Let S_1 be the sum of the first 2n terms of an arithmetic progression, and S_2 the sum of …

Question

Let $S_1$ be the sum of the first $2n$ terms of an arithmetic progression, and $S_2$ the sum of the first $4n$ terms of the same A.P. If $S_2-S_1 = 1000$, then the sum of the first $6n$ terms is equal to:

Accepted Answer

$S_{4n}-S_{2n} = 1000$: expanding gives $n\big[2a+(6n-1)d\big]=1000$.

$$S_{6n} = \frac{6n}{2}\big[2a+(6n-1)d\big] = 3\times1000 = 3000.$$

Solutions · 1