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

Q 1 :

Let $S_{k} = \frac{1 + 2 + . . . + k}{k}$ and $\sum_{j = 1}^{n} S_{j}^{2} = \frac{n}{A} (B n^{2} + C n + D)$ , where $A, B, C, D \in N$ and A has least value. Then [2023]

A + B is divisible by D
A + B + C + D is divisible by 5
A + C + D is not divisible by B
A + B = 5(D - C)

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(1)

We have, $S_{k} = \frac{1 + 2 + \dots + k}{k}$

$S_{k} = \frac{k (k + 1)}{2 k} = \frac{k + 1}{2}$

Now, $\sum_{j = 1}^{n} {(S_{j})}^{2} = \sum_{j = 1}^{n} {(\frac{j + 1}{2})}^{2} = \frac{1}{4} \sum_{j = 1}^{n} {(j + 1)}^{2}$

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

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

$= \frac{1}{4} [\frac{(n + 1) (n + 2) (2 n + 3)}{6} - 1]$

$= \frac{1}{4} (\frac{2 n^{3} + 9 n^{2} + 13 n}{6}) = \frac{n}{24} (2 n^{2} + 9 n + 13)$

$= \frac{n}{A} (B n^{2} + C n + D)$

$∴$ $A = 24, B = 2, C = 9, D = 13$

$So, A + B = 24 + 2 = 26$

$∴$ $A + B is divisible by D .$

Q 2 :

Let $a_{n}$ be the $n^{t h}$ term of the series 5 + 8 + 14 + 23 + 35 + 50 + ... and $S_{n} = \sum_{k = 1}^{n} a_{k} .$ Then $S_{30} - a_{40}$ is equal to [2023]

11280
11290
11310
11260

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(2)

Let $S = 5 + 8 + 14 + 23 + 35 + 50 + \dots + a_{n}$

$S = 5 + 8 + 14 + 23 + \dots + a_{n - 1} + a_{n}$

Subtracting the above two equations, we get

$0 = 5 + 3 + 6 + 9 + \dots up to (n - 1) terms - a_{n}$

$\Rightarrow a_{n} = 5 + (\frac{n - 1}{2}) [6 + (n - 2) \times 3]$

$= 5 + \frac{3 n (n - 1)}{2} = \frac{1}{2} (3 n^{2} - 3 n + 10)$

Now, $S_{n} = \sum_{k = 1}^{n} a_{k} = \frac{1}{2} \sum_{k = 1}^{n} (3 n^{2} - 3 n + 10)$

$= \frac{1}{2} [\frac{3 n (n + 1) (2 n + 1)}{6} - \frac{3 n (n + 1)}{2} + 10 n]$

$= \frac{1}{2} [\frac{n (n + 1) (2 n + 1)}{2} - \frac{3 n (n + 1)}{2} + \frac{20 n}{2}]$

$= \frac{1}{2} [\frac{2 n^{3} + 3 n^{2} + n - 3 n^{2} - 3 n + 20 n}{2}]$

$= \frac{1}{2} [\frac{2 n^{3} + 18 n}{2}] = \frac{1}{2} n (n^{2} + 9)$

Now, $S_{30} - a_{40} = \frac{1}{2} \times 30 (30^{2} + 9) - \frac{1}{2} (3 \times 40^{2} - 3 \times 40 + 10)$

$= 13635 - 2345 = 11290$

Q 3 :

Let $< a_{n} >$ be a sequence such that $a_{1} + a_{2} + . . . + a_{n} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a_{k}} = p_{1} p_{2} p_{3} . . . p_{m},$ where $p_{1}, p_{2}, . . . ., p_{m}$ are the first $m$ prime numbers, then $m$ is equal to [2023]

5
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8

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(2)

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

$S_{n} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)}$

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

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

$a_{n} = \frac{4}{n (n + 1) (n + 2)}$

$28 \sum_{k = 1}^{10} \frac{1}{a_{k}} = 28 \sum_{k = 1}^{10} \frac{k (k + 1) (k + 2)}{4} = 7 \sum_{k = 1}^{10} k (k + 1) (k + 2)$

