Question

In an A.P., if $t_7=13$ and $S_{14}=203$, find $S_8$.

Answer 1

$\because t_n=a+(n-1)d$

$\Rightarrow t_7=13\Rightarrow a+(7-1)d=13$

$\Rightarrow a+6d=\mathrm{13.......}\left(1\right)$

$\because S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{14}=203$

$\frac{14}{2}[2a+(14-1\left)d\right]=203$

$7(2a+13d)=203$

$2a+13d=\frac{203}{7}$

$2a+13d=\mathrm{29......}\left(2\right)$

Multiplying equation $\left(1\right)$, subtracting from equation $\left(2\right)$

$2a+13d=29$

$\underset{–}{-2a-12d=-26}$

$d=3$

Putting the value of $d$ in equation $\left(1\right),a+6\times 3=13$

$a=13-18=-5$

Thus, $S_8=\frac{8}{2}[2\times (-5)+(8-1)\times 3]$

$=4[-10+7\times 3]$

$=4[-10+21]$

$=4\times 11$

$=44$