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

Let 7^th term of the AP series be a₁₇, 8^th term be a₈, 11^th term be a₁₁ and n^thterm be a_n.

According to the question,

17^th term of an AP is 5 more than twice its 8^th term.

a₁₇ = 5 + 2a₈ …(i)

and the a₁₁=43 [∵, 11^th term is 43] …(ii)

Now a_n is given by a_n = a + (n – 1)d, where a is first term of the AP, n is total number in the series and d is common difference between adjacent numbers in the series.

∴, a₁₇ = a + (17 – 1)d = a + 16d. Similarly, a₈ = a + 7d and a₁₁ = a + 10d

From equation (i),

(a + 16d) = 5 + 2(a + 7d)

⇒ a + 16d = 5 + 2a + 14d

⇒ 16d – 14d = 2a – a + 5

⇒ 2d = a + 5

⇒ 2d – a = 5 …(iii)

From equation (ii),

(a + 10d) = 43

10d + a = 43 …(iv)

Adding equations (iii) and (iv), we get

(2d – a) + (10d + a) = 5 + 43

⇒ 12d = 48

⇒ d = 48/12 = 4

⇒ d = 4

Putting d = 4 in equation (iii), we get

2d – a = 5

⇒ a = 2d – 5

⇒ a = 2(4) – 5 = 8 – 5 = 3

Substituting a = 3 and d= 4 in a_n = a + (n – 1)d,

a_n = 3 + (n – 1)(4) = 3 + 4n – 4 = 4n – 1

Thus, we have got n^th term, that is, 4n – 1.