JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Sequences and Series

Let

S_{1}

be the sum of first

2 n

terms of an arithmetic progression. Let

S_{2}

be the sum of first

4 n

terms of the same arithmetic progression. If

(S_{2} - S_{1})

is 1000, then the sum of the first

6 n

terms of the arithmetic progression is equal to :

Answer: (D)

Solution

For an AP with first term

a

and common difference

d

S_{1} = n (2 a + (2 n - 1) d), S_{2} = 2 n (2 a + (4 n - 1) d)

S_{2} - S_{1} = n [4 a + (8 n - 2) d - 2 a - (2 n - 1) d] = n (2 a + (6 n - 1) d) = 1000

The sum of the first

6 n

terms is

S_{6 n} = 3 n (2 a + (6 n - 1) d) = 3 \times n (2 a + (6 n - 1) d) = 3 \times 1000

so the sum of the first

6 n

terms is

3000

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option