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

Q 1 :

Let the first term $a$ 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 [2023]

231
241
210
220

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3

4

(1)

$We have, first term = a$

$Common ratio = r$

$According to question, a^{2} + {(a r)}^{2} + {(a r^{2})}^{2} = 33033$

$\Rightarrow a^{2} (1 + r^{2} + r^{4}) = 33033$

$\Rightarrow a = 11, r = 4$

$∴$ $Required sum = a + a r + a r^{2}$

$= 11 + 11 \times 4 + 11 {(4)}^{2} = 11 + 44 + 176 = 231$

Q 2 :

Let $a_{1}, a_{2}, a_{3}, . . . . . .$ be a G.P. of increasing positive numbers. Let the sum of its $6^{t h}$ and $8^{t h}$ terms be 2 and the product of its $3^{r d}$ and $5^{t h}$ terms be $\frac{1}{9} .$ Then $6 (a_{2} + a_{4}) (a_{4} + a_{6})$ is equal to [2023]

$2 \sqrt{2}$
2
3
$3 \sqrt{3}$

1

2

3

4

(3)

$Given, a_{6} + a_{8} = 2$

$\Rightarrow a r^{5} (1 + r^{2}) = 2 ...(i)$

$a_{3} \times a_{5} = \frac{1}{9} (Given)$

$\Rightarrow a r^{2} \times a r^{4} = \frac{1}{9} \Rightarrow {(a r^{3})}^{2} = \frac{1}{9} \Rightarrow a r^{3} = \pm \frac{1}{3} ...(ii)$

$Now, by (i) and (ii), we get$

$a r^{3} r^{2} (1 + r^{2}) = 2 \Rightarrow (\pm \frac{1}{3}) r^{2} (1 + r^{2}) = 2 \Rightarrow \frac{r^{2} (1 + r^{2})}{3} = 2$

$\Rightarrow r^{4} + r^{2} - 6 = 0 \Rightarrow r^{2} = 2 \Rightarrow r = \sqrt{2}$

$a (2 \sqrt{2}) = \frac{1}{3}$ $(By (ii))$

$a = \frac{1}{6 \sqrt{2}}$

$∴$ $6 (a_{2} + a_{4}) (a_{4} + a_{6}) = 6 (a r + a r^{3}) (a r^{3} + a r^{5})$

$= 6 a r (1 + r^{2}) a r^{3} (1 + r^{2}) = 6 a^{2} r^{4} {(1 + r^{2})}^{2}$

$= 6 (\frac{1}{72}) \cdot 4 {(1 + 2)}^{2} = 3$

Q 3 :

For three positive integers $p, q, r, x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = p q + 1$ such that $3, 3 \log_{y} x, 3 \log_{z} y, 7 \log_{x} z$ are in A.P. with common difference 1/2. The $r - p - q$ is equal to [2023]

6
2
12
- 6

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3

4

(2)

$p q^{2} \log x = q r \log y = p^{2} r \log z$

$Now, \frac{3 \log x}{\log y} - 3 = \frac{1}{2} \Rightarrow \frac{\log x}{\log y} = \frac{7}{6} = \frac{q r}{p q^{2}} = \frac{r}{p q}$

$\Rightarrow 7 p q = 6 r \Rightarrow 7 (r - 1) = 6 r \Rightarrow r = 7$

$and \frac{3 \log y}{\log z} - 3 = 1 \Rightarrow \frac{\log y}{\log z} = \frac{4}{3} = \frac{p^{2} r}{q r} = \frac{p^{2}}{q} \Rightarrow 3 p^{2} = 4 q$

$Also, \frac{7 \log z}{\log x} - 3 = \frac{3}{2} \Rightarrow \frac{7 \log z}{\log x} = \frac{9}{2} \Rightarrow \frac{7 p q^{2}}{p^{2} r} = \frac{7 q^{2}}{p r}$

$\Rightarrow \frac{7 q^{2}}{p \times 7} = \frac{9}{2} \Rightarrow \frac{q^{2}}{p} = \frac{9}{2} \Rightarrow 2 q^{2} = 9 p \Rightarrow 4 q^{4} = 81 p^{2}$

$⇒ 4 q^{4} = 27 \times 3 p^{2} = 27 \times 4 q \Rightarrow q^{3} = 27 \Rightarrow q = 3$

$We have, r = p q + 1 \Rightarrow 7 = 3 p + 1 \Rightarrow p = 2$

$∴$ $r - p - q = 7 - 2 - 3 = 2$

Q 4 :

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 [2023]

343
216
$\frac{343}{8}$
$\frac{125}{8}$

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3

4

(2)

