Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn, then the ratio S3n/ Sn is equal to

Question

Let S_n denote the sum of the first n terms of an A.P. If S_2n = 3S_n, then the ratio \(\frac{S_{3n}}{S_n}\) is equal to

(a) 4 

(b) 6 

(c) 8

(d) 10

Answer 1

Answer : (b) 6

Let a be the first term and d the common difference of the given A.P.

Then, 

\( S_n = \frac{n}{2}[2a+(n-1)d]\)  

\( S_{2n} = \frac{2n}{2}[2a+(2n-1)d]\) 

\( S_{3n} = \frac{3n}{2}[2a+(3n-1)d]\)

Given, S_2n = 3S_n

⇒ \( \frac{2n}{2}[2a+(2n-1)d]\) = \( \frac{3n}{2}[2a+(n-1)d]\)

⇒ 4a + 4nd – 2d = 6a + 3nd – 3d 

⇒ d + nd = 2a

⇒ a = \(\frac{d(n+1)}{2}\) ... (i) 

Now,