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

Q 21 :
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

(2)

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 22 :
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

(4)

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 23 :
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 24 :
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, .....

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

Q 25 :
Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

0%

(6699)

A.P. : 3, 7, 11, 15, ....., 403

$d_{1} = 4$

A.P. : 2, 5, 8, 11, ....., 404

$d_{2} = 3$

L.C.M. $(d_{1}, d_{2}) = 12$

Common terms = 11, 23, 35 ... 395

$a = 11, d = 12, a_{n} = 395$

$a_{n} = a + (n - 1) d \Rightarrow 395 = 11 + (n - 1) (12)$

$\Rightarrow \frac{384}{12} = (n - 1) \Rightarrow n = 33$

$S = \frac{33}{2} (11 + 395) = \frac{33}{2} (406) = 6699$

Q 26 :
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

17
18
10
11

1

2

3

4

(4)

Let First term, $a_{1} = a$ , common difference = d

Given, $\sum_{k = 1}^{12} a_{2 k - 1} = \frac{- 72}{5} a, a_{1} \neq 0$

$\Rightarrow a_{1} + a_{3} + a_{5} + . . . . + a_{23} = - \frac{72}{5} a$

$\Rightarrow a + (a + 2 d) + (a + 4 d) + . . . . + (a + 22 d) = \frac{- 72}{5} a$

Now, $\sum_{k = 1}^{12} a_{2 k - 1}$ is an AP with common difference as 2d.

$∴ \frac{12}{2} [2 a + 11 \times 2 d] = - \frac{72}{5} a$

$\Rightarrow 12 a + 132 d = - \frac{72}{5} a$

$\Rightarrow 132 a + 132 \times 5 d = 0$

$\Rightarrow a = - 5 d$

Also, given $\sum_{k = 1}^{n} a_{k} = 0 \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 0$

$\Rightarrow n (- 10 d + n d - d) = 0 \Rightarrow n d (n - 11) = 0 \Rightarrow n = 11$ .

Q 27 :
The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is : [2025]

10
8
4
6

1

2

3

4

(2)

Let n be the number of terms of an A.P., $a$ be the first term and $d$ be the common difference.

Let n = 2m, so terms are a, a + d, a + 2d, a + (2m – 1)d

Odd terms = a, a + 2d, a + 4d, ...., a + (2m – 2)d

Even terms = a + d, a + 3d, a + 5d, .... a + (2m – 1)d

Now, $S_{even} - S_{odd}$ = (a + d – a) + (a + 3d – a – 2d) + .... + (a + (2m – 1)d – a – (2m – 2)d)

$\Rightarrow 30 - 24 = m d \Rightarrow m d = 6$ ... (i)

Now, $a + (2 m - 1) d - a = \frac{21}{2} \Rightarrow 2 m d - d = \frac{21}{2}$

$\Rightarrow 12 - d = \frac{21}{2} \Rightarrow d = 12 - \frac{21}{2} = \frac{3}{2}$ [From (i)]

Since, $m d = 6 \Rightarrow m \times \frac{3}{2} = 6 \Rightarrow m = 4$

Now, $n = 2 m = 2 \times 4 = 8$

So, number of terms = 8.

Q 28 :
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

7240
8124
7130
6982

1

2

3

4

(1)

Let $S_{n}$ = 1 + 3 + 11 + 25 + 45 + 71 + ... + $T_{n}$

Here, series of differences i.e., (3 – 1), (11 – 3), (25 – 11) ..... i.e., 2, 8, 14, .... is in A.P.

