Question

If $S_n$ denotes the sum of $n$ terms of an arithmetic progression, then the value of $\left(S_{2n}-S_n\right)$ is equal to

• A. $2S_n$
• B. $S_{3n}$
• C. $\frac{S_{3n}}{3}$
• D. $\frac{S_n}{2}$

Answer 1

The correct option is C

$\frac{S_{3n}}{3}$

Explanation for the correct option

Step $1$: The sum of $n$ terms of an arithmetic progression,

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

Where, number of terms $n$, first term $a$, and common differences $d$

Then,

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

Step 2: Subtract equation $\left(1\right)$ from equation $\left(2\right)$,

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

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

$\left(S_{2n}-S_n\right)=\left[\left(2an-an\right)+d\left(2n^2-n-\frac{n^2}{2}+\frac{n}{2}\right)\right]$

$\left(S_{2n}-S_n\right)=\left[\left(an\right)+d\left(\frac{3n^2}{2}-\frac{n}{2}\right)\right]$

$\left(S_{2n}-S_n\right)=\left[\left(an\right)+\frac{n}{2}\left(3n-1\right)d\right]$

$\left(S_{2n}-S_n\right)=\left[\frac{2an+n\left(3n-1\right)d}{2}\right]$

$\left(S_{2n}-S_n\right)=\frac{n}{2}\left[2a+\left(3n-1\right)d\right]$

Step 3: Now by multiplying both sides by $3$, we get,

$3\left(S_{2n}-S_n\right)=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]$

$3\left(S_{2n}-S_n\right)=S_{3n}$

$\left(S_{2n}-S_n\right)=\frac{S_{3n}}{3}$

Hence, the correct option is ‘C’.