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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(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]

(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]

(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]

(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]

(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]

(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

Q 11 :

The sum $1^{2} - 2 \cdot 3^{2} + 3 \cdot 5^{2} - 4 \cdot 7^{2} + 5 \cdot 9^{2} - . . . . . + 15 \cdot 29^{2}$ is _______. [2023]

(6952)

$Separating odd and even placed terms:$

$S = (1 \cdot 1^{2} + 3 \cdot 5^{2} + \dots + 15 \cdot {(29)}^{2}) - (2 \cdot 3^{2} + 4 \cdot 7^{2} + \dots + 14 \cdot {(27)}^{2})$

$S = \sum_{n = 1}^{8} (2 n - 1) {(4 n - 3)}^{2} - \sum_{n = 1}^{7} (2 n) {(4 n - 1)}^{2}$

$S = \sum_{n = 1}^{8} (32 n^{3} - 64 n^{2} + 42 n - 9) - \sum_{n = 1}^{7} (32 n^{3} - 16 n^{2} + 2 n)$

$S = 32 \sum_{n = 1}^{8} n^{3} - 64 \sum_{n = 1}^{8} n^{2} + 42 \sum_{n = 1}^{8} n - 9 \sum_{n = 1}^{8} (1) - 32 \sum_{n = 1}^{7} n^{3} + 16 \sum_{n = 1}^{7} n^{2} - 2 \sum_{n = 1}^{7} n$

$S = 32 {[\frac{n (n + 1)}{2}]}^{2} - 64 [\frac{n (n + 1) (2 n + 1)}{6}] + 42 [\frac{n (n + 1)}{2}] - 9 \times 8$ $- 32 {[\frac{n (n + 1)}{2}]}^{2} + 16 [\frac{n (n + 1) (2 n + 1)}{6}] - 2 [\frac{n (n + 1)}{2}]$

$S = 32 {[36]}^{2} - 64 [12 \times 17] + 42 [36] - 72 - 32 {(28)}^{2} + 16 (140) - 2 (28)$

$S = 6952$

Q 12 :

The value of $\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots + 100 \times {(101)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots + 100^{2} \times 101}$ is [2024]

$\frac{305}{301}$
$\frac{306}{305}$
$\frac{31}{30}$
$\frac{32}{31}$

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(1)

$\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots + 100 \times {(101)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots + 100^{2} \times 101}$

$\Rightarrow \frac{\sum_{n = 1}^{100} n {(n + 1)}^{2}}{\sum_{n = 1}^{100} n^{2} (n + 1)}$

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

$\Rightarrow \frac{{(\frac{100 (101)}{2})}^{2} + \frac{2 \cdot 100 (101) (201)}{6} + \frac{100 (101)}{2}}{{(\frac{100 (101)}{2})}^{2} + \frac{100 (101) (201)}{6}} = \frac{305}{301}$

Q 13 :

If $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$ and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{99 \cdot 100} = n,$ then the point $(m, n)$ lies on the line [2024]

$11 (x - 1) - 100 (y - 2) = 0$
$11 (x - 2) - 100 (y - 1) = 0$
$11 x - 100 y = 0$
$11 (x - 1) - 100 y = 0$

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(3)

$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$

and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{99 \cdot 100} = n$

$\Rightarrow \frac{\sqrt{2} - \sqrt{1}}{1} + \frac{\sqrt{3} - \sqrt{2}}{1} + \dots + \frac{\sqrt{100} - \sqrt{99}}{1} = m$

$\Rightarrow \sqrt{100} - 1 = m \Rightarrow m = 10 - 1 = 9 \dots (i)$

Also, $(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{99} - \frac{1}{100}) = n$

$\Rightarrow 1 - \frac{1}{100} = n \Rightarrow n = \frac{99}{100} \dots (i i)$

$∴$ $(m, n) = (9, \frac{99}{100})$

So, $(m, n)$ lies on $11 x - 100 y = 0$

Q 14 :

If the sum of the series $\frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) (1 + 2 d)} + \dots + \frac{1}{(1 + 9 d) (1 + 10 d)}$ is equal to 5, then $50 d$ is equal to : [2024]

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(1)

We have,

$\frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) (1 + 2 d)} + \dots + \frac{1}{(1 + 9 d) (1 + 10 d)} = 5$

$\Rightarrow \sum_{r = 1}^{10} [\frac{1}{{1 + (r - 1) d} {1 + r d}}] = 5$

$\Rightarrow \sum_{r = 1}^{10} \frac{1}{d} [\frac{1}{1 + (r - 1) d} - \frac{1}{1 + r d}] = 5$

$\Rightarrow \frac{1}{d} \sum_{r = 1}^{10} [\frac{1}{1 + (r - 1) d} - \frac{1}{1 + r d}] = 5$

