Question

Question 28
If the 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

Let a and d be the first term and the common difference of an AP respectively.

Sum of n terms of an AP,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$Given,S_6=36$
$\Rightarrow \frac{6}{2}[2a+(6-1\left)d\right]=36$
$\Rightarrow 2a+5d=12...\left(i\right)$

$Also,S_{16}=256$
$\Rightarrow \frac{16}{2}[2a+(16-1\left)\right]=256$
$\Rightarrow 2a+15d=32...\left(ii\right)$

On subtracting eq.(i) from eq.(ii), we get;
$10d=20$
$\Rightarrow d=\mathrm{2...}\left(iii\right)$
From eq.(ii),
$2a+5\left(2\right)=12$
$\Rightarrow 2a=12–10=2$
$\Rightarrow a=1$
$\therefore S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]$
$=5\left[2\right(1)+9(2\left)\right]=5(2+18)$
$=5\times 20=100$