The 16^th term of an AP is 1 more than twice its 8^th term. If the 12^th term of the AP is 47, then find its n^th term.

Let 16^th term of the AP series be a₁₆, 8^th term be a₈, 12^th term be a₁₂ and n^thterm be a_n.

According to the question,

16^th term of an AP is 1 more than twice its 8^th term.

a₁₆ = 1 + 2a₈ …(i)

and the a₁₂ = 47 [∵, 12^th term is 47] …(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.

∴, a₁₆ = a + (16 – 1)d = a + 15d. Similarly, a₈ = a + 7d and a₁₂ = a + 11d

From equation (i),

(a + 15d) = 1 + 2(a + 7d)

⇒ a + 15d = 1 + 2a + 14d

⇒ 15d – 14d = 2a – a + 1

⇒ d = a + 1

⇒ d – a = 1 …(iii)

From equation (ii),

(a + 11d) = 47

11d + a = 47 …(iv)

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

(d – a) + (11d + a) = 1 + 47

⇒ 12d = 48

⇒ d = 48/12 = 4

⇒ d = 4

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

4 – a = 1

⇒ a = 4 – 1

⇒ a = 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 .