$\Rightarrow [\frac{1}{1} - \frac{1}{1 + d}] + [\frac{1}{1 + d} - \frac{1}{1 + 2 d}] + \dots + [\frac{1}{1 + 9 d} - \frac{1}{1 + 10 d}] = 5 d$

$\Rightarrow 1 - \frac{1}{1 + 10 d} = 5 d \Rightarrow \frac{10 d}{1 + 10 d} = 5 d$

$\Rightarrow 5 + 50 d = 10 \Rightarrow 50 d = 5$

Q 15 :

The sum of the series $\frac{1}{1 - 3 . 1^{2} + 1^{4}} + \frac{2}{1 - 3 . 2^{2} + 2^{4}} + \frac{3}{1 - 3 . 3^{2} + 3^{4}} + \dots$ up to 10-terms is [2024]

$- \frac{45}{109}$
$\frac{55}{109}$
$- \frac{55}{109}$
$\frac{45}{109}$

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(3)

Given $\frac{1}{1 - 3 . 1^{2} + 1^{4}} + \frac{2}{1 - 3 . 2^{2} + 2^{4}} + \frac{3}{1 - 3 . 3^{2} + 3^{4}} + \dots$

$T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}}$

$\Rightarrow T_{r} = \frac{r}{(r^{4} - 2 r^{2} + 1) - r^{2}} = \frac{r}{{(r^{2} - 1)}^{2} - r^{2}}$

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

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

$\Rightarrow T_{r} = \frac{1}{2} [\frac{1}{r^{2} - r - 1} - \frac{1}{r^{2} + r - 1}]$

Sum of 10 terms

$\sum_{r = 1}^{10} T_{r} = \frac{1}{2} {[\frac{1}{- 1} - \frac{1}{1}]} + \frac{1}{2} [\frac{1}{1} - \frac{1}{5}] + \dots + \frac{1}{2} [\frac{1}{89} - \frac{1}{109}]$

Using telescopic,

$\Rightarrow \frac{1}{2} [- 1 - \frac{1}{109}] = \frac{1}{2} [\frac{- 109 - 1}{109}] = \frac{1}{2} [\frac{- 110}{109}] = \frac{- 55}{109}$

Q 16 :

If $1 + \frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{3}} + \frac{5 - 2 \sqrt{6}}{18} + \frac{9 \sqrt{3} - 11 \sqrt{2}}{36 \sqrt{3}} + \frac{49 - 20 \sqrt{6}}{180} + \dots$ upto $\infty = 2 + (\sqrt{\frac{b}{a}} + 1) \log_{e} (\frac{a}{b}),$ where $a$ and $b$ are integer with $g c d (a, b) = 1,$ then $11 a + 18 b$ is equal to _______. [2024]

(76)

Let $S = 1 + \frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{3}} + \frac{5 - 2 \sqrt{6}}{18} + \frac{9 \sqrt{3} - 11 \sqrt{2}}{36 \sqrt{3}} + \dots \infty$

$= 1 + \frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{3}} + \frac{{(\sqrt{3} - \sqrt{2})}^{2}}{{(\sqrt{3})}^{2} \cdot 6} + \frac{{(\sqrt{3} - \sqrt{2})}^{3}}{{(\sqrt{3})}^{3} \cdot 12} + \dots \infty$

$= 1 + \frac{t}{2} + \frac{t^{2}}{6} + \frac{t^{3}}{12} + \dots \infty$ $[where t = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}}]$

$= 1 + t (1 - \frac{1}{2}) + t^{2} (\frac{1}{2} - \frac{1}{3}) + t^{3} (\frac{1}{3} - \frac{1}{4}) + \dots \infty$

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

$= 2 + (\frac{1}{t} - 1) \log (1 - t)$

$= 2 + (\frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}} - 1) \log (1 - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}})$

$= 2 + (\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}) \log (\frac{\sqrt{2}}{\sqrt{3}}) = 2 + (\frac{\sqrt{6} + 2}{1}) \frac{1}{2} \log \frac{2}{3}$

$= 2 + (\sqrt{\frac{3}{2}} + 1) \log \frac{2}{3}$

$\Rightarrow$ $a = 2$ and $b = 3$

$∴$ $11 a + 18 b = 22 + 54 = 76$

Q 17 :

Let the first term of a series be $T_{1} = 6$ and its $r^{t h}$ term $T_{r} = 3 T_{r - 1} + 6^{r}, r = 2, 3, \dots, n .$ If the sum of the first $n$ terms of this series is $\frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1),$ then $n$ is equal to ______ . [2024]

(6)

