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

1

2

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

1

2

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

1

2

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

1

2

3

4

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(12)

Let the first term be $A$

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

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

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

$\Rightarrow m \leq 22$ $(∵ m \in ℕ)$

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

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

$Thus, 11 \leq m \leq 22$

$∴$ $m can take 12 values.$

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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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