Question

The sum of $4^{th}$ and $8^{th}$ terms of an A.P. is $24$ and the sum of the $6^{th}$ and ${10}^{th}$ terms is $34$. Find the first four terms of the A.P.

Answer 1

Sum of $4^{th}$ and $8^{th}$ terms$=24$

As the term are in $AP$ and $r^{th}$ term $=a+(r-1)d$

$a=1^{st}$ term $r=$term and $d=$difference.

$=a+(4-1)d+(8-1)d=242a+10d=24a+5d=12-\left(i\right)$

Sum of $6^{th}$ and ${10}^{th}$ term $=34$

$a+(6-1)d+a+(10-1)d=34=2a+14d=34a+7d=17-\left(ii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$

$2d=5d=\frac{5}{2}So,a+5d=12a+\frac{25}{2}=12a=\frac{-1}{2}$

Sum of first $4$ terms

$=\frac{n}{2}(2a+(n-1\left)d\right)where,n=2=\frac{4}{2}\left(2\right(\frac{-1}{2})+(4-1\left)\frac{5}{2}\right)=2(-1+\frac{15}{2})=\frac{2\left(13\right)}{2}=13$