$a^{3}, b^{3}, c^{3} are in A.P. and \log_{a} b, \log_{c} a, \log_{b} c are in G.P.$

$\frac{\log_{c} a}{\log_{a} b} = \frac{\log_{b} c}{\log_{c} a}$

$\Rightarrow \frac{\log_{e} a \times \log_{e} a}{\log_{e} c \times \log_{e} b} = \frac{\log_{e} c}{\log_{e} b} \times \frac{\log_{e} c}{\log_{e} a}$

$\Rightarrow {(\log_{e} a)}^{3} = {(\log_{e} c)}^{3} \Rightarrow \log_{e} a = \log_{e} c \Rightarrow a = c ...(i)$

$\Rightarrow b^{3} - a^{3} = c^{3} - b^{3} \Rightarrow b^{3} - a^{3} = a^{3} - b^{3}$

$\Rightarrow b^{3} = a^{3} \Rightarrow a = b \Rightarrow a = b = c$

$First term, a_{1} = \frac{a + 4 b + c}{3}, d = \frac{a - 8 b + c}{10}$

$S_{20} = - 444$

$\Rightarrow - 444 = 10 [2 \times \frac{6 a}{3} + 19 \times (\frac{- 6 a}{10})]$

$\Rightarrow - 444 = 10 [4 a - \frac{57}{5} a] \Rightarrow - 444 = 10 [\frac{20 a - 57 a}{5}]$

$\Rightarrow - 444 = 2 (- 37 a) \Rightarrow - 222 = - 37 a$

$\Rightarrow a = 6$ $∴$ $a b c = a^{3} = 6^{3} = 216$

Q 5 :

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 [2023]

$\frac{9}{2}$
3
7
14

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3

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(3)

$Let the terms be \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$

$According to question,$ $\frac{a}{r^{3}} \cdot \frac{a}{r} \cdot a r \cdot a r^{3} = 1296$ $\Rightarrow a = 6$

Now, $\frac{a}{r^{3}} + \frac{a}{r} + a r + a r^{3} = 126$ $\Rightarrow \frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} = 21$

$\Rightarrow {(r + \frac{1}{r})}^{3} - 2 (r + \frac{1}{r}) = 21$

$Let r + \frac{1}{r} = t \Rightarrow t^{3} - 2 t - 21 = 0$

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

$Sum of roots = r_{1}^{2} + r_{2}^{2} = 7$

$Hence, the positive sum is 7 .$

Q 6 :

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 _________ . [2023]

(150)

$\frac{2}{y} = \frac{1}{x} + \frac{1}{z} and 2 y^{2} = x z$

$\Rightarrow \frac{2}{y} = \frac{x + z}{x z} = \frac{x + z}{2 y^{2}} \Rightarrow x + z = 4 y$

Now, $x y + y z + z x = \frac{3}{\sqrt{2}} x y z \Rightarrow y (x + z) + z x = \frac{3}{\sqrt{2}} x z \cdot y$

$\Rightarrow 4 y^{2} + 2 y^{2} = \frac{3}{\sqrt{2}} y \cdot 2 y^{2} \Rightarrow 6 y^{2} = 3 \sqrt{2} y^{3} \Rightarrow y = \sqrt{2}$

$∴$ $x + y + z = 5 y = 5 \sqrt{2} \Rightarrow 3 {(x + y + z)}^{2} = 3 \times 50 = 150$

Q 7 :

Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$ be in an arithmetic geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$ , then $a_{4}$ is equal to _______ . [2023]

(16)

Let $\frac{a - 2 d}{4}, \frac{a - d}{2}, a, 2 (a + d), 4 (a + 2 d)$ be in A.G.P.

Given that, $a = 2$

$Now, (\frac{1}{2} + 1 + 2 + 4 + 8) + (\frac{- 1}{2} - \frac{1}{2} + 2 + 8) d = \frac{49}{2}$

$\Rightarrow \frac{31}{2} + 9 d = \frac{49}{2} \Rightarrow d = 1$

$Then, a_{4} = 4 (a + 2 d) = 4 (2 + 2) = 16$

Q 8 :

Let $S = 109 + \frac{108}{5} + \frac{107}{5^{2}} + . . . + \frac{2}{5^{107}} + \frac{1}{5^{108}}$ . Then the value of $(16 S - {(25)}^{- 54})$ is equal to _________ . [2023]

(2175)

$S = 109 + \frac{108}{5} + \frac{107}{5^{2}} + \dots + \frac{2}{5^{107}} + \frac{1}{5^{108}}$ ...(i)

