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

Q 1 :
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

(B)

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 2 :
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

(C)

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 3 :
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

(C)

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 4 :
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

(B)

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 5 :
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

(D)

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 6 :
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

(A)

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 $∴$ $d = \frac{2 - (- 6)}{5} = \frac{8}{5}$

Q 7 :
If $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P. and $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are also in an A.P., then $a : b : c$ is equal to [2024]

16 : 4 : 1
9 : 6 : 4
6 : 3 : 2
25 : 10 : 4

1

2

3

4

(B)

We have, $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P.

$\Rightarrow 2 \log_{e} b = \log_{e} a + \log_{e} c \Rightarrow b^{2} = a c$ ...(i)

Also, $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are in an A.P.

$\Rightarrow 2 (\log_{e} 2 b - \log_{e} 3 c) = \log_{e} a - \log_{e} 2 b + \log_{e} 3 c - \log_{e} a$

$\Rightarrow \log_{e} {(\frac{2 b}{3 c})}^{2} = \log_{e} (\frac{3 c}{2 b}) \Rightarrow \frac{4 b^{2}}{9 c^{2}} = \frac{3 c}{2 b}$

$\Rightarrow \frac{4 b^{2}}{9 c^{2}} = \frac{3 c}{2 b}$ ...(ii)

From (i) and (ii), we have

$a = \frac{9 c^{2}}{4 c} \Rightarrow a = \frac{9}{4} c$ $∴$ $a : b : c = \frac{9}{4} : \frac{3}{2} : 1 = 9 : 6 : 4$

Q 8 :
Let $S_{n}$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145,$ then $S_{15} - S_{5}$ is [2024]

390
405
410
395

1

2

3

4

(D)

Given, $S_{20} = 790$

$\Rightarrow \frac{20}{2} (2 a + 19 d) = 790 \Rightarrow 2 a + 19 d = 79$ ...(i)

Given, $S_{10} = 145$

$\frac{10}{2} (2 a + 9 d) = 145 \Rightarrow 2 a + 9 d = 29$ ...(ii)

From (i) and (ii), we get $10 d = 50 \Rightarrow d = 5$

From (i), we get $a = \frac{79 - 95}{2} = - 8$

So, $S_{15} - S_{5} = \frac{15}{2} [- 16 + 70] - \frac{5}{2} [- 16 + 20] = 395$

Q 9 :
Let $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms.

Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2} .$

If $A_{3} = - 153, A_{5} = - 435$ and $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66,$ then $a_{17} - A_{7}$ is equal to _______ . [2024]

0%

(910)

$a_{1}, a_{2}, a_{3}, \dots$ is an A.P.

Let $d$ be the common difference

$∴$ $a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{2 k} - a_{2 k - 1} = d$

Now, $a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$

$= (a_{1} - a_{2}) (a_{1} + a_{2}) + \dots + (a_{2 k - 1} - a_{2 k}) (a_{2 k - 1} + a_{2 k})$

$= - d [a_{1} + a_{2} + \dots + a_{2 k - 1} + a_{2 k}]$

$= - d \cdot [\frac{2 k}{2} [a_{1} + a_{2 k}]] = - d k (a_{1} + a_{2 k})$

$∴$ $A_{k} = - d k (a_{1} + a_{2 k})$

So, $A_{3} = - 3 d (a_{1} + a_{6}) = - 153$

$\Rightarrow - 3 d (a_{1} + a_{1} + 5 d) = - 153$

$\Rightarrow - 3 d (2 a_{1} + 5 d) = - 153 \Rightarrow 2 a_{1} + 5 d = \frac{51}{d}$ ...(i)

Similarly $A_{5} = - 435$

$\Rightarrow - 5 d (2 a_{1} + 9 d) = - 435 \Rightarrow 2 a_{1} + 9 d = \frac{87}{d}$ ...(ii)

