Question

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

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

Answer 1

The correct option is A

$3000$

Explanation for the correct option:

Finding the sum of first $6n$ terms of the arithmetic progression

Given- ${\text{S}}_2{\text{-S}}_1\text{=1000}$ ( where $S_2$ is sum of first $4n$ terms and $S_1$ is sum of the first $2n$ terms)

$$\begin{gathered} {\text{S}}_2{\text{-S}}_1\text{=1000} \\ \Rightarrow \left(\frac{4n}{2}\right)(2a+(4n–1\left)d\right)–\left(\frac{2n}{2}\right)(2a+(2n–1\left)d\right)=1000 \\ \Rightarrow 2an+6n^{2}d–nd=1000 \\ \Rightarrow \left(\frac{6n}{2}\right)(2a+(6n–1\left)d\right)=3000 \\ \Rightarrow S_{6n}=3000 \end{gathered}$$

Thus the sum of first $6n$ terms is $3000$

Hence, the correct option is (A)