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Sequences and Series · Question 24

JEE Main 2026 (Online) 24th January Evening ShiftMCQMedium📈 Topic in 55% papers+4−1
1

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

Answer

(1) 27

Solution

a1 + a4 = a2 + a3 = 24 and a3 - a2 = l ⇒ a2 = 12 - l2

Since a2 is an integer, l is even. Let l = 2k (k integer). The terms are

12 - 3k, 12 - k, 12 + k, 12 + 3k

⇒ (144 - 9k2)(144 - k2) + 16k4 = 361

⇒ 20736 - 1440k2 + 25k4 = 361

⇒ 5k4 - 288k2 + 4075 = 0

⇒ k2 = 288 3810 = 25 or 32.6

Since k is an integer, k2 = 25 ⇒ k = 5

∴ Terms are -3, 7, 17, 27 (or in reverse order); largest term = 27

Answer: (1)

JEE Main 2026 (Online) 24th January Evening Shift

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2

( 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
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3

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

  1. 3
  2. 5
  3. 7
  4. 12
JEE Main 2026 (Online) 23rd January Evening Shift
4

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
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5

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 .

JEE Main 2026 (Online) 22nd January Evening Shift
6

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  1. 3256
  2. 3252
  3. 3248
  4. 3244
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7

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JEE Main 2026 (Online) 21st January Morning Shift
8

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
JEE Main 2026 (Online) 21st January Evening Shift
9

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)
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10

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  1. 23
  2. 14
  3. 18
  4. 15
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11

Let x1, x2, x3, x4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1, x2, x3, x4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1 x2 x3 x4) is:

  1. 18
  2. 216
  3. 36
  4. 72
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