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

Q 1 :
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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4

(A)

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

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

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

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

Q 2 :
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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4

(C)

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

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 3 :
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]

5
15
10
20

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4

(A)

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 4 :
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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4

(C)

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 5 :
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]

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(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 6 :
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]

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(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 7 :
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]

0%

(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 8 :
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]

0%

(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 9 :
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]

0%

(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 10 :
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]

0%

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

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

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

$∴ n = 9$

Topic Question Set

Q 1 :
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 2 :
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 3 :
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 4 :
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]

0%

Q 6 :
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]

0%

Q 7 :
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]

0%

Q 8 :
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]

0%

Q 9 :
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]

0%

Q 10 :
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]

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Topic Question Set

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

Q 2 : 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 3 : 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 4 : The sum of the series 11-3.12+14+21-3.22+24+31-3.32+34+…up to 10-terms is [2024]

Q 5 : 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]

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Q 6 : 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]

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Q 7 : 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]

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Q 8 : If (1α+1+1α+2+⋯+1α+1012)-(12·1+14·3+16·5+⋯+12024·2023)=12024, Then α is equal to _______ . [2024]

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Q 9 : 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]

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Q 10 : 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]

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Q 1 :
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 2 :
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 3 :
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 4 :
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 6 :
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 7 :
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 8 :
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 9 :
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 10 :
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]