Question

Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 - S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

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

Hint:

If $S_k$ represents the sum of first $k$ terms, then for equal intervals of $m$ terms, the blocks $S_m, (S_{2m}-S_m), (S_{3m}-S_{2m})$ form an A.P.

Answer 1

The Correct Option is C

Step 1: Let $A$ be the sum of first $2n$ terms, $B$ be the sum of next $2n$ terms, and $C$ be the sum of next $2n$ terms. In an A.P., $A, B, C$ are also in an A.P. with common difference $D' = (2n)^2 d$.
Step 2: $S_1 = A$. $S_2 = A + B$. Given $S_2 - S_1 = 1000 \implies B = 1000$.
Step 3: $S_3$ (sum of $6n$ terms) $= A + B + C$. Since $A, B, C$ are in A.P., $B = \frac{A+C}{2} \implies A+C = 2B$.
Step 4: $S_3 = (A+C) + B = 2B + B = 3B = 3(1000) = 3000$.