Question

If the sum of first $n$ terms of an A.P is $\frac{1}{2}(3n^2+7n)$, then find its $n$th term. Hence write its $20$th term.

• A. $a_n=5n+2$, $a_{20}=12$
• B. $a_n=7n+2$, $a_{20}=62$
• C. $a_n=3n+2$, $a_{20}=62$
• D. None of these

Answer 1

The correct option is C $a_n=3n+2$, $a_{20}=62$

Given,

$S_n=\frac{1}{2}[3n^2+7n]$

$S_1=\frac{1}{2}\left[3\right(1{)}^2+7\left(1\right)]=5=a_1$

$S_2=\frac{1}{2}\left[3\right(2{)}^2+7\left(2\right)]=13=a_1+a_2$

$a_2=13-5=8$

$d=a_2-a_1=8-5=3$

$a_n=5+(n-1)3$

$\therefore a_n=3n+2$

$\therefore a_{20}=3\left(20\right)+2=62$