We have, $T_{1} = 6$ and $T_{r} = 3 T_{r - 1} + 6^{r}$

Now, $T_{2} = 3 T_{1} + 6^{2} = 3 \times 6 + 6^{2}$

$T_{3} = 3 T_{2} + 6^{3} = 3 (3 \times 6 + 6^{2}) + 6^{3} = 3^{2} \times 6 + 3 \times 6^{2} + 6^{3}$

$T_{4} = 3 T_{3} + 6^{4} = 3 (3^{2} \times 6 + 3 \times 6^{2} + 6^{3}) + 6^{4}$

$= 3^{3} \times 6 + 3^{2} \times 6^{2} + 3 \times 6^{3} + 6^{4}$

$∴$ $T_{r} = 3^{r - 1} \times 6 + 3^{r - 2} \times 6^{2} + \dots + 6^{r}$

$= 3^{r - 1} \times 6 (1 + \frac{6}{3} + {(\frac{6}{3})}^{2} + \dots {(\frac{6}{3})}^{r - 1})$

$= 3^{r - 1} \times 6 \times (1 + 2 + 2^{2} + \dots + 2^{r - 1})$

$= 3^{r - 1} \times 6 (\frac{2^{r} - 1}{1}) = \frac{6 \times 3^{r}}{3} (2^{r} - 1)$

$= 2 \times 3^{r} (2^{r} - 1) = 2 (6^{r} - 3^{r})$

Sum of $n$ terms $S_{n} = \sum_{n} 2 (6^{r} - 3^{r})$

$= 2 (\sum_{n} 6^{r} - \sum_{n} 3^{r}) = 2 (\frac{6 (6^{n} - 1)}{5} - \frac{3 (3^{n} - 1)}{2})$

$= \frac{1}{5} (12 \times 6^{n} - 12 - 15 \times 3^{n} + 15)$

$= \frac{3}{5} (4 \times 6^{n} - 5 \times 3^{n} + 1) \Rightarrow n^{2} - 12 n + 39 = 3$

$\Rightarrow n^{2} - 12 n + 36 = 0 \Rightarrow n = 6$

Q 18 :

Let the positive integers be written in the form:

1

   2 3

  4 5 6

7 8 9 10

If the $k^{t h}$ row contains exactly $k$ numbers for every natural number $k,$ then the row in which the number 5310 will be, is ______.     [2024]

(103)

First element of $k^{t h}$ row $= \frac{k (k - 1)}{2} + 1$

Last element of $k^{t h}$ row = $= \frac{k (k + 1)}{2}$

$\Rightarrow \frac{k (k - 1)}{2} + 1 \leq 5310 \leq \frac{k (k + 1)}{2}$

$\Rightarrow k (k - 1) + 2 \leq 10620 \leq k (k + 1)$

$\Rightarrow k^{2} - k + 2 \leq 10620 \leq k^{2} + k$

$\Rightarrow - k + 2 \leq 10620 - k^{2} \leq k \Rightarrow 2 \leq 10620 - k^{2} + k \leq 2 k$

$\Rightarrow 1 \leq 5310 - \frac{k^{2}}{2} + \frac{k}{2} \leq k \Rightarrow 5310 - \frac{k^{2}}{2} - \frac{k}{2} \leq 0$

$\Rightarrow k^{2} + k - 5310 \times 2 \geq 0 \Rightarrow k^{2} + k - 10620 \geq 0$

For $k = 100, k^{2} + k - 10620 = 10100 - 10620 < 0$

For $k = 103, k^{2} + k - 10620 = 10609 + 103 - 10620 > 0$

So, the value of $k = 103$

Q 19 :

If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024},$ Then $α$ is equal to _______ . [2024]

(1011)

Given $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024} ...(i)$

Now, $\sum_{r = 1}^{1012} \frac{1}{2 r (2 r - 1)} = \sum_{r = 1}^{1012} [\frac{1}{2 r - 1} - \frac{1}{2 r}]$

$= [(1 - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{2023} - \frac{1}{2024})]$

$= (1 + \frac{1}{3} + \dots + \frac{1}{2023}) - (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2024})$

$= (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2023}) - \frac{1}{2} (1 + \frac{1}{2} + \dots + \frac{1}{1012}) - \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{1011})$

$\Rightarrow \frac{1}{1012} + \frac{1}{1013} + \dots + \frac{1}{2023} - \frac{1}{2024}$

$\Rightarrow \frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012} = \frac{1}{2024} - \sum_{r = 1}^{1012} \frac{1}{2 r (2 r - 1)} = \frac{1}{1012} + \dots + \frac{1}{2023}$

$\Rightarrow α + 1012 = 2023 \Rightarrow α = 1011$

Q 20 :

Let $α = 1^{2} + 4^{2} + 8^{2} + 13^{2} + 19^{2} + 26^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4} .$ If $4 α - β = 55 k + 40,$ then $k$ is equal to ______ . [2024]

(353)

$α = 1^{2} + 4^{2} + 8^{2} + 13^{2} \dots$ upto 10 terms, and $β = 1^{4} + 2^{4} + 3^{4} + \dots$ upto 10 terms.

