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 a 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.