Question

Let $S _{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of first 4n terms of the same arithmetic progression. If $\left( S _{2}- S _{1}\right)$ is 1000, then the sum of the first $6 n$ terms of the arithmetic progression is equal to:

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

Answer 1

Answer: (D)

$S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d], S_{4 n}=\frac{4 n}{2}[2 a+(4 n-1)d]$
$\Rightarrow S _{2}- S _{1}=\frac{4 n }{2}[2 a +(4 n -1) d ]-\frac{2 n }{2}[2 a +(2 n -1)d] $
$=4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n$
$=2 na + nd [8 n -2-2 n +1]$
$\Rightarrow 2 na + nd [6 n -1]=1000$
$2 a+(6 n-1) d=\frac{1000}{n}$
Now, $S_{6 n }=\frac{6 n }{2}[2 a +(6 n -1) d ]$
$=3 n \cdot \frac{1000}{ n }=3000$