Question

If the ${12}^{th}$ term of an AP is -13 and the sum of first four terms is 24, what is the sum the first 10 terms ?

Answer 1

$T_{12}=-13$

$S_4=24$

$S_{10}=?$

$a+11d=-13$

$\Rightarrow a=-13-11d$

$\frac{4}{2}[2a+3d]=24$

$\Rightarrow 2a+3d=12$ ………..$\left(1\right)$

Substituting the value of a in equation $\left(1\right)$

$-26-22d+3d=12$

$\Rightarrow -19d=38$

$\Rightarrow d=-2$

$a=-13+22=9$

$S_{10}=\frac{10}{2}\left[2\right(9)+9(-2\left)\right]=0$.