Question

If the sum of $n$ terms of an A.P. is given by $S_n=3n^2-4n,$ then its ${50}^{th}$ term is

• A. $293$
• B. $353$
• C. $391$
• D. $412$

Answer 1

The correct option is A $293$

Let $T_n$ be the $n^{th}$ term of A.P.
Then, $T_n=S_n-S_{n-1}$
$=3n^2-4n-\left[3\right(n-1{)}^2-4(n-1)]=6n-7$

Hence, $T_{50}=6\times 50-7=293$