Question

The ${10}^{th}$ terms of AP is $-37$ and the sum of its first six terms is $-27.$ find the sum of its first $8$ terms.

Answer 1

Let $a$ be the first term and $d$ be the common difference.

Now, $a_{10}=-37$

$\Rightarrow a+9d=-37\dots \dots \left(i\right)$

We know that sum of first n terms of AP is given by

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

Now, $S_6=\frac{6}{2}[2a+(6-1\left)d\right]$

$\Rightarrow -27=32a+5d$

$\Rightarrow 2a+5d=-9\dots \dots \left(ii\right)$

Multiply eq. (i) by $2,$ we get

$2a+18d=-74\dots \dots \left(iii\right)$

Subtracting eq.(ii) from eq.(iii), we get

$13d=-65$

$\Rightarrow d=-5$

Now from eq.(i), we get

$a+9\times (-5)=-37$

$\Rightarrow a=8$

$\therefore S_8=\frac{8}{2}[2\times 8+(8-1\left)\right(-5\left)\right]=-76$