Question

The $2^{nd},{31}^{st}$, and last terms of an A.P. are $7\frac{3}{4},\frac{1}{2}$ and $-6\frac{1}{2}$ respectively; find the first term and the number of terms.

Answer 1

$a_n=a+(n-1)d{a}_2=a+(2-1)d7\frac{3}{4}=a+da+d=\frac{31}{4}......\left(i\right)a_{31}=a+(31-1)da+30d=\frac{1}{2}.....\left(ii\right)$

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

$\Rightarrow a=\frac{15}{2},d=\frac{1}{4}$