If the second order differences of a square are in A.P. then general term is given by

$T_{n} = a n^{2} + b n + c$

Put n = 1, 2, 3, we get

$T_{1}$ = 1 = a + b + c ... (i)

$T_{2}$ = 3 = 4a + 2b + c ... (ii)

$T_{3}$ = 11 = 9a + 3b + c ... (iii)

On solving equation (i), (ii) and (iii), we get the general term of given series as

$T_{n} = 3 n^{2} - 7 n + 5$

Hence, $\sum_{n = 1}^{20} (3 n^{2} - 7 n + 5)$

$= 3 (\frac{20 \cdot 21 \cdot 41}{6}) - 7 (\frac{20 \cdot 21}{2}) + 5 (20) = 7240$ .

Q 29 :
Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n (A \cup B)$ is [2025]

4027
3761
4003
3814

1

2

3

4

(2)

Given A = {1, 6, 11, 16, ...}, B = {9, 16, 23, 30, ...} are two A.P. and n(A) = 2025, n(B) = 2025

$2025^{th} term of set A = 1 + 2024 \times 5 = 10121$

$2025^{th} term of set B = 9 + 2024 \times 7 = 14177$

$\Rightarrow$ A = {1, 6, 11, 16, ..., 10121} and B = {9, 16, 23, ...,14177}

Now $A \cap B$ = 16, 51, 86, .... which is an A.P. with a = 16, d = 35

then number of terms in $(A \cap B)$ = 16 + (n – 1) 35 < 10121

$\Rightarrow$ n – 1 < 288.7

$\Rightarrow$ n < 289.7

$\Rightarrow$ n = 289 [ $∵$ n is natural number]

$\Rightarrow n (A \cup B) = n (A) + n (B) - n (A \cap B)$

= 2025 + 2025 – 289 = 3761.

Q 30 :
Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP's in A and B respectively such that D = d + 3, d > 0. If $\frac{p + q}{p - q} = \frac{19}{5}$ , then p – q is equal to [2025]

450
630
540
600

1

2

3

4

(3)

Let A = (a – d, a, a + d) and B = (b – D, b, b + D)

For set A, Sum = 36, product = p

For set B, Sum = 36, product = q

$\Rightarrow a = 12, p = 12 (144 - d^{2})$

$\Rightarrow b = 12, q = 12 (144 - D^{2})$

$= 12 (144 - {(d + 3)}^{2}) = 12 (144 - d^{2} - 6 d - 9)$

Now, $\frac{p + q}{p - q} = \frac{19}{5} \Rightarrow \frac{p}{q} = \frac{12}{7}$

$\Rightarrow \frac{144 - d^{2}}{144 - d^{2} - 6 d - 9} = \frac{12}{7}$

$\Rightarrow 5 d^{2} + 72 d - 612 = 0$

$\Rightarrow d = \frac{- 72 \pm \sqrt{{(132)}^{2}}}{10}$

$\Rightarrow d = 6 and D = 9$ [ $∵ d > 0$ ]

$∴ p - q = 12 (D^{2} - d^{2}) = 12 (9^{2} - 6^{2}) = 540$ .

Topic Question Set

Q 21 :
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 22 :
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]

0%

0%

Q 25 :
Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

0%

Q 26 :
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

Q 27 :
The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is : [2025]

Q 28 :
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

Q 29 :
Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n (A \cup B)$ is [2025]

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Topic Question Set

Q 21 : 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 22 : 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 23 : 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 24 : 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 25 : Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

0%

Q 26 : Let a1,a2,a3, ..... be in an A.P. such that ∑k=112a2k–1=–725a1,a1≠0. If ∑k=1nak=0, then n is : [2025]

Q 27 : The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by 212. Then the number of terms which are integers in the A.P. is : [2025]

Q 28 : The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

Q 29 : Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B) is [2025]

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Q 21 :
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 22 :
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]

Q 25 :
Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, ....., 404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to _____. [2024]

Q 26 :
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

Q 27 :
The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$ . Then the number of terms which are integers in the A.P. is : [2025]

Q 28 :
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to 20 terms, is equal to [2025]

Q 29 :
Let A = {1, 6, 11, 16, ...} and B = {9, 16, 23, 30, ...} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n (A \cup B)$ is [2025]