# Mark the correct alternative in the following question:

Question:

Mark the correct alternative in the following question:

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

(a) 4

(b) 6

(c) 8

(d) 10

Solution:

As, $S_{2 n}=3 S_{n}$

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

$\Rightarrow 2[2 a+(2 n-1) d]=3[2 a+(n-1) d]$

$\Rightarrow 4 a+2(2 n-1) d=6 a+3(n-1) d$

$\Rightarrow 4 a+4 n d-2 d=6 a+3 n d-3 d$

$\Rightarrow 6 a-4 a+3 n d-3 d-4 n d+2 d=0$

$\Rightarrow 2 a-n d-d=0$

$\Rightarrow 2 a-(n+1) d=0$

$\Rightarrow 2 a=(n+1) d$    ....(i)

Now,

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

$=\frac{3[(n+1) d+(3 n-1) d]}{[(n+1) d+(n-1) d]}$    [ Using (1) ]

$=\frac{3[n d+d+3 n d-d]}{[n d+d+n d-d]}$

$=\frac{3[4 n d]}{[2 n d]}$

$=3 \times 2$

$=6$

Hence, the correct alternative is option (b).