Question

The $2nd,31st$, and last term of an a.p is $\frac{31}{4},\frac{1}{2}and-\frac{13}{2}$ find the first term and the number of term

Answer 1

$2^{nd}$ term of $A.P=\frac{31}{4}$

$\Rightarrow a+d=\frac{31}{4}$

$4a+4d=31\to ----\left(1\right)$

${31}^{st}$ term of $A.P=\frac{1}{2}$

$\Rightarrow a+30d=\frac{1}{2}$

$2a+60d=1\to ----\left(2\right)$

Simultaneous equation $\to \left(1\right)$ and $\left(2\right)$

$\begin{array}{c}2(4a+4d=31)\\ 4(2a+60=1)\end{array}=\frac{8a+8d=628a+240d=4---}{-232d=58d=\frac{58}{-232}=-\frac{1}{4}}$

$4a+4(-\frac{1}{4})=31$

$4a=32$

$a=8\Rightarrow$ First term $=8$

last term $=\frac{-13}{2}$

$\Rightarrow a+(n-1)d=\frac{-13}{2}$

$8+(n-1)-\frac{1}{4}=\frac{-13}{2}$

$8-\frac{n}{4}+\frac{1}{4}=\frac{-13}{2}$

$\frac{n}{4}=8+\frac{1}{4}+\frac{13}{2}\Rightarrow n=\frac{59}{4}\times 4=59$

Therefore number of terms $=59$