Question

The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.

Answer 1

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

a₁₃ = 4 × a₃    ...(Given)

⇒ a + 12d  = 4(a + 2d)   ...[a_n = a + (n – 1)d]

⇒ a + 12d = 4a + 8d

⇒ 3a = 4d   ...(1)

Also,

a₅ = 16   ...(Given) 

⇒ a + 4d = 16   ...(2)

Solving (1) and (2), we get

a + 3a = 16

⇒ 4a = 16 

⇒ a = 4

Putting a = 4 in (1), we get

4d = 3 × 4 = 12 

⇒ d = 3

Using the formula, `S_4 = n/2 [2a + (n - 1)d]`, we get

`S_10 = 10/2 [2 xx 4 + (10 - 1) xx 3]`

= 5 × (8 + 27)

= 5 × 35

= 175

Hence, the required sum is 175.