$\frac{S}{5} = \frac{109}{5} + \frac{108}{5^{2}} + \dots + \frac{2}{5^{108}} + \frac{1}{5^{109}}$ ...(ii)

Subtracting (ii) from (i), we get

$\frac{4 S}{5} = 109 - \frac{1}{5} - \frac{1}{5^{2}} \dots - \frac{1}{5^{108}} - \frac{1}{5^{109}}$

$= 109 - (\frac{1}{5} \frac{(1 - \frac{1}{5^{109}})}{(1 - \frac{1}{5})}) = 109 - \frac{1}{4} (1 - \frac{1}{5^{109}})$

$= 109 - \frac{1}{4} + \frac{1}{4} \times \frac{1}{5^{109}} \Rightarrow S = \frac{5}{4} (109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{109}})$

$\Rightarrow 16 S = 20 \times 109 - 5 + \frac{1}{5^{108}}$

$∴$ $16 S - {(25)}^{- 54} = 2180 - 5 = 2175$

Q 9 :

For $k \in ℕ$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + . . .$ is 10, then the value of k is __________ . [2023]

(2)

Given, $10 = 1 + \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + \dots up to \infty$

$\Rightarrow 9 = \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + \dots up to \infty$ ...(i)

$\Rightarrow \frac{9}{k} = \frac{4}{k^{2}} + \frac{8}{k^{3}} + \frac{13}{k^{4}} + \frac{19}{k^{5}} + \dots up to \infty$ ...(ii)

Subtract (ii) from (i), we get

$S = 9 (1 - \frac{1}{k}) = \frac{4}{k} + \frac{4}{k^{2}} + \frac{5}{k^{3}} + \frac{6}{k^{4}} + \dots up to \infty$ ...(iii)

$\frac{S}{k} = \frac{4}{k^{2}} + \frac{4}{k^{3}} + \frac{5}{k^{4}} + \frac{6}{k^{5}} + \dots up to \infty$ ...(iv)

Subtract (iv) from (iii), we get

$(1 - \frac{1}{k}) S = \frac{4}{k} + \frac{1}{k^{3}} + \frac{1}{k^{4}} + \frac{1}{k^{5}} + \dots up to \infty$ ...(v)

Putting $S = 9 (1 - \frac{1}{k})$ from (iii) in (v), we get

$9 {(1 - \frac{1}{k})}^{2} = \frac{4}{k} + \frac{\frac{1}{k^{3}}}{1 - \frac{1}{k}}$

$\Rightarrow 9 {(1 - \frac{1}{k})}^{3} = \frac{4}{k} (1 - \frac{1}{k}) + \frac{1}{k^{3}} \Rightarrow 9 \frac{{(k - 1)}^{3}}{k^{3}} = \frac{4}{k} (\frac{k - 1}{k}) + \frac{1}{k^{3}}$

$\Rightarrow 9 {(k - 1)}^{3} = 4 k (k - 1) + 1$

Put $k - 1 = x$ , then

$9 x^{3} = 4 (x + 1) x + 1 \Rightarrow 9 x^{3} = 4 x^{2} + 4 x + 1$

$\Rightarrow 9 x^{3} - 4 x^{2} - 4 x - 1 = 0$

$\Rightarrow (x - 1) (9 x^{2} + 5 x + 1) = 0 \Rightarrow (x - 1) = 0 \Rightarrow x = 1$

$∴$ $k - 1 = 1 \Rightarrow k = 2$

Q 10 :

The $4^{t h}$ term of GP is 500 and its common ratio is $1 / m, m \in N .$ Let $S_{n}$ denote the sum of the first $n$ terms of this GP. If $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ , then the number of possible values of $m$ is _______ . [2023]

(12)

Let the first term be $A$

$T_{4} = 500; \frac{A}{m^{3}} = 500$

$S_{6} - S_{5} > 1 \Rightarrow T_{6} > 1 \Rightarrow \frac{A}{m^{5}} > 1$

$\Rightarrow \frac{500}{m^{2}} > 1 \Rightarrow m^{2} < 500 \Rightarrow m < \sqrt{500}$

$\Rightarrow m \leq 22$ $(∵ m \in ℕ)$

$S_{7} - S_{6} < \frac{1}{2} \Rightarrow T_{7} < \frac{1}{2} \Rightarrow \frac{A}{m^{6}} < \frac{1}{2} \Rightarrow \frac{500}{m^{3}} < \frac{1}{2}$

$\Rightarrow m^{3} > 1000 \Rightarrow m \geq 11$

$Thus, 11 \leq m \leq 22$

$∴$ $m can take 12 values.$

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]

(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]

(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]

(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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3

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(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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2

3

4

(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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2

