JEE PYQ: Sequence And Series Question 7
Question 7 - 2021 (18 Mar Shift 2)
Let $S_1$ be the sum of first $2n$ terms of an arithmetic progression. Let $S_2$ be the sum of 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:
(1) 1000
(2) 7000
(3) 5000
(4) 3000
Show Answer
Answer: (4)
Solution
$S_{2n} = \frac{2n}{2}[2a + (2n-1)d]$, $S_{4n} = \frac{4n}{2}[2a + (4n-1)d]$
$\Rightarrow S_2 - S_1 = \frac{4n}{2}[2a + (4n-1)d] - \frac{2n}{2}[2a + (2n-1)d]$
$= 2na + nd[8n - 2 - 2n + 1]$
$\Rightarrow 2na + 2n[6n - 1] = 1000$
$2a + (6n - 1)d = \frac{1000}{n}$
Now, $S_{6n} = \frac{6n}{2}[2a + (6n-1)d]$
$= 3n \cdot \frac{1000}{n} = 3000$
BETA
