Find the number of terms in each of the following AP's :

Solution:

(i) $\mathrm{a}=7, \mathrm{~d}=6$,

$t_{n}=205$

$\Rightarrow a+(n-1) d=205$

$\Rightarrow 7+(n-1) \times 6=205 \quad \Rightarrow 6 n+1=205$

$\Rightarrow 6 n=204 \quad \Rightarrow n=34$

Hence, 34 terms

(ii) $a=18$

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=15 \frac{1}{2}-18$

$\mathrm{d}=\frac{31-36}{2}=-\frac{5}{2}$

Let there are n terms in this A.P.

Therefore, $a_{n}=-47$ and we know that

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

$-47=18+(n-1)\left(-\frac{5}{2}\right)$

$-47=18+(n-1)\left(-\frac{5}{2}\right)$

$-65=(n-1)\left(-\frac{5}{2}\right)$

$(n-1)=\frac{-130}{-5}$

$(n-1)=26$

$\mathrm{n}=27$

Therefore, this given A.P. has 27 terms in it.