3

4

(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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2

3

4

(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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2

3

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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

1

2

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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3

4

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

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

Q 21 :

Let $a$ and $b$ be two distinct positive real numbers. Let the 11th term of a GP, whose first term is $a$ and third term is $b,$ is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b .$ Then $p$ is equal to [2024]

21
24
20
25

1

2

3

4

(1)

Given, $T_{1} = a, T_{3} = a r_{1}^{2} = b$

$\Rightarrow r_{1} = {(\frac{b}{a})}^{1 / 2}$ ...(i)

Also, $T_{1} = a, T_{5} = b$

$\Rightarrow a r_{2}^{4} = b \Rightarrow r_{2} = {(\frac{b}{a})}^{1 / 4}$ ...(ii)

And $11^{t h}$ term of first GP = $p^{t h}$ term of second GP

Now, $a r_{1}^{10} = a r_{2}^{p - 1}$

$\Rightarrow a {(\frac{b}{a})}^{5} =$ $a$ ${{(\frac{b}{a})}^{1 / 4}}^{p - 1}$ (Using (i) and (ii))

$\Rightarrow {(\frac{b}{a})}^{5} = {(\frac{b}{a})}^{\frac{p - 1}{4}} \Rightarrow 5 = \frac{p - 1}{4} \Rightarrow p - 1 = 20 \Rightarrow p = 21$

Q 22 :

For $0 < c < b < a,$ let $(a + b - 2 c) x^{2} + (b + c - 2 a) x + (c + a - 2 b) = 0$ and $α \neq 1$ be one of its root. Then, among the two statements [2024]

(I) If $α \in (- 1, 0),$ then $b$ cannot be the geometric mean of $a$ and $c$

(II) If $α \in (0, 1),$ then $b$ may be the geometric mean of $a$ and $c$

only (II) is true
Both (I) and (II) are true
only (I) is true
Neither (I) nor (II) is true

1

2

3

4

(2)

We have, $(a + b - 2 c) x^{2} + (b + c - 2 a) x + (c + a - 2 b) = 0$

Put $x = 1$

$a + b - 2 c + b + c - 2 a + c + a - 2 b = 0$

$0 = 0$

$∴$ $x = 1$ is another root

$∴$ $α \cdot 1 = \frac{c + a - 2 b}{a + b - 2 c} \Rightarrow α = \frac{c + a - 2 b}{a + b - 2 c}$

If $- 1 < α < 0,$ then $- 1 < \frac{c + a - 2 b}{a + b - 2 c} < 0$

$\Rightarrow 0 < \frac{c + a - 2 b + a + b - 2 c}{a + b - 2 c} < 1 \Rightarrow 0 < \frac{2 a - b - c}{a + b - 2 c} < 1$

Since, $a > b > c > 0 \Rightarrow a + b > c + c$

$\Rightarrow a + b > 2 c \Rightarrow a + b - 2 c > 0$

$∴$ $2 a - b - c > 0 \Rightarrow 2 a > b + c \Rightarrow a > \frac{b + c}{2}$

$∴$ $b$ can not be the geometric mean of $a$ and $c$

If $0 < α < 1,$ then $0 < \frac{c + a - 2 b}{a + b - 2 c} < 1$

$\Rightarrow 0 < c + a - 2 b and c + a - 2 b < a + b - 2 c$

$\Rightarrow c < b$

$2 b < c + a, b < \frac{c + a}{2}$

$∴$ $b$ may be the geometric mean of $a$ and $c .$

Q 23 :

Let $2^{nd}$ , $8^{th}$ and $44^{th}$ terms of a non-constant A.P. be respectively the $1^{st}$ , $2^{nd}$ and $3^{rd}$ 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 [2024]

990
980
970
960

1

2

3

4

(3)

We have, first term of an A.P. =1.

Let the $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of a G.P. are $\frac{a}{r}, a, a r$ respectively.

Now, $T_{2} = \frac{a}{r} \Rightarrow 1 + (2 - 1) d = \frac{a}{r} \Rightarrow 1 + d = \frac{a}{r}$ ...(i)

$T_{8} = a \Rightarrow 1 + 7 d = a$ ...(ii)

$T_{44} = a r \Rightarrow 1 + 43 d = a r$ ...(iii)

Using equations (i) and (ii), we get

$\frac{1 + d}{1 + 7 d} = \frac{1}{r}$ ...(iv)

From (ii) and (iii), we get

$\frac{1 + 7 d}{1 + 43 d} = \frac{1}{r}$ ...(v)

