The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.

Question

Philoid · Accepted Answer

Let a be the first term and d be the common difference.

Given: a₁₀ = 41

a₁₆ = 5 a₃

Now, Consider a₁₀ = 41

⇒ a + (10 - 1)d = 41

⇒ a + 9d = 41 ………………….(1)

Consider a₁₆ = 5 a₃

⇒ a + 15d = 5(a + 2d)

⇒ a + 15d = 5a + 10d

⇒ 4a - 5d = 0 ………………….(2)

Now, subtracting equation (2) from 4 times of equation (1), we get,

41d = 164

⇒ d = 4

∴ from equation (2), we get,

4a= 5d

⇒ 4a = 20

⇒ a = 5

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 15 terms is given by:

S₁₅ = [2(5) + (15 - 1)(4)]

= (15/2) × [10 + 56]

= 15 × 33

= 495

∴ S₁₅ = 495