Determine the nth term of the AP whose 7th term is −1 and 16th term is 17.

Let a be the first term and d be the common difference of the AP. Then,

$a_{7}=-1$

$\Rightarrow a+(7-1) d=-1 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+6 d=-1 \quad \ldots$ (1)

Also,

$a_{16}=17$

$\Rightarrow a+15 d=17 \quad \ldots(2)$

From (1) and (2), we get

$-1-6 d+15 d=17$

$\Rightarrow 9 d=17+1=18$

$\Rightarrow d=2$

Putting d = 2 in (1), we get

$a+6 \times 2=-1$

$\Rightarrow a=-1-12=-13$

$\therefore a_{n}=a+(n-1) d$

$=-13+(n-1) \times 2$

$=2 n-15$

Hence, the nth term of the AP is (2n − 15).