$∴$ $\frac{1 + d}{1 + 7 d} = \frac{1 + 7 d}{1 + 43 d}$

$\Rightarrow 1 + 43 d + d + 43 d^{2} = 1 + 14 d + 49 d^{2}$

$\Rightarrow 6 d^{2} - 30 d = 0 \Rightarrow 6 d (d - 5) = 0 \Rightarrow d = 0, d = 5$

$∴ d = 5 (∵ d \neq 0)$

$∴ S_{20} = \frac{20}{2} [2 + 19 \times 5] = 10 [2 + 95] = 970$

Q 24 :

If the range of $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}, θ \in R$ is $[α, β],$ then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{α}{β},$ is equal to _______. [2024]

(96)

We have, $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}$

$= 1 + \frac{2 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ} = 1 + \frac{2 \cos^{2} θ}{1 + \cos^{4} θ - \cos^{2} θ}$

$= 1 + \frac{2}{\frac{1}{\cos^{2} θ} + \cos^{2} θ - 1}$

Now, $\cos^{2} θ + \frac{1}{\cos^{2} θ} \geq 2$ $[∵ A . M . \geq G . M .]$

$\Rightarrow \frac{1}{\cos^{2} θ} + \cos^{2} θ - 1 \geq 1 \Rightarrow \cos^{2} θ + \frac{1}{\cos^{2} θ} - 1 \in [1, \infty)$

$\Rightarrow \frac{1}{\cos^{2} θ + \frac{1}{\cos^{2} θ} - 1} \in (0, 1]$

When $\cos θ = 0, f (θ) = 1$

$∴$ $f (θ) \in [1, 3] \Rightarrow α = 1, β = 3$

Sum of infinite G.P. with first term 64 and common ratio

$\frac{1}{3} = \frac{64}{1 - \frac{1}{3}} = 32 \times 3 = 96$

Q 25 :

If three successive terms of a G.P. with common ratio $r (r > 1)$ 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 ______. [2024]

(1)

Let $a, a r$ and $a r^{2}$ be the three sides of the triangle.

Now, $a + a r > a r^{2}$

$\Rightarrow r^{2} - r - 1 < 0 \Rightarrow r \in (\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2})$

So, $3 [r] + [- r] = 3 + (- 2) = 1$

Q 26 :

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4^{2}} (3 + 2 p) + \frac{1}{4^{3}} (3 + 3 p) + \dots \infty,$ then the value of $p$ is _______ . [2024]

(9)

$8 = 3 + \frac{1}{(4)} (3 + p) + \frac{1}{{(4)}^{2}} (3 + 2 p) + \frac{1}{{(4)}^{3}} (3 + 3 p) + \dots \infty$ ...(i)

Multiplying both sides by $\frac{1}{4},$ we get $2 = \frac{3}{4} + \frac{3 + p}{{(4)}^{2}} + \frac{3 + 2 p}{{(4)}^{3}} + \dots + \infty$ ...(ii)

Subtracting (ii) from (i), we get $6 = 3 + \frac{p}{4} + \frac{p}{4^{2}} + \dots$

$\Rightarrow 3 = p [\frac{1}{4} + \frac{1}{{(4)}^{2}} + \frac{1}{{(4)}^{3}} + \dots + \infty]$

$\Rightarrow 3 = p [\frac{\frac{1}{4}}{1 - \frac{1}{4}}]$ $[∵ S_{\infty} = \frac{a}{1 - r}$ for infinite geometric series $]$

$\Rightarrow 3 = p [\frac{1}{4} \times \frac{4}{3}] \Rightarrow p = 9$

Q 27 :

Let the coefficient of $x^{r}$ in the expansion of ${(x + 3)}^{n - 1} + {(x + 3)}^{n - 2} (x + 2) + {(x + 3)}^{n - 3} {(x + 2)}^{2} + \dots + {(x + 2)}^{n - 1}$ be $α_{r} .$ If $\sum_{r = 0}^{n} α_{r} = β^{n} - γ^{n}, β, γ \in N,$ then the value of $β^{2} + γ^{2}$ equals ______ . [2024]

(25)

We have,

${(x + 3)}^{n - 1} + {(x + 3)}^{n - 2} (x + 2) + {(x + 3)}^{n - 3} {(x + 2)}^{2} + \dots + {(x + 2)}^{n - 1}$

$∴$ $\sum_{r = 0}^{n} α_{r} = 4^{n - 1} + 4^{n - 2} \times 3 + 4^{n - 3} \times 3^{2} + \dots + 3^{n - 1}$

$= 4^{n - 1} [1 + \frac{3}{4} + {(\frac{3}{4})}^{2} + \dots + {(\frac{3}{4})}^{n - 1}]$

