Question

Let $S_n$ denote sum of first $n$ terms of an A.P. If $S_{3n}=3S_n$, then $\frac{S_{3n}}{S_n}$ is

Answer 1

Given

Sum of first n terms of an AP$=S_n$

$S_{3n}=3S_n$

$\frac{S_{3n}}{S_n}=\frac{3}{1}$

$\frac{\frac{3n}{2}(2a+(3n-1)d}{\frac{n}{2}(2a+(n-1\left)d\right)}=\frac{3}{1}$

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

$2a+(3n-1)d=2a+(n-1)d$

$a+(3n-1)d=a+(n-1)d$

therefore,$a_{3n}=a_n$