$= 7 [{(\frac{10 \times 11}{2})}^{2} + \frac{3 \times 10 \times 11 \times 21}{6} + \frac{2 \times 10 \times 11}{2}]$

$= 7 [{(5 \times 11)}^{2} + 5 \times 11 \times 21 + 10 \times 11]$

$= 7 \times 5 \times 11 [55 + 21 + 2] = 7 \times 5 \times 11 \times 78$

$= 2 \times 3 \times 5 \times 7 \times 11 \times 13 = p_{1} \cdot p_{2} \cdot p_{3} \dots p_{m} \Rightarrow m = 6$

Q 4 :

The sum to 10 terms of the series $\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + . . .$ is [2023]

$\frac{56}{111}$
$\frac{58}{111}$
$\frac{55}{111}$
$\frac{59}{111}$

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(3)

$\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + \dots + up to 10 terms$

$\sum_{r = 1}^{10} \frac{r}{1 + r^{2} + r^{4}} = \frac{1}{2} \sum_{r = 1}^{10} \frac{2 r}{1 + r^{2} + r^{4}}$

$= \frac{1}{2} \sum_{r = 1}^{10} \frac{(r^{2} + r + 1) - (r^{2} - r + 1)}{(r^{2} + r + 1) (r^{2} - r + 1)}$

$= \frac{1}{2} [\sum_{r = 1}^{10} (\frac{1}{r^{2} - r + 1} - \frac{1}{r^{2} + r + 1})]$ $= \frac{1}{2} [1 - \frac{1}{10^{2} + 10 + 1}]$

$= \frac{1}{2} [1 - \frac{1}{100 + 10 + 1}] = \frac{1}{2} [1 - \frac{1}{111}] = \frac{1}{2} \times \frac{110}{111} = \frac{55}{111}$

Q 5 :

The sum $\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$ is equal to [2023]

$\frac{11 e}{2} + \frac{7}{2 e} - 4$
$\frac{11 e}{2} + \frac{7}{2 e}$
$\frac{13 e}{4} + \frac{5}{4 e} - 4$
$\frac{13 e}{4} + \frac{5}{4 e}$

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(3)

$\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$

$= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{2 n (2 n - 1) + 8 n + 8}{(2 n)!}$

$= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{(2 n - 2)!} + 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)!} + 4 \sum_{n = 1}^{\infty} \frac{1}{(2 n)!}$

$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$

$e^{- 1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

$(e + \frac{1}{e}) = 2 (1 + \frac{1}{2!} + \frac{1}{4!} + \dots)$

$e - \frac{1}{e} = 2 (1 + \frac{1}{3!} + \frac{1}{5!} + \dots)$

$Now, \frac{1}{2} (\sum_{n = 1}^{\infty} \frac{1}{(2 n - 2)!}) + 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)!} + 4 \sum_{n = 1}^{\infty} \frac{1}{(2 n)!}$

$= \frac{1}{2} [\frac{e + \frac{1}{e}}{2}] + 2 [\frac{e - \frac{1}{e}}{2}] + 4 [\frac{e + \frac{1}{e} - 2}{2}]$

$= \frac{(e + \frac{1}{e})}{4} + e - \frac{1}{e} + 2 e + \frac{2}{e} - 4$

$= \frac{e}{4} + \frac{1}{4 e} + 3 e + \frac{1}{e} - 4 = \frac{e + 12 e}{4} + \frac{1 + 4}{4 e} - 4$

$= \frac{13 e}{4} + \frac{5}{4 e} - 4$

Q 6 :

If ${(20)}^{19} + 2 (21) {(20)}^{18} + 3 {(21)}^{2} {(20)}^{17} + . . . . + 20 {(21)}^{19} = k {(20)}^{19},$ then $k$ is equal to _________ . [2023]

