Question

The 16^th term of an AP is 5 times its 3^rd term. If its 10^th 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 + 15 d = 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.