The sum of first m terms of an AP is (4m2 − m). If its nth term is 107, find the value of n.


The sum of first m terms of an AP is (4m2 − m). If its nth term is 107, find the value of n. Also, find the 21st term of this AP.   


Let Sm denote the sum of the first m terms of the AP. Then,

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

$\Rightarrow S_{m-1}=4(m-1)^{2}-(m-1)$

$=4\left(m^{2}-2 m+1\right)-(m-1)$

$=4 m^{2}-9 m+5$

Suppose am denote the mth term of the AP.

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

$=\left(4 m^{2}-m\right)-\left(4 m^{2}-9 m+5\right)$

$=8 m-5 \quad \ldots(1)$


$a_{n}=107$                              (Given)

$\Rightarrow 8 n-5=107$                     [From (1)]

$\Rightarrow 8 n=107+5=112$

$\Rightarrow n=14$

Thus, the value of n is 14.

Putting m = 21 in (1), we get

$a_{21}=8 \times 21-5=168-5=163$

Hence, the 21st term of the AP is 163 .

Leave a comment

Free Study Material