Let us find general terms for $a$

$S_{n} = 1 + 4 + 8 + . . . + a_{n}$

Again $S_{n} =$ $1 + 4 + . . . + a_{n - 1} + a_{n}$

________________________________

$O = 1 + 3 + 4 + . . . (n - 1) t e r m - a_{n}$

$\Rightarrow a_{n} = 1 + 3 + 4 + \dots$

$\Rightarrow a_{n} = 1 + (3 + 4 + 5 + \dots + (n - 1) terms)$

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

$\frac{n^{2} + 4 n - 4 - n + 2}{2} = \frac{n^{2} + 3 n - 2}{2}$

So, $α = \sum_{n = 1}^{10} {(\frac{n^{2} + 3 n - 2}{2})}^{2} and β = \sum_{n = 1}^{10} n^{4}$

$\Rightarrow 4 α = \sum_{n = 1}^{10} (n^{4} + 9 n^{2} + 4 + 6 n^{3} - 12 n - 4 n^{2})$ and $β = \sum_{n = 1}^{10} n^{4}$

Now, $4 α - β = \sum_{n = 1}^{10} (5 n^{2} + 6 n^{3} - 12 n + 4)$

$\Rightarrow 4 α - β = 5 \sum_{n = 1}^{10} n^{2} + 6 \sum_{n = 1}^{10} n^{3} - 12 \sum_{n = 1}^{10} n + 4 \times 10$

$= 5 \times \frac{10 (10 + 1) (20 + 1)}{6} + 6 {(\frac{10 \times 11}{2})}^{2} - 12 (\frac{10 \times 11}{2}) + 40$

$= 353 \times 55 + 40$

So, $k = 353$

Q 21 :

Let $S_{n}$ be the sum to $n$ -terms of an arithmetic progression 3, 7, 11,..... If $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42,$ then $n$ equals _____ . [2024]

(9)

Let $S_{k} = 3 + 7 + 11 + \dots$ upto $k$ terms

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

$= k (2 k + 1) = 2 k^{2} + k$

$∴$ $\sum_{k = 1}^{n} (2 k^{2} + k) = \frac{2 n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}$

Since, $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42$

$\Rightarrow 40 < 2 (2 n + 1) + 3 < 42 \Rightarrow 40 < 4 n + 5 < 42$

$\Rightarrow 35 < 4 n < 37 \Rightarrow 8.75 < n < 9.25$

$∴ n = 9$

Q 22 :

If $S (x) = (1 + x) + 2 {(1 + x)}^{2} + 3 {(1 + x)}^{3} + \dots + 60 {(1 + x)}^{60}$ , $x \neq 0$ and ${(60)}^{2} S (60) = a {(b)}^{b} + b,$ where $a, b \in N,$ then $(a + b)$ is equal to _______. [2024]

(3660)

$= \frac{(1 + x) ({(1 + x)}^{60} - 1)}{1 + x - 1} - 60 {(1 + x)}^{61}$

$\Rightarrow x^{2} S (x) = 60 x \cdot {(1 + x)}^{61} - (1 + x) ({(1 + x)}^{60} - 1)$

For $x = 60,$ we have

${(60)}^{2} S (60) = 3600 {(61)}^{61} - 61 ({(61)}^{60} - 1)$

$= {(61)}^{61} [3600 - 1] + 61 = 3599 {(61)}^{61} + 61$

$\Rightarrow a = 3599, b = 61$

$∴ a + b = 3599 + 61 = 3660$

Q 23 :

The sum $1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + . . .$ upto $\infty$ terms, is equal to [2025]

6e
3e
2e
4e

1

2

3

4

(3)

Let $S = 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + . . .$

$= \sum_{r = 1}^{\infty} \frac{r^{2}}{r!} = \sum_{r = 1}^{\infty} \frac{r}{(r - 1)!}$

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

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

Q 24 :

$1 + 3 + 5^{2} + 7 + 9^{2} + . . .$ upto 40 terms is equal to [2025]

41880
33980
40870
43890

1

2

3

4

(1)

$1 + 3 + 5^{2} + 7 + 9^{2} + . . . 40$ terms

$= (1^{2} + 5^{2} + 9^{2} + . . . 20 terms) + (3 + 7 + 11 + . . . 20 terms)$

$= \sum_{r = 1}^{20} {(4 r - 3)}^{2} + \sum_{r = 1}^{20} (4 r - 1)$

$= \sum_{r = 1}^{20} (16 r^{2} - 20 r + 8)$

$= 16 \sum_{r = 1}^{20} r^{2} - 20 \sum_{r = 1}^{20} r + 8 \sum_{r = 1}^{20} 1$

$= \frac{16 \times 20 \times 21 \times 41}{6} - \frac{20 \times 20 \times 21}{2} + 8 \times 20 = 41880$ .

