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

Q 11 :
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]

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

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

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

Now, is an AP with common difference as 2d.

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

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

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

$\Rightarrow a = - 5 d$

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 13 :
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 14 :
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to20 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

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

Put n = 1 + 2 + 3

$T_{1}$ = 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 15 :
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 $(A \cup B)$ 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 = 15

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 16 :
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, $\frac{p + q}{p - q} = \frac{19}{5} \Rightarrow \frac{p}{q} = \frac{12}{7}$

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

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

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

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

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

Q 17 :
Let $a_{n}$ be the $n^{th}$ term an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

65
70
56
64

1

2

3

4

(4)

We have, $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$

Now, $a_{6} = 7 \Rightarrow a + 5 d = 7$ ... (i)

$S_{7} = 7 \Rightarrow \frac{7}{2} (2 a + 6 d) = 7$

$\Rightarrow a + 3 d = 1$ ... (ii)

On solving (i) and (ii), we get

d = 3 and a = – 8

So, $700 = \frac{n}{2} [- 16 + (n - 1) 3]$ $[∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]]$

$\Rightarrow 1400 = - 16 n + 3 n^{2} - 3 n$

$\Rightarrow 3 n^{2} - 19 n - 1400 = 0$

$\Rightarrow (3 n + 56) (n - 25) = 0 \Rightarrow n = 25$

$∴ a_{25} = a + 24 d = - 824 \times 3 = 64$ .

Q 18 :
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

6
4
5
8

1

2

3

4

(3)

Let the A.P. be $a_{1}, a_{2}, a_{3} . . . a_{2 k}$

Given, $\sum_{r = 1}^{k} a_{2 r - 1} = 40, \sum_{r = 1}^{k} a_{2 r} = 55 and a_{2 k} - a_{1} = 27$

$\Rightarrow \frac{k}{2} [2 a_{1} + (k - 1) 2 d] = 40$ ... (i)

and $\frac{k}{2} [2 a_{2} + (k - 1) 2 d] = 55$ ... (ii)

and $a_{1} + (2 k - 1) d - a_{1} = 27$ ... (iii)

After solving, we get

From (i), $a_{1} = \frac{40}{k} - (k - 1) d$

From (ii), $a_{2} = \frac{55}{k} - (k - 1) d$

and from (iii), $d = \frac{27}{2 k - 1}$

(ii) –(i)

$\Rightarrow d = \frac{15}{k} \Rightarrow \frac{27}{2 k - 1} = \frac{15}{k} \Rightarrow 9 k = 10 k - 5$

$∴$ k = 5.

Q 19 :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

–120
–1200
–1020
–1080

1

2

3

4

(4)

Let first term be 'a' and common difference be 'd'.

Also, $S_{4} = \frac{1}{5} (S_{8} - S_{4})$

$\Rightarrow 5 S_{4} = S_{8} - S_{4} \Rightarrow S_{8} = 6 S_{4}$

$\Rightarrow \frac{8}{2} [2 \times 3 + 7 \times d] = 6 \times \frac{4}{2} [2 \times 3 + 3 \times d]$ [ $∵ a = 3$ ]

$\Rightarrow 6 + 7 d = 18 + 9 d \Rightarrow 2 d = - 12 \Rightarrow d = - 6$

$∴ S_{20} = \frac{20}{2} [2 \times 3 + (19) \times - 6]$

= 10[– 108] = – 1080.

Q 20 :
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]

505
525
510
515

1

2

3

4

(4)

Given $S_{40} = 1030$ and $S_{12} = 57$

Let a be the first term and d be the common difference. We have,

$S_{40} = \frac{40}{2} [2 a + 39 d] = 100; S_{12} = \frac{12}{2} [2 a + 11 d] = 57$

On simplifying, we get

$2 a + 39 d = \frac{103}{2}$ ... (i)

$2 a + 11 d = \frac{19}{2}$ ... (ii)

Subtracting (ii) from (i), we get

$28 d = 42 \Rightarrow a = \frac{3}{2}$

Using equation (ii), we get

$2 a + 11 \times \frac{3}{2} = \frac{19}{2} \Rightarrow a = \frac{- 7}{2}$

Now, we have

$S_{30} - S_{10} = \frac{30}{2} [2 a + 29 d] - \frac{10}{2} [2 a + 9 d]$

. $= 20 a + 390 d = 20 \times (- \frac{7}{2}) + 390 \times \frac{3}{2} = 515$ .

Topic Question Set

Q 11 :
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]

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Q 12 :
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 13 :
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 14 :
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to20 terms, is equal to [2025]

Q 15 :
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 $(A \cup B)$ is [2025]

Q 17 :
Let $a_{n}$ be the $n^{th}$ term an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

Q 18 :
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

Q 19 :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

Q 20 :
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]

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Topic Question Set

Q 11 : 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]

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Q 12 : 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 13 : 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 14 : The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to20 terms, is equal to [2025]

Q 15 : 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 (A∪B) is [2025]

Q 17 : Let an be the nth term an A.P. If Sn=a1+a2+a3+...+an=700, a6=7 and S7=7, then an is equal to [2025]

Q 18 : Suppose that the number of terms in an A.P. is 2k, k∈N. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

Q 19 : If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

Q 20 : In an arithmetic progression, if S40=1030 and S12=57, then S30–S10 is equal to : [2025]

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Q 11 :
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 12 :
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 13 :
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 14 :
The sum 1 + 3 + 11 + 25 + 45 + 71 + ... up to20 terms, is equal to [2025]

Q 15 :
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 $(A \cup B)$ is [2025]

Q 17 :
Let $a_{n}$ be the $n^{th}$ term an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

Q 18 :
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

Q 19 :
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to [2025]

Q 20 :
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]