# The sum of first m terms of an A.P. is 4m2 − m.

Question:

The sum of first $m$ terms of an A.P. is $4 m^{2}-m$. If its nth term is 107 . find the value of $n$. Also, find the 21 st term of this A.P.

Solution:

$S_{m}=4 m^{2}-m$

We know

$a_{m}=S_{m}-S_{m-1}$

$\therefore a_{m}=4 m^{2}-m-4(m-1)^{2}+(m-1)$

$a_{m}=8 m-5$

Now,

$a_{n}=107$

$\Rightarrow 8 n-5=107$

$\Rightarrow 8 n=112$

$\Rightarrow n=14$

$a_{21}=8(21)-5=163$