Question

If the 5^th term of an A.P. is 31 and 25^th term is 140 more than the 5^th term, find the A.P.

Answer 1

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

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₅ = 31
⇒ a + (5 − 1)d = 31
⇒ a + 4d = 31
⇒ a = 31 − 4d .... (1)

Also, a₂₅ = 140 + a₅
⇒ a + (25 − 1)d = 140 + 31
⇒ a + 24d = 171 .... (2)

On substituting the values of (1) in (2), we get
31 − 4d + 24d = 171
⇒ 20d = 171 − 31
⇒ 20d = 140
⇒ d = 7
⇒ a = 31 − 4 × 7 [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 10, 17, 24, .... .