Q. If sum of $n$ terms of an AP is $2n + 3n^{2}$, then $r$th term is
A) $2r + 3r^{2}$
B) $3r^{2} - 4r + 1$
C) $6r - 1$
D) $4r + 1$
Explanation
(c)
Let $S_{n} = 2n + 3n^{2}$
$\therefore\ \ S_{n - 1} = 2(n - 1) + 3(n - 1)^{2}$
$= 2n - 2 + 3n^{2} + 3 - 6n$
$= 3n^{2} - 4n + 1$
$\therefore\ T_{n} = S_{n} - S_{n - 1}$
$= 2n + 3n^{2} - (3n^{2} - 4n + 1)$
$= 6n - 1$
$\therefore\ T_{n} = 6r - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.