Question

If $S_1,S_2,S_3$ are the sum of the $n,2n,3n$ terms respectively of an AP, then

• A. $S_3=3(S_2-S_1)$
• B. $S_3=S_2+S_1$
• C. $S_3=2(S_2+S_1)$
• D. $S_3=2S_2-S_1$

Answer 1

The correct option is A $S_3=3(S_2-S_1)$

Since, $\text{sum of n terms}=S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Given, $S_1,S_2,S_3$ are the sum of $n,2n,3n$ terms of an AP.
$\therefore S_1=\frac{n}{2}[2a+(n-1\left)d\right]$ ......(i)
$S_2=\frac{2n}{2}[2a+(2n-1\left)d\right]$ ......(ii)
$S_3=\frac{3n}{2}[2a+(3n-1\left)d\right]$ .......(iii)
Now, $S_2-S_1=\frac{2n}{2}[2a+(2n-1\left)d\right]-\frac{n}{2}[2a+(n-1\left)d\right]$

$S_2-S_1=\frac{n}{2}[4a+2(2n-1)d-(2a+(n-1)d]$

$S_2-S_1=\frac{n}{2}[4a+4nd-2d-(2a+nd-d]$

$S_2-S_1=\frac{n}{2}[4a+4nd-2d-2a-nd+d]$

$S_2-S_1=\frac{n}{2}[2a+3nd-d]$

$S_2-S_1=\frac{n}{2}[2a+(3n-1\left)d\right]$
By multiplying by $3$ on both sides we get,
$3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1\left)d\right]$
$3(S_2-S_1)=S_3$
Option A is correct.