Question

The 17^th term of an A.P. is 5 more than twice its 8^th term. If the 11^th term of the A.P. is 43. find the n^th term.

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,

a₁₇ = 5 + 2a₈
⇒ a + (17 − 1)d = 5 + 2(a + (8 − 1)d)
⇒ a + 16d = 5 + 2a + 14d
⇒ 16d − 14d = 5 + 2a − a
⇒ 2d = 5 + a
⇒ a = 2d − 5 .... (1)

Also, a₁₁ = 43
⇒ a + (11 − 1)d = 43
⇒ a + 10d = 43 ....(2)

On substituting the values of (1) in (2), we get
2d − 5 + 10d = 43
⇒ 12d = 5 + 43
⇒ 12d = 48
⇒ d = 4
⇒ a = 2 × 4 − 5 [From (1)]
⇒ a = 3

∴ a_n = a + (n − 1)d
= 3 + (n − 1)4
= 3 + 4n − 4
= 4n − 1

Thus, the n^th term of the given A.P. is 4n − 1.