Question

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.

Answer 1

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

a₁₆ = 5 × a₃    ...(Given)

⇒ a + 15d = 5(a + 2d)   ...[a_n = a + (n – 1)d] 

⇒ a + 15d = 5a + 10d

⇒ 4a = 5d

Also,

a₁₀ = 41   ...(Given) 

⇒ a + 9d – 41   ...(2)

Solving (1) and (2), we get

`a + 9 xx (4a)/5 = 41`

⇒ `(5a + 36a)/5 = 41`

⇒ `(41a)/5 = 41`

⇒ a = 5 

Putting a = 5 in (1), we get 

5d = 4 × 5 = 20

⇒ d = 4

Using the formula, `S_n = n/2 [2a + (n - 1)d]`, we get

`S_15 = 15/2 [2 xx 5 + (15 - 1) xx 4]`

= `15/2 xx (10 + 56)`

= `15/2 xx 66`

= 495

Hence, the required sum is 495.