Question

In a sequence, if $S_n$ is the sum of the first n terms and $S_{n-1}$ is the sum of the first (n-1) terms, then the $n^{th}$ term is

• A. $S_{n-2}$
• B. $S_n-S_{n-1}$
• C. $S_{n+1}-S_n$
• D. $S_{n+1}-S_{n-1}$

Answer 1

The correct option is B $S_n-S_{n-1}$

Let $a$ be the first term and $d$ be their common difference of the AP.
Then, Sum to $n$ terms of an AP $=\frac{n}{2}(2a+(n-1\left)d\right)$
Clearly, we get $S_n$ by adding the $nth$ term to the sum $S_{n-1}$
So $S_n=S_{n-1}+T_n$

$\therefore n^{th}termis{S}_n-S_{n-1}$