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 $21 s t$ 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,
$a_{n}=107$
$\Rightarrow 8 n-5=107$
$\Rightarrow 8 n=112$
$\Rightarrow n=14$
$a_{21}=8(21)-5=163$
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