Question

If $S_1,S_2,\text{and}{S}_3$ are the sum of $n,2n,\text{and}3n$ terms respectively of an arithmetic progression, then

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

Answer 1

The correct option is C

$S_3=3\left(S_2-S_1\right)$

Finding the relations between terms:

Step 1:Solve using sum of nth term formula of an AP

We know that formula to find the sum of an A.P is

${\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left(2a+\left(n-1\right)d\right)$, where $S_n$ is the sum of the $n$ terms, $a$ is the initial term, and $d$ is a common difference.

Then $S_1$ with $n$ terms will be

$S_1=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\dots \left(1\right)$

For $2n$ terms, the sum will be,

$S_2=\frac{2n}{2}\left(2a+\left(2n-1\right)d\right)\dots \left(2\right)$

For $3n$ terms, the sum will be,

$S_3=\frac{3n}{2}\left(2a+\left(3n-1\right)d\right)\dots \left(3\right)$

Step 2: Solve the equations to find relationship between the above three sums

Subtract $\left(1\right)\text{from}\left(2\right)$, then we get

$\begin{array}{rcl}{S}_2-S_1& =& \frac{2n}{2}\left(2a+\left(2n-1\right)d\right)-\frac{n}{2}\left(2a+\left(n-1\right)d\right)\\ & =& \frac{n}{2}\left(4a+4nd-2d-2a-nd+d\right)\\ & =& \frac{n}{2}\left(2a+3nd-d\right)\\ & =& \frac{n}{2}\left(2a+\left(3n-1\right)d\right)\end{array}$

Now, if we multiply both the sides $3$, we have

$3\left(S_2-S_1\right)=\frac{3n}{2}\left(2a+\left(3n-1\right)d\right)$

From $\left(3\right)$,

$3\left(S_2-S_1\right)=S_3$

Hence, the correct option is (C).