Question

The sum of $4^{th}$ and $8^{th}$ form of an $AP$ is $24$ and the sum of its $6^{th}$ and ${10}^{th}$ terms is $44$. Find the sum of first ${10}^{th}terms$ of $AP$.

Answer 1

$T_n=a+(n-1)d$

$T_4=a+(4-1)d=a+3d$

Similarly

$T_8=a+7d,T_6=a+5d,T_{10}=a+9d$

Given,

$a+3d+a+7d=24$

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$...(1)

$a+5d+a+9d=44$

$\Rightarrow 2a+14d=44$

$\Rightarrow a+7d=22$...(2)

(2) - (1) gives

$2d=10\Rightarrow d=5$

From (2)

$a=22-7\left(5\right)=-13$

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

$S_{10}=\frac{10}{2}\left[2\right(-13)+(10-1\left)5\right]$

$=5[-26+45]$

$=95$