$= 4^{n - 1} \times \frac{1 - {(\frac{3}{4})}^{n}}{1 - \frac{3}{4}} = 4^{n - 1} (1 - {(\frac{3}{4})}^{n}) (4)$

$= 4^{n} - 3^{n} = β - γ \Rightarrow b = 4, γ = 3$

$∴$ $β^{2} + γ^{2} = 16 + 9 = 25$

Q 28 :

Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then: [2024]

$P^{2} = 6 \sqrt{3} Q$
$P^{2} = 72 \sqrt{3} Q$
$P^{2} = 36 \sqrt{3} Q$
$P = 36 \sqrt{3} Q^{2}$

1

2

3

4

(3)

$∆ A B C is an equilateral triangle of side' a' unit.$

$Perimeter of ∆ A B C = 3 a$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} a^{2}$

Now, $∆ D E F$ is made by joining midpoints of the sides of $∆ A B C$

$∴$ $D E = E F = D F = \frac{a}{2}$

$Perimeter of ∆ D E F = \frac{3 a}{2}$

$Area of ∆ D E F = \frac{\sqrt{3}}{4} \times \frac{a^{2}}{4} = \frac{\sqrt{3}}{16} a^{2}$

$∴$ $P = 3 a + \frac{3 a}{2} + \frac{3 a}{4} + \dots$

$= 3 a [1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots] = \frac{3 a}{1 - \frac{1}{2}} = 6 a$

Now, $Q = \frac{\sqrt{3}}{4} a^{2} [1 + \frac{1}{4} + \frac{1}{4^{2}} + \dots] = \frac{\frac{\sqrt{3}}{4} a^{2}}{1 - \frac{1}{4}} = \frac{4}{3} \times \frac{\sqrt{3}}{4} a^{2}$

$= \frac{1}{\sqrt{3}} a^{2} = \frac{1}{\sqrt{3}} {(\frac{P}{6})}^{2}$ $[∵ P = 6 a]$

$\Rightarrow 36 \sqrt{3} Q = P^{2}$

Q 29 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be a G.P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to [2025]

131
129
128
130

1

2

3

4

(2)

Let $a$ be the first term and $r$ be the common ratio of GP respectively.

Given, $a_{3} a_{5} = 729 \Rightarrow a r^{2} \cdot a r^{4} = 729$

$\Rightarrow a^{2} r^{6} = 729 \Rightarrow a r^{3} = 27$ ... (i)

and $a_{2} + a_{4} = a r + a r^{3} = \frac{111}{4}$

$\Rightarrow a r + 27 = \frac{111}{4} \Rightarrow a r = \frac{3}{4}$ ... (ii)

Now, divide equation (i) by equation (ii), we get

$\frac{a r^{3}}{a r} = \frac{27}{3 / 4}$

$\Rightarrow r^{2} = 36 \Rightarrow r = 6$ [ $∵$ G.P. is an increasing series]

Substitute $r$ = 6 in equation (ii), we get

$a = \frac{1}{8}$

Now, $24 (a_{1} + a_{2} + a_{3})$

$= 24 (a + a r + a r^{2})$

$= 24 a (1 + r + r^{2})$

$= 24 \times \frac{1}{8} (1 + 6 + 36)$

= 3(43) = 129.

Q 30 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ , then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is : [2025]

36
216
72
18

1

2

3

4

(2)

Given, $x_{1}, x_{2}, x_{3}, x_{4}$ be in a G.P.

$x_{1} = a, x_{2} = a r, x_{3} = a r^{2}, x_{4} = a r^{3}$

According to question

$a - 2, a r - 7, a r^{2} - 9, a r^{3} - 5$ are in A.P.

So, $2 (a r - 7) = a - 2 + a r^{2} - 9$ ... (i)

$2 (a r^{2} - 9) = a r - 7 + a r^{3} - 5$ ... (ii)

Solving (i) and (ii), we get r = 2, a = – 3

Product $= x_{1} x_{2} x_{3} x_{4} = a^{4} r^{6} = 81 \times 64 = 5184$

$∴$ The value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4}) = \frac{(5184)}{24} = 216$ .

