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

Q 11 :

Let $A_{1}, A_{2}, A_{3}$ be the three A.P. with the same common difference $d$ and having their first terms as A, A + 1, A + 2, respectively. Let $a, b, c$ be the $7^{t h}, 9^{t h}, 17^{t h}$ terms of $A_{1}, A_{2}, A_{3},$ respectively such that $| \begin{matrix} a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1 \end{matrix} |$ $+ 70 = 0 .$ If $a = 29,$ then the sum of first 20 terms of an A.P. whose first term is $c - a - b$ and common difference is $\frac{d}{12}$ , is equal to ________ . [2023]

0%

(495)

Given $| \begin{matrix} a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1 \end{matrix} |$ $+ 70 = 0$

$\Rightarrow$ $| \begin{matrix} A + 6 d & 7 & 1 \\ 2 (A + 1 + 8 d) & 17 & 1 \\ A + 2 + 16 d & 17 & 1 \end{matrix} |$ $+ 70 = 0$

$\Rightarrow (A + 6 d) (0) - 7 [(2 A + 2 + 16 d) - (A - 2 - 16 d)] + 1 [17 (2 A + 2 + 16 d - A - 2 - 16 d)] + 70 = 0$

$\Rightarrow - 7 (A) + 1 [17 (A)] + 70 = 0$

$\Rightarrow 10 A + 70 = 0 \Rightarrow A = - 7$

$\Rightarrow A + 6 d = 29 \Rightarrow - 7 + 6 d = 29 \Rightarrow 6 d = 36 \Rightarrow d = 6$

$c = A + 2 + 16 d = - 7 + 2 + 16 (6) = 91$

$b = A + 1 + 8 d = - 7 + 1 + 8 (6) = 42$

$\Rightarrow c - a - b = 91 - 29 - 42 = 20$

$S_{20} = \frac{20}{2} [2 \times 20 + 19 \times \frac{6}{12}] = 495$

Q 12 :

The $8^{t h}$ common term of the series

$S_{1} = 3 + 7 + 11 + 15 + 19 + . . .,$

$S_{2} = 1 + 6 + 11 + 16 + 21 + . . . . .$

is _________ . [2023]

0%

(151)

$S_{1} = 3 + 7 + 11 + 15 + 19 + \dots$

$a_{1} = 3, d_{1} = 4$

$S_{2} = 1 + 6 + 11 + 16 + 21 + \dots$

$a_{2} = 1, d_{2} = 5$

First common term, $a$ = 11

$Common difference, d = LCM (d_{1}, d_{2}) = LCM (4, 5) = 20$

$8^{t h}$ common term of the series

$a_{8} = a + 7 d = 11 + 7 \times 20 = 151$

Q 13 :

Let $a_{1}, a_{2}, . . . ., a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18,$ then $12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + . . . + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$ is equal to _______ . [2023]

0%

(8)

We have, $a_{1}, a_{2}, \dots, a_{n}$ are in A.P.

$a_{5} = 2 a_{7}$

$\Rightarrow a + 4 d = 2 (a + 6 d) \Rightarrow a + 8 d = 0 ...(1)$

$and a_{11} = 18 \Rightarrow a + 10 d = 18 ...(2)$

$On solving (1) and (2), we get 2 d = 18 \Rightarrow d = 9 \Rightarrow a = - 72$

$a_{10} = a + 9 d = - 72 + 81 = 9$

$a_{18} = a + 17 d = - 72 + 153 = 81$

$Now, 12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \dots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}}}{{(\sqrt{a_{10}})}^{2} - {(\sqrt{a_{11}})}^{2}} + \frac{\sqrt{a_{11}} - \sqrt{a_{12}}}{{(\sqrt{a_{11}})}^{2} - {(\sqrt{a_{12}})}^{2}} + \dots + \frac{\sqrt{a_{17}} - \sqrt{a_{18}}}{{(\sqrt{a_{17}})}^{2} - {(\sqrt{a_{18}})}^{2}})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}}}{a + 9 d - a - 10 d} + \frac{\sqrt{a_{11}} - \sqrt{a_{12}}}{a + 10 d - a - 11 d} + \dots + \frac{\sqrt{a_{17}} - \sqrt{a_{18}}}{a + 16 d - a - 17 d})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}} + \sqrt{a_{11}} - \sqrt{a_{12}} + \dots + \sqrt{a_{17}} - \sqrt{a_{18}}}{- d})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{18}}}{- 9}) = \frac{12 \times (\sqrt{9} - \sqrt{81})}{- 9} = \frac{12 \times (3 - 9)}{- 9} = 8$

Q 14 :

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to _________ . [2023]

0%

(710)

$Number divisible by 3,1002,1005,…,2799$

$T_{n} = 1002 + 3 (n - 1) = 2799 \Rightarrow n = 600$

$Number divisible by 11 are 164$

$Number divisible by 33 are 54$

$Total number = 600 + 164 - 54 = 710$

Q 15 :

For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

4
10
8
16

1

2

3

4

(2)

We have, $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are in A.P.

