The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.

Question

Philoid · Accepted Answer

Let a be first term and d be common difference of AP.

As we know nth term of an AP is

a_n = a + (n - 1)d

Given

a₁₆ = 5a₃

a + 15d = 5(a + 2d)

a + 15d = 5a + 10d

4a - 5d = 0

4a = 5d …[1]

Also given

a₁₀ = a + 9d

41 = a + 9d

Multiplying both side by 4

164 = 4a + 36d

164 = 5d + 36d [By 1 ]

41d = 164

d = 4

Also,

4a = 5d

4a = 5(4)

a = 5

As we know sum of first n terms of an AP is

S_{15 =} 15(33) = 495

The 16^th term of an AP is five times its third term. If its 10^th term is 41, then find the sum of its first fifteen terms.