Topic Question Set

Q 1 :

Let the first term $a$ 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 [2023]

Q 2 :

Let $a_{1}, a_{2}, a_{3}, . . . . . .$ be a G.P. of increasing positive numbers. Let the sum of its $6^{t h}$ and $8^{t h}$ terms be 2 and the product of its $3^{r d}$ and $5^{t h}$ terms be $\frac{1}{9} .$ Then $6 (a_{2} + a_{4}) (a_{4} + a_{6})$ is equal to [2023]

Q 3 :

For three positive integers $p, q, r, x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = p q + 1$ such that $3, 3 \log_{y} x, 3 \log_{z} y, 7 \log_{x} z$ are in A.P. with common difference 1/2. The $r - p - q$ is equal to [2023]

Q 5 :

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 [2023]

Q 6 :

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 _________ . [2023]

Q 7 :

Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$ be in an arithmetic geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$ , then $a_{4}$ is equal to _______ . [2023]

Q 8 :

Let $S = 109 + \frac{108}{5} + \frac{107}{5^{2}} + . . . + \frac{2}{5^{107}} + \frac{1}{5^{108}}$ . Then the value of $(16 S - {(25)}^{- 54})$ is equal to _________ . [2023]

Q 9 :

For $k \in ℕ$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + . . .$ is 10, then the value of k is __________ . [2023]

Q 10 :

The $4^{t h}$ term of GP is 500 and its common ratio is $1 / m, m \in N .$ Let $S_{n}$ denote the sum of the first $n$ terms of this GP. If $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ , then the number of possible values of $m$ is _______ . [2023]

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]

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]

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]

Q 21 :

Let $a$ and $b$ be two distinct positive real numbers. Let the 11th term of a GP, whose first term is $a$ and third term is $b,$ is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b .$ Then $p$ is equal to [2024]

Q 23 :

Let $2^{nd}$ , $8^{th}$ and $44^{th}$ terms of a non-constant A.P. be respectively the $1^{st}$ , $2^{nd}$ and $3^{rd}$ 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 [2024]

Q 24 :

If the range of $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}, θ \in R$ is $[α, β],$ then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{α}{β},$ is equal to _______. [2024]

Q 25 :

If three successive terms of a G.P. with common ratio $r (r > 1)$ 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 ______. [2024]

Q 26 :

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4^{2}} (3 + 2 p) + \frac{1}{4^{3}} (3 + 3 p) + \dots \infty,$ then the value of $p$ is _______ . [2024]

Q 28 :

Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then: [2024]

Q 29 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be a G.P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to [2025]

Q 30 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ , then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is : [2025]

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Topic Question Set

Q 1 : Let the first term a 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 [2023]

Q 2 : Let a1,a2,a3,...... be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be 2 and the product of its 3rd and 5th terms be 19. Then 6(a2+a4)(a4+a6) is equal to [2023]

Q 3 : For three positive integers p,q,r,xpq2=yqr=zp2r and r=pq+1 such that 3,3logyx,3logzy, 7logxz are in A.P. with common difference 1/2. The r-p-q is equal to [2023]

Q 4 : Let a,b,c>1,a3,b3 and c3 be in A.P., and logab,logca and logbc be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c3 and the common difference is a-8b+c10 is -444, then abc is equal to [2023]

Q 5 : 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 [2023]

Q 6 : Let 0<z<y<x be three real numbers such that 1x,1y,1z are in an arithmetic progression and x,2y,z are in a geometric progression. If xy+yz+zx=32xyz, then 3(x+y+z)2 is equal to _________ . [2023]

Q 7 : Suppose a1,a2,2,a3,a4 be in an arithmetic geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is 492, then a4 is equal to _______ . [2023]

Q 8 : Let S=109+1085+10752+...+25107+15108. Then the value of (16S-(25)-54) is equal to _________ . [2023]

Q 9 : For k∈ℕ, if the sum of the series 1+4k+8k2+13k3+19k4+... is 10, then the value of k is __________ . [2023]

Q 10 : The 4th term of GP is 500 and its common ratio is 1/m,m∈N. Let Sn denote the sum of the first n terms of this GP. If S6>S5+1 and S7<S6+12, then the number of possible values of m is _______ . [2023]

Q 11 : For the two positive numbers a,b, if a,b and 118 are in a geometric progression, while 1a, 10 and 1b are in an arithmetic progression, then 16a+12b is equal to ________ . [2023]

Q 12 : Let a1,a2,a3,..... 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 a1a9+a2a4a9+a5+a7 is equal to________ . [2023]

Q 13 : Let {ak} and {bk} k∈ℕ, be two G.P.s with common ratios r1 and r2 respectively such that a1=b1=4 and r1<r2. Let ck=ak+bk, k∈ℕ. If c2=5 and c3=134 then ∑k=1∞ck-(12a6+8b4) is equal to ________ . [2023]

