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Sequences and Series

JEE Main 2026 (Online) 24th January Morning ShiftMCQEasy📈 Topic in 57% papers+4−1
Ques. 1JEE Main 2026, 24 January (Morning Shift)

Consider an A.P.: a1, a2, , an; a1 > 0. If a2 - a1 = -34, an = 14a1, and i=1n ai = 5252, then i=117 ai is equal to

  1. 238
  2. 136
  3. 476
  4. 952

Answer

(1) 238

Solution

The common difference is d = a2 - a1 = - 34, and an = a1 + (n - 1)d = 14a1, so

34a1 = 34(n - 1) a1 = n - 1

The sum of n terms is i=1n ai = n2(a1 + an) = n2 54a1 = 5n(n - 1)8, hence

5n(n - 1)8 = 5252

n(n - 1) = 420 = 21 20

So n = 21 and a1 = 20, which gives

i=117 ai = 172 [2(20) + 16 (- 34 ) ] = 172(40 - 12)

i=117 ai = 17 14 = 238

Answer: (1)

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Ques. 2JEE Main 2026, 24 January (Evening Shift)

Let a1, a2, a3, a4 be an A.P. of four terms such that each term of the A.P. and its common difference l are integers. If a1 + a2 + a3 + a4 = 48 and a1a2a3a4 + l4 = 361, then the largest term of the A.P. is equal to

  1. 27
  2. 24
  3. 23
  4. 21
Ques. 3JEE Main 2026, 24 January (Evening Shift)

( 13 + 47 ) + ( 132 + 13 47 + 4272 ) + ( 133 + 132 47 + 13 4272 + 4373 ) + up to infinite terms, is equal to

  1. 74
  2. 43
  3. 65
  4. 52
Ques. 4JEE Main 2026, 23 January (Evening Shift)

Let k=1n ak = n2 + n. If a10 = 59 and a6 = 7a1, then + is equal to :

  1. 3
  2. 5
  3. 7
  4. 12
Ques. 5JEE Main 2026, 22 January (Morning Shift)

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is

  1. -26
  2. -20
  3. -24
  4. -22
Ques. 6JEE Main 2026, 22 January (Evening Shift)

Suppose a, b, c are in A.P. and a2, 2b2, c2 are in G.P. If a < b < c and a + b + c = 1, then 9(a2 + b2 + c2) is equal to .

Ques. 7JEE Main 2026, 21 January (Morning Shift)

Let a1, a2, a3, be a G.P. of increasing positive terms such that a2 a3 a4 = 64 and a1 + a3 + a5 = 8137. Then a3 + a5 + a7 is equal to :

  1. 3256
  2. 3252
  3. 3248
  4. 3244
Ques. 8JEE Main 2026, 21 January (Morning Shift)

Let a1 = 1 and for n 1, an+1 = 12an + n2 - 2n - 1n2(n + 1)2. Then | n=1 (an - 2n2 ) | is equal to .

Ques. 9JEE Main 2026, 21 January (Evening Shift)

The positive integer n, for which the solutions of the equation x(x + 2) + (x + 2)(x + 4) + + (x + 2n - 2)(x + 2n) = 8n3 are two consecutive even integers, is :

  1. 3
  2. 6
  3. 9
  4. 12
Ques. 10JEE Main 2026, 21 January (Evening Shift)

Let a1, a22, a322, , a1029 be a G.P. of common ratio 1 2. If a1 + a2 + + a10 = 62, then a1 is equal to:

  1. 2 - 1
  2. 2( 2 - 1)
  3. 2 - 2
  4. 2(2 - 2)
Ques. 11JEE Main 2025, 8 April (Evening Shift)

If 114 + 124 + 134 + = 490, 114 + 134 + 154 + = , 124 + 144 + 164 + = , then is equal to :

  1. 23
  2. 14
  3. 18
  4. 15

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