Let S1 be the sum of first 2n terms of an arithmetic progression.

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 :

Accepted Answer

S1 = $${{2n} \over 2}$$[2a + (2n - 1)d]

S2 = $${{4n} \over 2}$$[2a + (4n - 1)d]

(where a = T1 and d is common difference)

S2 - S1$$ \Rightarrow $$ 2n[2a + (4n - 1)d] - n[2a + (2n - 1)d] = 1000

$$ \Rightarrow $$ n[2a + d(8n - 2 - 2n + 1)] = 1000

$$ \Rightarrow $$ n[2a + (6n - 1)d] = 1000

S6 = $${{6n} \over 2}$$[2a + (6n - 1)d] = 3(S2 - S1) = 3000

Solutions · 1