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Sequences and Series

JEE Main 2025 (Online) 2nd April Evening ShiftMCQMedium📈 Topic in 57% papers+4−1
Ques. 1JEE Main 2025, 2 April (Evening Shift)

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 :

  1. 6
  2. 4
  3. 8
  4. 10

Answer

(2) 4

Solution

Let the A.P. have 2n terms, first term a and common difference d.

Each even term exceeds the odd term just before it by d, and there are n such pairs, so

Seven - Sodd = nd

nd = 30 - 24 = 6 ...(i)

Also last term − first term = (2n - 1)d = 212

2nd - d = 212

12 - d = 212 d = 32

From (i), n = 4, so the A.P. has 8 terms whose sum is 24 + 30 = 54:

82 [2a + 7 32 ] = 54

2a + 212 = 272 a = 32

∴ Terms: 32, 3, 92, 6, 152, 9, 212, 12

The integer terms are 3, 6, 9, 12, i.e. 4 terms.

Answer: (2)

Keep practising — Sequences and Series

Ques. 2JEE Main 2025, 2 April (Evening Shift)

If the sum of the first 10 terms of the series 4 11 + 4 14 + 4 21 + 4 24 + 4 31 + 4 34 + is mn, where (m, n) = 1, then m + n is equal to .

Ques. 3JEE Main 2025, 29 January (Morning Shift)

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

  1. 108
  2. 90
  3. 122
  4. 84
Ques. 4JEE Main 2025, 29 January (Morning Shift)

The value of n ( k=1n k3 + 6k2 + 11k + 5(k+3)! ) is :

  1. 5/3
  2. 2
  3. 4/3
  4. 7/3
Ques. 5JEE Main 2025, 29 January (Evening Shift)

Let a1,a2, ,a2024 be an Arithmetic Progression such that a1+(a5+a10+a15+ +a2020)+a2024=2233. Then a1+a2+a3+ +a2024 is equal to .

Ques. 6JEE Main 2025, 28 January (Morning Shift)

Let an be a sequence such that a0 = 0, a1 = 12 and 2an+2 = 5an+1 - 3an, n = 0, 1, 2, 3, . Then k=1100 ak is equal to

  1. 3a100 + 100
  2. 3a100 - 100
  3. 3a99 - 100
  4. 3a99 + 100
Ques. 7JEE Main 2025, 28 January (Morning Shift)

Let Tr be the rth term of an A.P. If for some m, Tm = 125, T25 = 120, and 20 r=125Tr = 13, then 5m r=m2mTr is equal to

  1. 98
  2. 126
  3. 112
  4. 142
Ques. 8JEE Main 2025, 28 January (Evening Shift)

For positive integers n, if 4an = (n2 + 5n + 6 ) and Sn = k=1n ( 1ak ), then the value of 507S2025 is :

  1. 540
  2. 675
  3. 1350
  4. 135
Ques. 9JEE Main 2025, 28 January (Evening Shift)

The interior angles of a polygon with n sides are in an A.P. with common difference 6 . If the largest interior angle of the polygon is 219 , then n is equal to .

Ques. 10JEE Main 2025, 24 January (Morning Shift)

Let Sn= 12+ 16+ 112+ 120+ upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is 2026 S2025, then the absolute difference between 20th and 15th terms of the A.P. is

  1. 20
  2. 45
  3. 90
  4. 25
Ques. 11JEE Main 2025, 24 January (Evening Shift)

In an arithmetic progression, if S40=1030 and S12=57, then S30-S10 is equal to :

  1. 525
  2. 505
  3. 510
  4. 515

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