Q 25 :

If the sum of the first 20 terms of the series $\frac{4 \cdot 1}{4 + 3 \cdot 1^{2} + 1^{4}} + \frac{4 \cdot 2}{4 + 3 \cdot 2^{2} + 2^{4}} + \frac{4 \cdot 3}{4 + 3 \cdot 3^{2} + 3^{4}} + \frac{4 \cdot 4}{4 + 3 \cdot 4^{2} + 4^{4}} + . . . is \frac{m}{n}$ , where m and n are coprime, then m + n is equal to : [2025]

423
421
420
422

1

2

3

4

(2)

$\frac{4 \cdot 1}{4 + 3 \cdot 1^{2} + 1^{4}} + \frac{4 \cdot 2}{4 + 3 \cdot 2^{2} + 2^{4}} + . . . .$ upto 20 terms

$= \sum_{r = 1}^{20} \frac{4 r}{4 + 3 r^{2} + r^{4}}$

$= \sum_{r = 1}^{20} \frac{4 r}{(r^{2} + r + 2) (r^{2} - r + 2)}$

$= 2 \sum_{r = 1}^{20} \frac{1}{r^{2} - r + 2} - \frac{1}{r^{2} + r + 2}$

$= 2 [(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{8}) + (\frac{1}{8} - \frac{1}{14}) + . . . + (\frac{1}{382} - \frac{1}{422})]$

$= 2 [\frac{1}{2} - \frac{1}{422}] = \frac{420}{422} = \frac{210}{211} = \frac{m}{n}$ , m and n are co-prime.

So, m + n = 210 + 211 = 421.

Q 26 :

If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . \infty = \frac{π^{4}}{90}$ , $\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . \infty = α$ , $\frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + . . . \infty = β$ , then $\frac{α}{β}$ is equal to [2025]

14
23
15
18

1

2

3

4

(3)

We have, $β = \frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + . . .$

$= \frac{1}{2^{4}} (1 + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . .) = \frac{1}{16} (\frac{π^{4}}{90})$ [Given]

$∴ β = \frac{π^{4}}{1440}$ ... (i)

Now, $\frac{π^{4}}{90} = \frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . .$

$= (\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . . . .) + (\frac{1}{2^{4}} + \frac{1}{4^{4}} + . . . . . .)$

$= α + β$

$\Rightarrow α = \frac{π^{4}}{90} - β = \frac{π^{4}}{90} - \frac{π^{4}}{1440}$ [Using (i)]

$= \frac{16 π^{4} - π^{4}}{1440} = \frac{15 π^{4}}{1440}$

$∴ \frac{α}{β} = \frac{15}{1} = 15$ .

Q 27 :

If $\sum_{r = 1}^{n} T_{r} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{64}$ , then $\lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}})$ is equal to : [2025]

0
$\frac{2}{3}$
$\frac{1}{3}$
1

1

2

3

4

(2)

Given, $S_{n} = \sum_{r = 1}^{n} T_{r} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{64}$

So, $S_{n - 1} = \frac{(2 n - 3) (2 n - 1) (2 n + 1) (2 n + 3)}{64}$

$∵ T_{n} = S_{n} - S_{n - 1}$

$\Rightarrow T_{n} = \frac{(2 n - 1) (2 n + 1) (2 n + 3)}{8}$

$∴ \sum_{r = 1}^{n} (\frac{1}{T_{n}}) = \frac{8}{4} [\frac{1}{(2 n - 1) (2 n + 1)} - \frac{1}{(2 n + 1) (2 n + 3)}]$

$\lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}}) = 2 \lim_{n \to \infty} (\frac{1}{(2 n - 1) (2 n + 1)} - \frac{1}{(2 n + 1) (2 n + 3)})$

$\Rightarrow \lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}}) = 2 (\frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} + \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} + . . .)$

$\lim_{n \to \infty} \frac{2}{3} - \frac{2}{(2 n + 1) (2 n + 3)} = \frac{2}{3}$ .

Q 28 :

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . .$ upto n terms. If the sum of the first six terms of an A.P. with first term –p and common difference p is $\sqrt{2026 S_{2025}}$ , then the absolute difference between $20^{th}$ and $15^{th}$ terms of the A.P. is [2025]

20
45
25
90

1

2

3

4

(3)

$S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . .$ upto n terms

$= \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + . . . + \frac{1}{n (n + 1)}$

$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + . . . (\frac{1}{n} - \frac{1}{n + 1})$

$= 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$

$∴ \sqrt{2026 S_{2025}} = \sqrt{2026 \times \frac{2025}{2025 + 1}} = 45$

$S_{6} = 45$ , where $S_{6}$ denotes the sum of first six terms of an A.P.

