Question

The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.

Answer 1

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

a₁₉ = 3a₆   ...(Given)

⇒ a + 18d = 3(a + 5d)   ...[a_n = a + (n – 1)d]

⇒ a + 18d = 3a + 15d

⇒ 3a – a = 18d – 15d

⇒ 2a = 3d   ...(1)

Also,

a₉ = 19   ...(Given)

⇒ a + 8d = 19   ...(2)

From (1) and (2), we get

`(3d)/2 + 8d = 19`

⇒ `(3d + 16d)/2 = 19`

⇒ 19d = 38

⇒ d = 2

Putting d = 2 in (1), we get

2a = 3 × 2 = 6

⇒ a = 3

So,

a₂ = a + d = 3 + 2 = 5

a₃ = a + 2d = 3 + 2 × 2 = 7

Hence, the AP is 3, 5, 7, 9,........