Question

If the sum of first $13$ terms of an $AP$ is $21$ and the sum of First $21$ terms is $13$, Show that sum of First $34$ terms is $-34$.

Answer 1

As we know that, the sum of $n$ terms of an A.P. is given as

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

Therefore,

$S_{13}=\frac{13}{2}[2a+(13-1)d]$

$\Rightarrow 21=13a+78d.....\left(1\right)$

$S_{21}=\frac{21}{2}[2a+(21-1)d]$

$\Rightarrow 13=21a+210d.....\left(2\right)$

On solving $\left(1\right)\&\left(2\right)$, we get

$a=\frac{283}{91}$ and $d=\frac{-68}{273}$

Therefore,

$S_{34}=\frac{34}{2}[2\times \frac{283}{91}+(34-1)\left(\frac{-68}{273}\right)]$

$\Rightarrow S_{34}=17(\frac{566}{91}-\frac{748}{91})$

$\Rightarrow S_{34}=17\left(\frac{566-748}{91}\right)$

$\Rightarrow S_{34}=17\times (-\frac{182}{91})=-34$

Hence proved.