Question

If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, then find the sum of first 10 terms.

Answer 1

We have the sum of first $n$ terms of an AP,

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

Given,

$36=\frac{6}{2}[2a+(6-1\left)d\right]$

$12=2a+5d$ ---------(1)

$256=\frac{16}{2}[2a+(16-1\left)d\right]$

$32=2a+15d$ ---------(2)

Subtracting, (1) from (2)

$32-12=2a+15d-(2a+5d)$

$20=10d⟹d=2$

Substituting for $d$ in (1),

$12=2a+5\left(2\right)=2(a+5)$

$6=a+5⟹a=1$

$\therefore$ The sum of first $10$ terms of an AP,

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

$S_{10}=5[2+18]$

$S_{10}=100$

This is the sum of the first $10$ terms