Question

If sum of first $6$ terms of an $A.P.$ is $36$ and that of the first $16$ terms is $256$, find the sum of first $10$ terms.

Answer 1

Let $S_n$ be the sum of n terms of an AP with the first term and the common difference be d. Then

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

It is given that sum of first $6$ terms is $36$ and the sum of the first $16$ terms is $256$.

$\Rightarrow S_6=36$ and $S_{16}=256$

putting $n=6$ in $S_n$, we get

$S_6=\frac{6}{2}(2a+5d)$

$\Rightarrow 36=3(2a+5d)$

$\Rightarrow 12=2a+5d$ ...($1$)

putting $n=16$ in $S_n$, we get,

$S_{16}=\frac{16}{2}(2a+15d)$

$\Rightarrow 256=8(2a+15d)$

$\Rightarrow 32=2a+15d$ ...($2$)

on subtracting $1$ from $2$ we get
$20=10d$

$\Rightarrow d=2$
putting d in 1, we get

$12=2a+10$

$\Rightarrow a=1$
again putting $a=1,d=2andn=10$ in $S_n$, we get

$S_{10}=\frac{10}{2}(2\times 1+9\times 2)$

$\Rightarrow S_{10}=100$