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Sequences and Series

JEE Main 2025 (Online) 28th January Evening ShiftNumericalEasy📈 Topic in 58% papers+4
Ques. 1JEE 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 .

Answer

20

Solution

Taking the angles from the largest, they form an A.P. with first term 219 and common difference -6 .

Sum = n2[2(219) - 6(n - 1)] = n(222 - 3n)

Sum of interior angles = (n - 2) 180

222n - 3n2 = 180n - 360

3n2 - 42n - 360 = 0

n2 - 14n - 120 = 0

(n - 20)(n + 6) = 0 n = 20

(Check: smallest angle = 219 - 19 6 = 105 > 0.)

Answer: 20

Keep practising — Sequences and Series

Ques. 2JEE 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. 3JEE 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
Ques. 4JEE Main 2025, 24 January (Evening Shift)

If 7=5+ 17(5+ )+ 172(5+2 )+ 173(5+3 )+ , then the value of is :

  1. 17
  2. 1
  3. 67
  4. 6
Ques. 5JEE Main 2025, 23 January (Morning Shift)

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

  1. -120
  2. -1200
  3. -1080
  4. -1020
Ques. 6JEE Main 2025, 23 January (Evening Shift)

The roots of the quadratic equation 3x2-px+q=0 are 10th and 11th terms of an arithmetic progression with common difference 32. If the sum of the first 11 terms of this arithmetic progression is 88, then q-2p is equal to .

Ques. 7JEE Main 2025, 22 January (Morning Shift)

Let a1,a2,a3, be a G.P. of increasing positive terms. If a1a5=28 and a2+a4=29, then a6 is equal to:

  1. 812
  2. 784
  3. 628
  4. 526
Ques. 8JEE Main 2025, 22 January (Morning Shift)

If r=1nTr= (2n-1)(2n+1)(2n+3)(2n+5)64, then n r=1n ( 1Tr ) is equal to :

  1. 23
  2. 13
  3. 1
  4. 0
Ques. 9JEE Main 2025, 22 January (Evening Shift)

Suppose that the number of terms in an A.P. is 2k, k . 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:

  1. 8
  2. 6
  3. 4
  4. 5

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