Question

Let $S_1$ be the sum of first $2n$ terms of an arithemetic progression. Let $S_2$ be the sum of first $4n$ terms of the same arithmetic progression. If $(S_2-S_1)$ is $1000$ then sum of the first $6n$ terms of the arithmetic progression is equal to :

• A. $5000$
• B. $1000$
• C. $7000$
• D. $3000$

Answer 1

The correct option is D $3000$

$S_{4n}-S_{2n}=1000\Rightarrow \frac{4n}{2}(2a+(4n-1\left)d\right)-\frac{2n}{2}(2a+(2n-1\left)d\right)=1000\Rightarrow n(2a+(6n-1\left)d\right)=1000\Rightarrow \frac{6n}{2}(2a+(6n-1\left)d\right)=3000\therefore S_{6n}=3000$