Question

If $S_n$ denotes the sum of $n$ terms of an A.P. $(n\ge 3)$ and $S_{2n}=3S_n,$ then $\frac{t_{10}}{t_{15}}$ is equal to

• A. $\frac{3}{2}$
  
• B. $\frac{44}{29}$
• C. $\frac{29}{44}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $\frac{29}{44}$

Let first term of A.P. be $a$ and common difference be $d.$
Given, $S_{2n}=3S_n$
$\Rightarrow \frac{2n}{2}[2a+(2n-1\left)d\right]=3\cdot \frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow 4a+(2n-1)2d=6a+(n-1)3d$
$\Rightarrow 2a=(n+1)d$

Now, $t_n=a+(n-1)d$
$=(n+1)\frac{d}{2}+(n-1)d$
$=d\left(\frac{3n-1}{2}\right)$
$\therefore \frac{t_{10}}{t_{15}}=\frac{\frac{d}{2}\left(29\right)}{\frac{d}{2}\left(44\right)}=\frac{29}{44}$