$∴ \frac{6}{2} (- 2 p + 5 p) = 45 \Rightarrow p = 5$

$| a_{20} - a_{15} | = | (- 5 + 19 \times 5) - (- 5 + 14 \times 5) |$

$= | 90 - 65 | = 25$ .

Q 29 :

If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$ , then the value of $α$ is : [2025]

$\frac{6}{7}$
$\frac{1}{7}$
6
1

1

2

3

4

(3)

Given: $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$

Let $k = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$

$\Rightarrow \frac{k}{7} = \frac{5}{7} + \frac{1}{7^{2}} (5 + α) + \frac{1}{7^{3}} (5 + 2 α) + . . . . . \infty$

$\Rightarrow k - \frac{k}{7} = 5 + \frac{1}{7} α (\frac{1}{1 - \frac{1}{7}})$

$\Rightarrow \frac{6 k}{7} = 5 + \frac{α}{7} \times \frac{7}{6}$

Now, $k - \frac{k}{7} = 5 + \frac{α}{7} + \frac{α}{7^{2}} + \frac{α}{7^{3}} + . . . \infty$

$\Rightarrow \frac{6 k}{7} = 5 + \frac{α}{6}$

$\Rightarrow \frac{6 \times 7}{7} = 5 + \frac{α}{6}$ [ $∵$ k = 7]

$\Rightarrow \frac{α}{6} = 1 \Rightarrow α = 6$ .

Q 30 :

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a_{k}})$ , then the value of $507 S_{2025}$ is : [2025]

675
1350
540
135

1

2

3

4

(1)

We have, $a_{n} = \frac{n^{2} + 5 n + 6}{4}$

Also, $S_{n} = \sum_{k = 1}^{n} \frac{1}{a_{k}}$

$= \sum_{k = 1}^{n} \frac{4}{k^{2} + 5 k + 6}$

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

$= 4 \sum_{k = 1}^{n} \frac{1}{k + 2} - \frac{1}{k + 3}$

$= 4 (\frac{1}{3} - \frac{1}{4}) + 4 (\frac{1}{4} - \frac{1}{5}) + . . . + 4 (\frac{1}{n + 2} - \frac{1}{n + 3})$

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

So, $507 S_{2025} = \frac{507 (4) (2025)}{3 [2025 + 3]} = 675$ .

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]

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]

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]

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]

Q 11 :

The sum $1^{2} - 2 \cdot 3^{2} + 3 \cdot 5^{2} - 4 \cdot 7^{2} + 5 \cdot 9^{2} - . . . . . + 15 \cdot 29^{2}$ is _______. [2023]

Q 12 :

The value of $\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots + 100 \times {(101)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots + 100^{2} \times 101}$ is [2024]

Q 13 :

If $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$ and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{99 \cdot 100} = n,$ then the point $(m, n)$ lies on the line [2024]

Q 14 :

If the sum of the series $\frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) (1 + 2 d)} + \dots + \frac{1}{(1 + 9 d) (1 + 10 d)}$ is equal to 5, then $50 d$ is equal to : [2024]

Q 15 :

The sum of the series $\frac{1}{1 - 3 . 1^{2} + 1^{4}} + \frac{2}{1 - 3 . 2^{2} + 2^{4}} + \frac{3}{1 - 3 . 3^{2} + 3^{4}} + \dots$ up to 10-terms is [2024]

Q 17 :

Let the first term of a series be $T_{1} = 6$ and its $r^{t h}$ term $T_{r} = 3 T_{r - 1} + 6^{r}, r = 2, 3, \dots, n .$ If the sum of the first $n$ terms of this series is $\frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1),$ then $n$ is equal to ______ . [2024]

Q 18 :

Let the positive integers be written in the form:

1

   2 3

  4 5 6

7 8 9 10

If the $k^{t h}$ row contains exactly $k$ numbers for every natural number $k,$ then the row in which the number 5310 will be, is ______.     [2024]

Q 19 :

If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024},$ Then $α$ is equal to _______ . [2024]

Q 20 :

Let $α = 1^{2} + 4^{2} + 8^{2} + 13^{2} + 19^{2} + 26^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4} .$ If $4 α - β = 55 k + 40,$ then $k$ is equal to ______ . [2024]

Q 21 :

Let $S_{n}$ be the sum to $n$ -terms of an arithmetic progression 3, 7, 11,..... If $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42,$ then $n$ equals _____ . [2024]

Q 22 :

