Question

If the arithmetic progression whose common difference is non zero, the sum of first $3n$ terms is equal to the sum of the next $n$ terms. Then the ratio of the sum of the first $2n$ terms to the next $2n$ terms is

• A. $\frac{1}{5}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{5}$

Let the series be $a,a+d,a+2d,...,a+(n-1)d,a+(3n-1)d,a+3nd.$

Also,given that the sum of first $3n$ terms is equal the sum of next $n$ terms.

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

$\Rightarrow 3[2a+3nd-d]=[2a+6nd+nd-d]$

$\Rightarrow 6a+9nd-3d=2a+7nd-d$

$\Rightarrow 6a+9nd-3d-2a-7nd+d=0$

$\Rightarrow 4a+2nd-2d=0$

$\Rightarrow 2(2a+(n-1\left)d\right)=0$

$\Rightarrow 2a+(n-1)d=0$ $...\left(1\right)$

Now, $\frac{S_{2n}}{S_{2n\left(next2nterms\right)}}=\frac{\frac{2n}{2}[2a+(2n-1\left)d\right]}{\frac{2n}{2}\left[2\right(a+2nd)+(2n-1\left)d\right]}$

$=\frac{2a+(2n-1)d}{2a+4nd+2nd-d}$

$=\frac{2a+nd+nd-d}{2a+nd+5nd-d}$

$=\frac{2a+(n-1)d+nd}{2a+(n-1)d+5nd}$

$=\frac{nd}{5nd}$ $...\left(from\right(1\left)\right)$
$=\frac{1}{5}$