Question

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

Answer 1

We have,
Sn = $\frac{1}{2}[3n^2+7n]$
Replacing n by (n−1),
we get
Sn−1 = $\frac{1}{2}\left[3\right(n-1{)}^2+7(n-1)]$
⇒Sn−1 = $\frac{1}{2}\left[3\right(n^2+1-2n)+7n-7]$
⇒Sn−1 =$\frac{1}{2}[3n^2+3-6n+7n-7]$
= $\frac{1}{2}[3n^2+n-4]$
Now, nth term
= Sn − Sn−1
⇒an = $\frac{1}{2}[3n^2+7n]-\frac{1}{2}[3n^2+n-4]$
⇒an=$\frac{1}{2}[3n^2+7n-3n^2-n+4]$
⇒an =$\frac{1}{2}[6n+4]$
= 3n+2
Now,
$a_{20}$ = 3 ×20 + 2
= 60 + 2 = 62