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8. Sequences and Series
hard
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
$3000$
(JEE MAIN-2021)
Solution
$S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d], S_{4 n}=\frac{4 n}{2}[2 a+(4 n-1)d]$
$\Rightarrow S _{2}- S _{1}=\frac{4 n }{2}[2 a +(4 n -1) d ]-\frac{2 n }{2}[2 a +(2 n -1)d]$
$=4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n$
$=2 n a+n d[8 n-2-2 n+1]$
$\Rightarrow 2 n a+n d[6 n-1]=1000$
$2 a+(6 n-1) d=\frac{1000}{n}$
Now, $S_{6 n}=\frac{6 n}{2}[2 a+(6 n-1) d]$
$=3 n \cdot \frac{1000}{ n }=3000$
Standard 11
Mathematics