0%

(400)

We have,

${(20)}^{19} k = {(20)}^{19} + 2 \cdot 21 \cdot {(20)}^{18} + 3 \cdot {(21)}^{2} \cdot {(20)}^{17} + \dots + 20 \cdot {(21)}^{19}$

$\Rightarrow k {(20)}^{19} = 20^{19} [1 + 2 \cdot (\frac{21}{20}) + 3 \cdot {(\frac{21}{20})}^{2} + \dots + 20 \cdot {(\frac{21}{20})}^{19}]$

$\Rightarrow k = 1 + 2 \cdot \frac{21}{20} + 3 \cdot {(\frac{21}{20})}^{2} + \dots + 20 {(\frac{21}{20})}^{19} ...(i)$

$\Rightarrow \frac{21}{20} k = \frac{21}{20} + 2 \cdot {(\frac{21}{20})}^{2} + \dots + 19 \cdot {(\frac{21}{20})}^{19} + 20 \cdot {(\frac{21}{20})}^{20} ...(ii)$

$Subtracting (ii) from (i), we get$

$- \frac{k}{20} = 1 + (\frac{21}{20}) + {(\frac{21}{20})}^{2} + \dots + {(\frac{21}{20})}^{19} - 20 \cdot {(\frac{21}{20})}^{20}$

$= 1 {\frac{{(\frac{21}{20})}^{20} - 1}{\frac{21}{20} - 1}} - 20 \cdot {(\frac{21}{20})}^{20}$ $= 20 \cdot {(\frac{21}{20})}^{20} - 20 - 20 \cdot {(\frac{21}{20})}^{20}$

$\Rightarrow \frac{- k}{20} = - 20 \Rightarrow k = 400$

Q 7 :

The sum to 20 terms of the series $2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - . . . . . . . . . .$ is equal to _______ . [2023]

0%

(1310)

$Let S = 2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} + \dots$

$= (2 \cdot 2^{2} + 2 \cdot 4^{2} + 2 \cdot 6^{2} + \dots 10 terms)$ $- (3^{2} + 5^{2} + 7^{2} + \dots + 10 terms)$

$= 8 (1^{2} + 2^{2} + 3^{2} + \dots + 10 terms)$ $- (1^{2} + 3^{2} + 5^{2} + 7^{2} + \dots + 11 terms) + 1$

$= 8 [\frac{10 \times 11 \times 21}{6}] - \frac{11}{3} [23 \times 21] + 1$

$= 3080 - 1771 + 1 = 1310$

Q 8 :

If the sum of the series $(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^{2}} - \frac{1}{2 \cdot 3} + \frac{1}{3^{2}}) + (\frac{1}{2^{3}} - \frac{1}{2^{2} \cdot 3} + \frac{1}{2 \cdot 3^{2}} - \frac{1}{3^{3}}) + (\frac{1}{2^{4}} - \frac{1}{2^{3} \cdot 3} + \frac{1}{2^{2} \cdot 3^{2}} - \frac{1}{2 \cdot 3^{3}} + \frac{1}{3^{4}}) + . . .$ is $\frac{α}{β},$ where $α$ and $β$ are co-prime, then $α + 3 β$ is equal to ________ . [2023]

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(7)

$Let S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^{2}} - \frac{1}{2 \cdot 3} + \frac{1}{3^{2}}) + (\frac{1}{2^{3}} - \frac{1}{2^{2} \cdot 3} + \frac{1}{2 \cdot 3^{2}} - \frac{1}{3^{3}}) + \dots$

$= (x - y) + (x^{2} - x y + y^{2}) + (x^{3} - x^{2} y + x y^{2} - y^{3}) + \dots, where x = \frac{1}{2}, y = \frac{1}{3}$

$Multiply both sides by x + y$

$(x + y) S = (x^{2} - y^{2}) + (x^{3} + y^{3}) + (x^{4} - y^{4}) + \dots$

$(x + y) S = (x^{2} + x^{3} + x^{4} + \dots) - (y^{2} + y^{4} + y^{6} + \dots) + (y^{3} + y^{5} + \dots)$

$= [\frac{\frac{1}{4}}{1 - \frac{1}{2}}] - [\frac{\frac{1}{9}}{1 - \frac{1}{9}}] + [\frac{\frac{1}{27}}{1 - \frac{1}{9}}]$ [infinite G.P.]

