How many terms are there in the A.P.

Solution:

We have:

$a=-14$ and $S_{n}=40 \ldots$ (i)

$a_{5}=2$

$\Rightarrow a+(5-1) d=2$

$\Rightarrow-14+4 d=2$

$\Rightarrow 4 d=16$

$\Rightarrow d=4$    ...(ii)

Also, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 40=\frac{n}{2}[2(-14)+(n-1) \times 4] \quad($ From $(\mathrm{i})$ and $(\mathrm{ii}))$

$\Rightarrow 80=n[-28+4 n-4]$

$\Rightarrow 80=4 n^{2}-32 n$

$\Rightarrow n^{2}-8 n-20=0$

$\Rightarrow(n-10)(n+2)=0$

$\Rightarrow n=10,-2$

But, $\mathrm{n}$ cannot be negative.

$\therefore \mathrm{n}=10$