How many terms are there in the A.P.

(i) 7, 10, 13...43

Here, we have:

a  = 7

$d=(10-7)=3$

$a_{n}=43$

Let there be n terms in the given A.P.

Also, $a_{n}=a+(n-1) d$

$\Rightarrow 43=7+(n-1) 3$

$\Rightarrow 36=(n-1) 3$

$\Rightarrow 12=(n-1)$

$\Rightarrow 13=n$

Thus, there are 13 terms in the given A.P.

(ii) $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2}, \ldots, \frac{10}{3}$

Here, we have:

$a=-1$

$d=\left(\frac{-5}{6}-(-1)\right)=\left(1-\frac{5}{6}\right)=\frac{1}{6}$

$a_{n}=\frac{10}{3}$

Let there be n terms in the given A.P.

Also, $a_{n}=a+(n-1) d$

$\Rightarrow \frac{10}{3}=-1+(n-1) \frac{1}{6}$

$\Rightarrow \frac{13}{3}=(n-1) \frac{1}{6}$

$\Rightarrow 26=(n-1)$

$\Rightarrow 27=n$

Thus, there are 27 terms in the given A.P.