$= \frac{1}{2} - \frac{1}{8} + \frac{1}{24}$

$(\frac{1}{2} + \frac{1}{3}) \cdot S = \frac{5}{12}$

$S = \frac{5}{12} \times \frac{6}{5} = \frac{1}{2}$ $∴$ $α = 1, β = 2$ , so $(α + 3 β) = 1 + 3 \times 2 = 7$

Q 9 :

If $\frac{1^{3} + 2^{3} + 3^{3} + . . . u p t o n t e r m s}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + . . . u p t o n t e r m s} = \frac{9}{5},$ then the value of $n$ is _______. [2023]

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(5)

$Given,$ $\frac{1^{3} + 2^{3} + 3^{3} + \dots up to n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots up to n terms} = \frac{9}{5}$

$\Rightarrow \frac{{(\frac{n (n + 1)}{2})}^{2}}{\sum_{r = 1}^{n} r (2 r + 1)} = \frac{9}{5}$ $\Rightarrow \frac{{(\frac{n (n + 1)}{2})}^{2}}{2 \sum_{r = 1}^{n} r^{2} + \sum_{r = 1}^{n} r} = \frac{9}{5}$

$\Rightarrow \frac{{(\frac{n (n + 1)}{2})}^{2}}{2 (\frac{n (n + 1) (2 n + 1)}{6}) + \frac{n (n + 1)}{2}} = \frac{9}{5}$

$\Rightarrow \frac{{[\frac{n (n + 1)}{2}]}^{2}}{\frac{n (n + 1)}{2} [\frac{2 (2 n + 1)}{3} + 1]} = \frac{9}{5} \Rightarrow \frac{\frac{n (n + 1)}{2}}{\frac{4 n + 5}{3}} = \frac{9}{5}$

$\Rightarrow 5 n^{2} - 19 n - 30 = 0 \Rightarrow (5 n + 6) (n - 5) = 0 \Rightarrow n = \frac{- 6}{5}, 5$

$Hence, n = 5$

Q 10 :

Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2 .$ If $S = \sum_{n = 1}^{10} \frac{b_{n}}{2^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2^{n - 1}},$ then $2^{7} (2 S - T)$ is equal to _________ . [2023]

0%

(461)

$Given,$ $S = \sum_{n = 1}^{10} \frac{b_{n}}{2^{n}} = \frac{b_{1}}{2} + \frac{b_{2}}{2^{2}} + \dots + \frac{b_{9}}{2^{9}} + \frac{b_{10}}{2^{10}}$

$\Rightarrow \frac{S}{2} = \frac{b_{1}}{2^{2}} + \frac{b_{2}}{2^{3}} + \dots + \frac{b_{9}}{2^{10}} + \frac{b_{10}}{2^{11}}$

$Subtracting, we get$

$\frac{S}{2} = \frac{b_{1}}{2} + (\frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \dots + \frac{a_{9}}{2^{10}}) - \frac{b_{10}}{2^{11}}$

$\Rightarrow S = b_{1} - \frac{b_{10}}{2^{10}} + (\frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + \dots + \frac{a_{9}}{2^{9}})$

$\Rightarrow \frac{S}{2} = \frac{b_{1}}{2} - \frac{b_{10}}{2^{11}} + (\frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \dots + \frac{a_{9}}{2^{10}})$

