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

Let a be the first term and d be the common difference of the given A.P.
According to the given question, 16th term of the AP = 2 × 8th term of the AP + 1
i.e., a16 = 2a8 + 1
We know that an = a + (n-1)d
a + (16-1)d = 2[a + (8-1)d] + 1
⇒ a + 15d = 2[a + 7d] + 1
⇒ a + 15d = 2a + 14d + 1
⇒ d = a + 1 — (1)
Also, 12th term, a12 = 47
⇒ a + (12-1)d = 47
⇒ a + 11d = 47
⇒ a + 11(a + 1) = 47
⇒ a + 11a + 11 = 47
⇒ 12a = 36
⇒ a = 3
On Putting the value of a in (1), we get
d = 3 + 1 = 4
Thus, nth term of the AP, an = a + (n-1)d
On putting the respective values of a and d, we get
an = 3 + (n-1)4 = 3 + 4n - 4 = 4n - 1
Hence, nth term of the given AP is 4n - 1.