Question

Let $ {{S}_{n}} $ denote the sum of first n terms of an AP, if $ {{S}_{2n}}=3{{S}_{n}}, $ then the ratio $\frac {S_{3n}}{S_n}$ is equal to

• A. 4
• B. 6
• C. $ \pm 3 $
• D. $ \pm 4 $

Answer 1

Answer: (B)

Given, $ {{S}_{2n}}=3{{S}_{n}} $
$ \Rightarrow $ $ \frac{2n}{2}[2a+(2n-1)d]=3.\frac{n}{2}[2a+(n-1)d] $
where a is the first term and d is the common difference of given AP.
$ \Rightarrow $ $ 2a=(n+1)d $
$ \therefore $ $ \frac{{{S}_{3n}}}{{{S}_{n}}}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]} $
$ =\frac{12nd}{2nd}=6 $