Question

In an A.P., $S_1=6,S_7=105,$ then $S_n$: $S_{n-3}$ is same as

• A. $(n+3):(n-3)$
• B. $(n+3):n$
• C. $n:(n-3)$
• D. None of these

Answer 1

The correct option is A $(n+3):(n-3)$

Let a and d be the first term and common difference of the given AP respectively.
Given, $S_1=a=6$ and
$S_7=105\Rightarrow \frac{7}{2}[2a+(7-1\left)d\right]=105$
$\Rightarrow \frac{7}{2}[2\times 6+6d]=105$
$\Rightarrow 6+3d=15$
$\Rightarrow d=3$
Now, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}[12+(n-1\left)3\right]=\frac{3n}{2}[n+3]$ ...(1)
$S_{n-3}=\frac{n-3}{2}[2a+((n-3)-1\left)d\right]=\frac{n-3}{2}[12+(n-4\left)3\right]=\frac{n-3}{2}\left[3n\right]$ ...(2)
From (1) and (2), we get
$\frac{S_n}{S_{n-3}}=\frac{\frac{3n}{2}[n+3]}{\frac{n-3}{2}\left[3n\right]}=\frac{n+3}{n-3}$
Therefore, $S_n:S_{n-3}::(n+3):(n-3)$