Question

If the sums of n, 2n and 3n terms of an A.P. are $S_1,S_2$ and $S_3$ respectively, then $\frac{s_3}{(s_2-s_1)}$ is _____.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is D 3

Let the first term of the A.P. be $a$ and common difference be $d$.

The Sum of $n$ terms in an A.P. is calculated using the formula $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

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

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

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

Now consider,$S_{2n}-S_n$

$S_{2n}-S_n$=$\frac{n}{2}(4a+(4n-2)d-2a-(n-1\left)d\right)$

$S_{2n}-S_n$=$\frac{n}{2}(4a+4nd-2d-2a-nd+d)$

$S_{2n}-S_n$=$\frac{n}{2}(2a+3nd-d)$

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

$S_{2n}-S_n=\frac{S_{3n}}{3}$

$\therefore$ $\frac{s_3}{s_2-s_1}=3$

$\therefore$ Option D is correct.