$\Rightarrow 2 \frac{K}{2} = (4^{1 + x} + 4^{1 - x}) + (16^{x} + 16^{- x})$

$\Rightarrow K = 4 \cdot 4^{x} + \frac{4}{4^{x}} + 4^{2 x} + \frac{1}{4^{2 x}}$

$= 4 (4^{x} + \frac{1}{4^{x}}) + (4^{2 x} + \frac{1}{4^{2 x}})$ $\geq 4 \cdot 2 + 2$ $[∵ A . M . \geq G . M .]$

$= 10 \Rightarrow K \geq 10$

So, least value of $K$ is 10

Q 16 :

A software company sets up $m$ number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $m$ is equal to: [2024]

160
180
150
125

1

2

3

4

(3)

Let the work done by each computer $= k$

$∴$ Total work $= 17$ $m k$

Work done on 1 day by $m$ computers $= m k$

Work done on day 2 by $m - 4$ computers $= (m - 4) k$

Work done on day 3 by $m - 8$ computers $= (m - 8) k$

Work done on day 25 $= (m - (24 \times 4)) k$

$= (m - 96) k$

$∴$ $m k + (m - 4) k + \dots + (m - 96) k = 17 m k$

$\Rightarrow 25 m - (4 + 8 + \dots + 96) = 17 m$

$\Rightarrow 8 m = \frac{24}{2} (4 + 96) \Rightarrow m = 150$

Q 17 :

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is 15 : 7, then $S_{15} - S_{5}$ is equal to: [2024]

800
890
790
690

1

2

3

4

(3)

We have, $S_{10} = 390$

$\Rightarrow 5 (2 a + 9 d) = 390 \Rightarrow 2 a + 9 d = 78$ ...(i)

Also, $\frac{a_{10}}{a_{5}} = \frac{15}{7} \Rightarrow \frac{a + 9 d}{a + 4 d} = \frac{15}{7}$

$\Rightarrow 7 a + 63 d = 15 a + 60 d \Rightarrow 8 a = 3 d$

$\Rightarrow a = \frac{3}{8} d$ ...(ii)

From (i) & (ii), we get $d = 8$ and $a = 3$

Now, $S_{15} - S_{5} = \frac{15}{2} [6 + 112] - \frac{5}{2} [6 + 32] = 790$

Q 18 :

The number of common terms in the progressions
4, 9, 14, 19, ........…, up to 25th term and 3, 6, 9, 12, ....…, up to 37th term is: [2024]

8
7
5
9

1

2

3

4

(2)

4, 9, 14, 19, ... are in A.P. with $d_{1} = 5, n_{1} = 25$

3, 6, 9, 12, ... are in A.P. with $d_{2} = 3, n_{2} = 37$

L.C.M. $(d_{1}, d_{2}) = 15$

Common terms = 9, 24, 39, 54, 69, 84, 99

Q 19 :

The $20^{th}$ term from the end of the progression $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is: [2024]

$- 118$
$- 100$
$- 110$
$- 115$

1

2

3

4

(4)

Given, $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is in A.P. with first term $(a) = - 129 \frac{1}{4}$

Common difference $(d) = 20 - 19 \frac{1}{4} = \frac{80 - 77}{4} = \frac{3}{4}$

$20^{th}$ term from the end $= a_{20} = a + (20 - 1) d$

$= - 129 \frac{1}{4} + 19 \times \frac{3}{4} = \frac{- 517}{4} + \frac{57}{4} = \frac{- 460}{4} = - 115$

$∴$ $a_{20} = - 115$

Q 20 :

In an A.P., the sixth term $a_{6} = 2 .$ If the product $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P. is equal to [2024]

$\frac{8}{5}$
$\frac{5}{8}$
$\frac{3}{2}$
$\frac{2}{3}$

1

2

3

4

(1)

We have, $a_{6} = 2 \Rightarrow a + 5 d = 2 \Rightarrow d = \frac{2 - a}{5}$

Let $a_{1} a_{4} a_{5} = λ$

$\Rightarrow λ = a (a + 3 d) (a + 4 d)$ $= a (a^{2} + 7 a d + 12 d^{2})$ $= a^{3} + 7 a^{2} d + 12 a d^{2}$

$= a^{3} + 7 a^{2} (\frac{2 - a}{5}) + 12 a {(\frac{2 - a}{5})}^{2}$ $[∵ d = \frac{2 - a}{5}]$

$= a^{3} + \frac{14}{5} a^{2} - \frac{7}{5} a^{3} + \frac{12 a}{25} (4 - 4 a + a^{2})$

$= a^{3} + \frac{14}{5} a^{2} - \frac{7}{5} a^{3} + \frac{48}{25} a - \frac{48}{25} a^{2} + \frac{12}{25} a^{3}$

$= \frac{2}{25} a^{3} + \frac{22}{25} a^{2} + \frac{48}{25} a = \frac{2}{25} (a^{3} + 11 a^{2} + 24 a)$

$∴$ $\frac{d λ}{d a} = \frac{2}{25} (3 a^{2} + 22 a + 24)$

$λ$ to be greatest

$∴$ $\frac{d λ}{d a} = 0 \Rightarrow 3 a^{2} + 22 a + 24 = 0 \Rightarrow a = - 6, - \frac{4}{3}$

$\frac{d^{2} λ}{d a^{2}} = \frac{2}{25} (6 a + 22), For a = - 6, \frac{d^{2} λ}{d a^{2}} < 0$

So, $λ$ is maximum

For $a = \frac{- 4}{3}, \frac{d^{2} λ}{d a^{2}} > 0, λ$ is minimum $∴$ $d = \frac{2 - (- 6)}{5} = \frac{8}{5}$

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