min value of k = 1 (at n = 1)

Thus, $k x^{2} - 3 x + k^{2} = 0$

Sum of roots $= - \frac{(- 3)}{k} = \frac{3}{k}$

$\Rightarrow$ Maximum value of sum of roots $= \frac{3}{1} = 3$

$\Rightarrow β = 3$

Product of roots $= \frac{k^{2}}{k} = k$ .

$\Rightarrow$ Minimum value of product of roots = 1

$\Rightarrow α = 1$

Since, $α = 1$

$∴$ Common ratio $(r) = \frac{α}{β} = \frac{1}{3}$

$∵ S_{n} = \frac{a (1 - r^{n})}{1 - r}$

$∴ S_{6} = (\frac{1 (1 - {(\frac{1}{3})}^{6})}{1 - \frac{1}{3}})$

$\Rightarrow S_{6} = \frac{1 - \frac{1}{729}}{\frac{2}{3}} = \frac{364}{243}$

Topic Question Set

Q 11 :

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 12 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

Q 13 :

Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

Q 15 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64 and a_{1} + a_{3} + a_{5} = \frac{813}{7} .$ Then $a_{3} + a_{5} + a_{7}$ is equal to: [2026]

Q 16 :

Let 729, 81, 9, 1, ... be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence.

If $2 \sum_{n = 1}^{40} {(P_{n})}^{\frac{1}{n}} = \frac{3^{α} - 1}{3^{β}}$ and $\gcd (α, β) = 1$ , then $α + β$ is equal to: [2026]

Q 17 :

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^{2}}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3}$ is equal to: [2026]

Q 18 :

$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}) + (\frac{1}{3^{3}} + \frac{1}{3^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^{2}}{7^{2}} + \frac{4^{3}}{7^{3}}) + \dots$ upto infinite terms, is equal to [2026]

Q 19 :

Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \dots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to: [2026]

Q 20 :

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ then $9 (a^{2} + b^{2} + c^{2})$ is equal to______. [2026]

Q 21 :

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $ℝ - (a, b), then a^{2} + b^{2}$ is equal to_____. [2026]

Q 22 :

If the third term of a G.P is 4 then the product of its first 5 terms is

Q 23 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be an A.P. and $g_{1} = a_{1}, g_{2}, g_{3}, . . .$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ , then $a_{10} + g_{5}$ is equal to: [2026]

Q 24 :

Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ and $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty$ . Then the value of ${(0.2)}^{\log_{\sqrt{5}} (α)} + {(0.04)}^{\log_{5} (β)}$ is equal to: [2026]

Q 25 :

Let $α, β$ be the roots of the equation $x^{2} - x + p = 0$ and $γ, δ$ be the roots of the equation $x^{2} - 4 x + q = 0$ ; $p, q \in Z$ . If $α, β, γ, δ$ are in G.P., then |p + q| equals [2026]

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Topic Question Set

Q 11 : If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 12 : Let a1,a2,a3, ..... be a G.P. of increasing positive terms. If a1a5=28 and a2+a4=29, then a6 is equal to : [2025]

Q 13 : Let <an> be a sequence such that a0=0,a1=12 and 2an+2=5an+1–3an, n=0,1,2,3,.... Then ∑k=1100ak is equal to . [2025]

Q 14 : Let the coefficients of three consecutive terms, Tr, Tr+1 and Tr+2 in the binomial expansion of (a+b)12 be in a G.P. and let p be the number of all possible values or r. Let q be the sum of all rational terms in the binomial expansion of (34+43)12. Then p + q is equal to : [2025]

Q 15 : Let a1,a2,a3,… be a G.P. of increasing positive terms such that a2·a3·a4=64 and a1+a3+a5=8137. Then a3+a5+a7 is equal to: [2026]

Q 16 : Let 729, 81, 9, 1, ... be a sequence and Pn denote the product of the first n terms of this sequence. If 2∑n=140(Pn)1n=3α-13β and gcd(α,β)=1, then α+β is equal to: [2026]

Q 17 : 6326+10·1325+10·2324+10·22323+⋯+10·2243 is equal to: [2026]

Q 18 : (13+47)+(132+13×47+4272)+(133+132×47+13×4272+4373)+⋯ upto infinite terms, is equal to [2026]

Q 19 : Let a1,a22,a322,…,a1029 be a G.P. of common ratio 12. If a1+a2+⋯+a10=62, then a1 is equal to: [2026]

Q 20 : Suppose a,b,c are in A.P. and a2, 2b2, c2 are in G.P. If a<b<c and a+b+c=1 then 9(a2+b2+c2) is equal to______. [2026]

Q 21 : In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is ℝ-(a,b), then a2+b2 is equal to_____. [2026]

Q 22 : If the third term of a G.P is 4 then the product of its first 5 terms is

Q 23 : Let a1, a2, a3, ... be an A.P. and g1=a1, g2, g3, ... be an increasing G.P. If a1=a2+g2=1 and a3+g3=4, then a10+g5 is equal to: [2026]

Q 24 : Let α=14+18+116+...∞ and β=13+19+127+...∞. Then the value of (0.2)log5(α)+(0.04)log5(β) is equal to: [2026]

Q 25 : Let α,β be the roots of the equation x2–x+p=0 and γ,δ be the roots of the equation x2–4x+q=0; p,q∈Z. If α,β,γ,δ are in G.P., then |p + q| equals [2026]

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Q 11 :

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 12 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

Q 13 :

Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

Q 15 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64 and a_{1} + a_{3} + a_{5} = \frac{813}{7} .$ Then $a_{3} + a_{5} + a_{7}$ is equal to: [2026]

Q 16 :

Let 729, 81, 9, 1, ... be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence.

If $2 \sum_{n = 1}^{40} {(P_{n})}^{\frac{1}{n}} = \frac{3^{α} - 1}{3^{β}}$ and $\gcd (α, β) = 1$ , then $α + β$ is equal to: [2026]

Q 17 :

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^{2}}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3}$ is equal to: [2026]

Q 18 :

$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}) + (\frac{1}{3^{3}} + \frac{1}{3^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^{2}}{7^{2}} + \frac{4^{3}}{7^{3}}) + \dots$ upto infinite terms, is equal to [2026]

Q 19 :

Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \dots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to: [2026]

Q 20 :

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ then $9 (a^{2} + b^{2} + c^{2})$ is equal to______. [2026]

Q 21 :

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $ℝ - (a, b), then a^{2} + b^{2}$ is equal to_____. [2026]

Q 22 :

If the third term of a G.P is 4 then the product of its first 5 terms is

Q 23 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be an A.P. and $g_{1} = a_{1}, g_{2}, g_{3}, . . .$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ , then $a_{10} + g_{5}$ is equal to: [2026]

Q 24 :

Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ and $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty$ . Then the value of ${(0.2)}^{\log_{\sqrt{5}} (α)} + {(0.04)}^{\log_{5} (β)}$ is equal to: [2026]

Q 25 :

Let $α, β$ be the roots of the equation $x^{2} - x + p = 0$ and $γ, δ$ be the roots of the equation $x^{2} - 4 x + q = 0$ ; $p, q \in Z$ . If $α, β, γ, δ$ are in G.P., then |p + q| equals [2026]