Question

The sum of 'r' terms of an AP is $(2r^2+3r)$. Find its $n^{th}$​ term.

• A. 2n - 1
• B. 4n + 1
• C. 3n - 1
• D. 3n + 1

Answer 1

The correct option is B

4n + 1

Given, $S_r=(2r^2+3r)$

where $S_r$ is the sum of r terms of the given AP

Similarly, $S_n$ is the sum of n terms

$S_{n-1}$ is the sum of (n-1) terms.
$t_n=S_n-S_{n-1}$

where $t_n$ is the $n^{th}$​ term of the AP.

Since, $S_r=(2r^2+3r)$, we can substitute 'n' in this equation to get $S_n$

$\therefore$ $S_n=2n^2+3n$
Similarly, $S_{n-1}=2{(n-1)}^2+3(n-1)$
$t_n=S_n-S_{n-1}$
$\therefore$ $t_n$$=2n^2+3n-\left[2\right(n^2-2n+1)+3n-3]$
$=2n^2+3n-2n^2+4n-2-3n+3=4n+1$