371.

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 a and d respectively be the first term and the common difference of the A.P.

we know that the n^th term of an AP is given by  a_n = a + (n-1) d

According to the given information,

a₁₆ = 1 + 2a₈

$\Rightarrow$  a + (1 6- 1) d = 1 + 2[a + (8 - 1) d]

$\Rightarrow a\; +\; 15d\; =\; 1\; +\; 2a\; +\; 14d$

$\Rightarrow \; -a\; +\; d\; =\; 1 \; \; \; \; \; \; \; \; \; \; ........(1)$

Also, it is given that, a₁₂ = 47

$\Rightarrow \; a\; +\; (\; 12\; -\; 1\; )\; d\; =\; 47$

$\Rightarrow \; a\; +\; 11d\; =\; 47 \; \; \; \; \; \; \; \; .......(2)$

Adding  (1) and (2), we have:

12d = 48

$\Rightarrow \; d\; =\; 4$

From (1),

-a + 4 = 1  

$\Rightarrow \; a\; =\; 3$

Hence,  a_n = a + (n - 1) d

                = 3 + (n - 1) (4)

                = 3 + 4n - 4

                = 4n - 1  

Hence, the n^th term of the AP is  4n - 1.