If $S (x) = (1 + x) + 2 {(1 + x)}^{2} + 3 {(1 + x)}^{3} + \dots + 60 {(1 + x)}^{60}$ , $x \neq 0$ and ${(60)}^{2} S (60) = a {(b)}^{b} + b,$ where $a, b \in N,$ then $(a + b)$ is equal to _______. [2024]

Q 23 :

The sum $1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + . . .$ upto $\infty$ terms, is equal to [2025]

Q 24 :

$1 + 3 + 5^{2} + 7 + 9^{2} + . . .$ upto 40 terms is equal to [2025]

Q 26 :

If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . \infty = \frac{π^{4}}{90}$ , $\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . \infty = α$ , $\frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + . . . \infty = β$ , then $\frac{α}{β}$ is equal to [2025]

Q 27 :

If $\sum_{r = 1}^{n} T_{r} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{64}$ , then $\lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}})$ is equal to : [2025]

Q 29 :

If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$ , then the value of $α$ is : [2025]

Q 30 :

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a_{k}})$ , then the value of $507 S_{2025}$ is : [2025]

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Topic Question Set

Q 1 : Let Sk=1+2+...+kk and ∑j=1nSj2=nA(Bn2+Cn+D), where A,B,C,D∈N and A has least value. Then [2023]

Q 2 : Let an be the nth term of the series 5 + 8 + 14 + 23 + 35 + 50 + ... and Sn=∑k=1nak. Then S30-a40 is equal to [2023]

Q 3 : Let <an> be a sequence such that a1+a2+...+an=n2+3n(n+1)(n+2). If 28∑k=1101ak=p1p2p3...pm, where p1,p2,....,pm are the first m prime numbers, then m is equal to [2023]

Q 4 : The sum to 10 terms of the series 11+12+14+21+22+24+31+32+34+... is [2023]

Q 5 : The sum ∑n=1∞2n2+3n+4(2n)! 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]

Q 7 : The sum to 20 terms of the series 2·22-32+2·42-52+2·62-.......... is equal to _______ . [2023]

Q 8 : If the sum of the series (12-13)+(122-12·3+132)+(123-122·3+12·32-133)+(124-123·3+122·32-12·33+134)+... is αβ, where α and β are co-prime, then α+3β is equal to ________ . [2023]

Q 9 : If 13+23+33+... upto n terms1·3+2·5+3·7+... upto n terms=95, then the value of n is _______. [2023]

Q 10 : Let a1=b1=1 and an=an-1+(n-1),bn=bn-1+an-1,∀n≥2. If S=∑n=110 bn2n and T=∑n=18n2n-1, then 27(2S-T) is equal to _________ . [2023]

Q 11 : The sum 12-2·32+3·52-4·72+5·92-.....+15·292 is _______. [2023]

Q 12 : The value of 1×22+2×32+…+100×(101)212×2+22×3+…+1002×101 is [2024]

Q 13 : If 11+2+12+3+⋯+199+100=m and 11·2+12·3+⋯+199·100=n, then the point (m,n) lies on the line [2024]

Q 14 : If the sum of the series 11·(1+d)+1(1+d)(1+2d)+⋯+1(1+9d)(1+10d) is equal to 5, then 50d is equal to : [2024]

Q 15 : The sum of the series 11-3.12+14+21-3.22+24+31-3.32+34+…up to 10-terms is [2024]

Q 16 : If 1+3-223+5-2618+93-112363+49-206180+⋯ upto ∞=2+(ba+1)loge(ab), where a and b are integer with gcd(a,b)=1, then 11a+18b is equal to _______. [2024]

Q 17 : Let the first term of a series be T1=6 and its rth term Tr=3Tr-1+6r, r=2,3,…,n. If the sum of the first n terms of this series is 15(n2-12n+39)(4·6n-5·3n+1), then n is equal to ______ . [2024]

Q 18 : Let the positive integers be written in the form: 1 2 3 4 5 6 7 8 9 10 If the kth row contains exactly k numbers for every natural number k, then the row in which the number 5310 will be, is ______. [2024]

Q 19 : If (1α+1+1α+2+⋯+1α+1012)-(12·1+14·3+16·5+⋯+12024·2023)=12024, Then α is equal to _______ . [2024]

Q 20 : Let α=12+42+82+132+192+262+… upto 10 terms and β=∑n=110n4. If 4α-β=55k+40, then k is equal to ______ . [2024]

Q 21 : Let Sn be the sum to n-terms of an arithmetic progression 3, 7, 11,..... If 40<(6n(n+1)∑k=1nSk)<42, then n equals _____ . [2024]

Q 22 : If S(x)=(1+x)+2(1+x)2+3(1+x)3+…+60(1+x)60, x≠0 and (60)2S(60)=a(b)b+b, where a,b∈N, then (a+b) is equal to _______. [2024]