Q 14 : Let the first three terms 2,p and q, with q≠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 nth 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 703 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,ar,ar2,… be an infinite G.P. If ∑n=0∞arn=57 and ∑n=0∞a3r3n=9747, then a+18r 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 px2+qx-r=0, where p≠0. If p,q and r be the consecutive terms of a non constant G.P. and 1α+1β=34, 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 a1,a2,a3,… with a1=18 and a2≠a1, is the arithmetic mean of the next two terms and Sn=a1+a2+…+an, then S20-S18 is equal to [2024]

Q 21 : Let a and b be two distinct positive real numbers. Let the 11th term of a GP, whose first term is a and third term is b, is equal to pth term of another GP, whose first term is a and fifth term is b. Then p is equal to [2024]

Q 22 : For 0<c<b<a, let (a+b-2c)x2+(b+c-2a)x+(c+a-2b)=0 and α≠1 be one of its root. Then, among the two statements [2024] (I) If α∈(-1,0), then b cannot be the geometric mean of a and c (II) If α∈(0,1), then b may be the geometric mean of a and c

Q 23 : Let 2nd, 8th and 44th terms of a non-constant A.P. be respectively the 1st, 2nd and 3rd 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 [2024]

Q 24 : If the range of f(θ)=sin4θ+3cos2θsin4θ+cos2θ, θ∈R is [α,β], then the sum of the infinite G.P., whose first term is 64 and the common ratio is αβ, is equal to _______. [2024]

Q 25 : If three successive terms of a G.P. with common ratio r(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 ______. [2024]

Q 26 : If 8=3+14(3+p)+142(3+2p)+143(3+3p)+⋯∞, then the value of p is _______ . [2024]

Q 27 : Let the coefficient of xr in the expansion of (x+3)n-1+(x+3)n-2(x+2)+(x+3)n-3(x+2)2+…+(x+2)n-1 be αr. If ∑r=0nαr=βn-γn, β,γ∈N, then the value of β2+γ2 equals ______ . [2024]

Q 28 : Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then: [2024]

Q 29 : Let a1,a2,a3,... be a G.P. of increasing positive numbers. If a3a5=729 and a2+a4=1114, then 24(a1+a2+a3) is equal to [2025]

Q 30 : Let x1,x2,x3,x4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1,x2,x3,x4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4) is : [2025]

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Q 1 :

Let the first term $a$ 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 [2023]

Q 2 :

Let $a_{1}, a_{2}, a_{3}, . . . . . .$ be a G.P. of increasing positive numbers. Let the sum of its $6^{t h}$ and $8^{t h}$ terms be 2 and the product of its $3^{r d}$ and $5^{t h}$ terms be $\frac{1}{9} .$ Then $6 (a_{2} + a_{4}) (a_{4} + a_{6})$ is equal to [2023]

Q 3 :

For three positive integers $p, q, r, x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = p q + 1$ such that $3, 3 \log_{y} x, 3 \log_{z} y, 7 \log_{x} z$ are in A.P. with common difference 1/2. The $r - p - q$ is equal to [2023]

Q 5 :

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 [2023]

Q 6 :

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 _________ . [2023]

Q 7 :

Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$ be in an arithmetic geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$ , then $a_{4}$ is equal to _______ . [2023]

Q 8 :

Let $S = 109 + \frac{108}{5} + \frac{107}{5^{2}} + . . . + \frac{2}{5^{107}} + \frac{1}{5^{108}}$ . Then the value of $(16 S - {(25)}^{- 54})$ is equal to _________ . [2023]

Q 9 :

For $k \in ℕ$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + . . .$ is 10, then the value of k is __________ . [2023]

Q 10 :

The $4^{t h}$ term of GP is 500 and its common ratio is $1 / m, m \in N .$ Let $S_{n}$ denote the sum of the first $n$ terms of this GP. If $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ , then the number of possible values of $m$ is _______ . [2023]

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]

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]

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]

Q 21 :

Let $a$ and $b$ be two distinct positive real numbers. Let the 11th term of a GP, whose first term is $a$ and third term is $b,$ is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b .$ Then $p$ is equal to [2024]

Q 23 :

Let $2^{nd}$ , $8^{th}$ and $44^{th}$ terms of a non-constant A.P. be respectively the $1^{st}$ , $2^{nd}$ and $3^{rd}$ 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 [2024]

Q 24 :

If the range of $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}, θ \in R$ is $[α, β],$ then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{α}{β},$ is equal to _______. [2024]

Q 25 :

If three successive terms of a G.P. with common ratio $r (r > 1)$ 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 ______. [2024]

Q 26 :

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4^{2}} (3 + 2 p) + \frac{1}{4^{3}} (3 + 3 p) + \dots \infty,$ then the value of $p$ is _______ . [2024]

Q 29 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be a G.P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to [2025]

Q 30 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ , then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is : [2025]