Solving (i) and (ii) we get $4 d = \frac{36}{d} \Rightarrow d^{2} = 9 \Rightarrow d = 3$ [ $∵$ Given A.P. is a progression of positive terms]

$\Rightarrow a_{1} = 1$

Now, $a_{17} = a_{1} + 16 d = 1 + 48 = 49$

and $A_{7} = - 3 \times 7 (a_{1} + a_{14})$

$= - 21 (1 + 1 + 13 \times 3) = - 21 (41) = - 861$

$∴$ $a_{17} - A_{7} = 49 + 861 = 910$

Q 10 :
An arithmetic progression is written in the following way

2

5 8

11 14 17

20 23 26 29

--------------------------------------------------------------------

---------------------------------------------------------------------------

The sum of all the terms of the $10^{th}$ row is ______. [2024]

0%

(1505)

First term of the $10^{th}$ row is $10^{th}$ term of sequence 2, 5, 11, 20, 32, ...

$S = 2 + 5 + 11 + 20 + 32 + \dots + T_{10}$

$S =$ $2 + 5 + 11 + 20 + 32 + \dots + T_{9} + T_{10}$

On subtracting, we get

$0 = 2 + \underset{9 terms}{\underset{⏟}{3 + 6 + 9 + 12 + \dots} - T_{10}}$

$\Rightarrow T_{10} = 2 + \underset{9 terms}{\underset{⏟}{3 + 6 + 9 + 12 + \dots}}$

$= 2 + \frac{9}{2} [6 + 8 \times 3] = 2 + 135 = 137$

Terms in $10^{th}$ row with common difference 3 are 137, 140, 143, .....

$S_{10} = \frac{10}{2} [2 \times 137 + 9 \times 3] = 5 \times 301 = 1505$

Topic Question Set

Q 1 :
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]

Q 3 :
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]

Q 4 :
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]

Q 5 :
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]

Q 6 :
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]

Q 7 :
If $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P. and $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are also in an A.P., then $a : b : c$ is equal to [2024]

Q 8 :
Let $S_{n}$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145,$ then $S_{15} - S_{5}$ is [2024]

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Topic Question Set

Q 1 : For x≥0, the least value of K, for which 41+x+41-x, K2, 16x+16-x are three consecutive terms of an A.P., is equal to: [2024]

Q 3 : Let Sn denote the sum of the first n terms of an arithmetic progression. If S10=390 and the ratio of the tenth and the fifth terms is 15 : 7, then S15-S5 is equal to: [2024]

Q 4 : 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]

Q 5 : The 20th term from the end of the progression 20,1914,1812,1734,…,-12914 is: [2024]

Q 6 : In an A.P., the sixth term a6=2. If the product a1a4a5 is the greatest, then the common difference of the A.P. is equal to [2024]

Q 7 : If logea,logeb,logec are in an A.P. and logea-loge2b,loge2b-loge3c,loge3c-logea are also in an A.P., then a:b:c is equal to [2024]

Q 8 : Let Sn denote the sum of first n terms of an arithmetic progression. If S20=790 and S10=145, then S15-S5 is [2024]

Q 9 : Let a1,a2,a3,… be in an arithmetic progression of positive terms. Let Ak=a12-a22+a32-a42+…+a2k-12-a2k2. If A3=-153, A5=-435 and a12+a22+a32=66, then a17-A7 is equal to _______ . [2024]

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Q 10 : An arithmetic progression is written in the following way 2 5 8 11 14 17 20 23 26 29 -------------------------------------------------------------------- --------------------------------------------------------------------------- The sum of all the terms of the 10th row is ______. [2024]

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Q 1 :
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]

Q 3 :
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]

Q 4 :
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]

Q 5 :
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]

Q 6 :
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]

Q 7 :
If $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P. and $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are also in an A.P., then $a : b : c$ is equal to [2024]

Q 8 :
Let $S_{n}$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145,$ then $S_{15} - S_{5}$ is [2024]