Q 23 : The sum 1+1+32!+1+3+53!+1+3+5+74!+... upto ∞ terms, is equal to [2025]

Q 24 : 1+3+52+7+92+... upto 40 terms is equal to [2025]

Q 25 : If the sum of the first 20 terms of the series 4·14+3·12+14+4·24+3·22+24+4·34+3·32+34+4·44+3·42+44+... is mn, where m and n are coprime, then m + n is equal to : [2025]

Q 26 : If 114+124+134+... ∞=π490, 114+134+154+... ∞=α, 124+144+164+... ∞=β, then αβ is equal to [2025]

Q 27 : If ∑r=1nTr=(2n–1)(2n+1)(2n+3)(2n+5)64, then limn→∞ ∑r=1n(1Tr) is equal to : [2025]

Q 28 : Let Sn=12+16+112+120+... upto n terms. If the sum of the first six terms of an A.P. with first term –p and common difference p is 2026S2025, then the absolute difference between 20th and 15th terms of the A.P. is [2025]

Q 29 : If 7=5+17(5+α)+172(5+2α)+173(5+3α)+.....∞, then the value of α is : [2025]

Q 30 : For positive integers n, if 4an=(n2+5n+6) and Sn=∑k=1n(1ak), then the value of 507S2025 is : [2025]

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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]

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]

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]

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]

Q 11 :

The sum $1^{2} - 2 \cdot 3^{2} + 3 \cdot 5^{2} - 4 \cdot 7^{2} + 5 \cdot 9^{2} - . . . . . + 15 \cdot 29^{2}$ is _______. [2023]

Q 12 :

The value of $\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots + 100 \times {(101)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots + 100^{2} \times 101}$ is [2024]

Q 13 :

If $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$ and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{99 \cdot 100} = n,$ then the point $(m, n)$ lies on the line [2024]

Q 14 :

If the sum of the series $\frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) (1 + 2 d)} + \dots + \frac{1}{(1 + 9 d) (1 + 10 d)}$ is equal to 5, then $50 d$ is equal to : [2024]

Q 15 :

The sum of the series $\frac{1}{1 - 3 . 1^{2} + 1^{4}} + \frac{2}{1 - 3 . 2^{2} + 2^{4}} + \frac{3}{1 - 3 . 3^{2} + 3^{4}} + \dots$ up to 10-terms is [2024]

Q 17 :

Let the first term of a series be $T_{1} = 6$ and its $r^{t h}$ term $T_{r} = 3 T_{r - 1} + 6^{r}, r = 2, 3, \dots, n .$ If the sum of the first $n$ terms of this series is $\frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1),$ then $n$ is equal to ______ . [2024]

Q 18 :

Let the positive integers be written in the form:

1

2 3

4 5 6

7 8 9 10

If the $k^{t h}$ row contains exactly $k$ numbers for every natural number $k,$ then the row in which the number 5310 will be, is ______. [2024]

Q 19 :

If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024},$ Then $α$ is equal to _______ . [2024]

Q 20 :

Let $α = 1^{2} + 4^{2} + 8^{2} + 13^{2} + 19^{2} + 26^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4} .$ If $4 α - β = 55 k + 40,$ then $k$ is equal to ______ . [2024]

Q 21 :

Let $S_{n}$ be the sum to $n$ -terms of an arithmetic progression 3, 7, 11,..... If $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42,$ then $n$ equals _____ . [2024]

Q 22 :

If $S (x) = (1 + x) + 2 {(1 + x)}^{2} + 3 {(1 + x)}^{3} + \dots + 60 {(1 + x)}^{60}$ , $x \neq 0$ and ${(60)}^{2} S (60) = a {(b)}^{b} + b,$ where $a, b \in N,$ then $(a + b)$ is equal to _______. [2024]

Q 23 :

The sum $1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + . . .$ upto $\infty$ terms, is equal to [2025]

Q 24 :

$1 + 3 + 5^{2} + 7 + 9^{2} + . . .$ upto 40 terms is equal to [2025]

Q 26 :

If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . \infty = \frac{π^{4}}{90}$ , $\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . \infty = α$ , $\frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + . . . \infty = β$ , then $\frac{α}{β}$ is equal to [2025]

Q 27 :

If $\sum_{r = 1}^{n} T_{r} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{64}$ , then $\lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T_{r}})$ is equal to : [2025]

Q 29 :

If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7^{2}} (5 + 2 α) + \frac{1}{7^{3}} (5 + 3 α) + . . . . . \infty$ , then the value of $α$ is : [2025]

Q 30 :

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a_{k}})$ , then the value of $507 S_{2025}$ is : [2025]