Question

The $4^{th},{42}^{nd}$, and last terms of an A.P. are $0,-95$ and $-125$ respectively; find the first term and the number of terms.

Answer 1

$a_4=0,a_{42}=-95$ and $l=a_n=-125$

$a_n=a+(n-1)d\Rightarrow a_4=a+3d\Rightarrow 0=a+3d\Rightarrow a+3d=\mathrm{0......}\left(i\right)a_{42}=a+41d\Rightarrow -125=a+41d\Rightarrow a+41d=-\mathrm{95.....}\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow d=-\frac{5}{2},a=\frac{15}{2}l=a+(n-1)d-125=\frac{15}{2}+(n-1)(-\frac{5}{2})-\frac{265}{2}=(n-1)(-\frac{5}{2})$

So the first term is $\frac{15}{2}$ and number of terms are $54$