$Subtracting, we get:$

$\frac{S}{2} = \frac{b_{1}}{2} - \frac{b_{10}}{2^{11}} + (\frac{a_{1}}{2} - \frac{a_{9}}{2^{10}}) + (\frac{1}{2^{2}} + \frac{2}{2^{3}} + \dots + \frac{8}{2^{9}})$

$\Rightarrow \frac{S}{2} = \frac{a_{1} + b_{1}}{2} - \frac{(b_{10} + 2 a_{9})}{2^{11}} + \frac{T}{4}$

$\Rightarrow 2 S = 2 (a_{1} + b_{1}) - \frac{b_{10} + 2 a_{9}}{2^{9}} + T$

$\Rightarrow 2^{7} (2 S - T) = 2^{8} (a_{1} + b_{1}) - \frac{(b_{10} + 2 a_{9})}{4}$

$Given a_{n} - a_{n - 1} = n - 1$

$∴$ $\begin{matrix} a_{2} & - & a_{1} & = & 1 \\ a_{3} & - & a_{2} & = & 2 \\ ⋮ \\ a_{9} & - & a_{8} & = & 8 \end{matrix}$

_________________

$a_{9} - a_{1} = 1 + 2 + \dots 8 = 36$

$\Rightarrow a_{9} = 37$

$Also, b_{n} - b_{n - 1} = a_{n - 1}$ $∴$ $b_{10} - b_{1} = a_{1} + a_{2} + \dots + a_{9}$

$= 1 + 2 + 4 + 7 + 11 + 16 + 22 + 29 + 37$

$\Rightarrow b_{10} = 130$

$∴$ $2^{7} (2 S - T) = 2^{8} (1 + 1) - \frac{(130 + 2 \times 37)}{4}$ = 461

Topic Question Set

Q 1 :

Let $S_{k} = \frac{1 + 2 + . . . + k}{k}$ and $\sum_{j = 1}^{n} S_{j}^{2} = \frac{n}{A} (B n^{2} + C n + D)$ , where $A, B, C, D \in N$ and A has least value. Then [2023]

Q 2 :

Let $a_{n}$ be the $n^{t h}$ term of the series 5 + 8 + 14 + 23 + 35 + 50 + ... and $S_{n} = \sum_{k = 1}^{n} a_{k} .$ Then $S_{30} - a_{40}$ is equal to [2023]

Q 3 :

Let $< a_{n} >$ be a sequence such that $a_{1} + a_{2} + . . . + a_{n} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a_{k}} = p_{1} p_{2} p_{3} . . . p_{m},$ where $p_{1}, p_{2}, . . . ., p_{m}$ are the first $m$ prime numbers, then $m$ is equal to [2023]

Q 4 :

The sum to 10 terms of the series $\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + . . .$ is [2023]

Q 5 :

The sum $\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$ is equal to [2023]

Q 6 :

If ${(20)}^{19} + 2 (21) {(20)}^{18} + 3 {(21)}^{2} {(20)}^{17} + . . . . + 20 {(21)}^{19} = k {(20)}^{19},$ then $k$ is equal to _________ . [2023]

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Q 7 :

The sum to 20 terms of the series $2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - . . . . . . . . . .$ is equal to _______ . [2023]

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Q 9 :

If $\frac{1^{3} + 2^{3} + 3^{3} + . . . u p t o n t e r m s}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + . . . u p t o n t e r m s} = \frac{9}{5},$ then the value of $n$ is _______. [2023]

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Q 10 :

Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2 .$ If $S = \sum_{n = 1}^{10} \frac{b_{n}}{2^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2^{n - 1}},$ then $2^{7} (2 S - T)$ is equal to _________ . [2023]

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Topic Question Set

Q 4 : The sum to 10 terms of the series 11+12+14+21+22+24+31+32+34+... is [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 5 : The sum ∑n=1∞2n2+3n+4(2n)! is equal to [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 4 :

The sum to 10 terms of the series $\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + . . .$ is [2023]

Q 5 :

The sum $\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$ is equal to [2023]