Question

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. [CBSE 2013]

Answer 1

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

$$\begin{gathered} S_m=4m^2-m \\ \Rightarrow S_{m-1}=4{\left(m-1\right)}^2-\left(m-1\right) \\ =4\left(m^2-2m+1\right)-\left(m-1\right) \\ =4m^2-9m+5 \end{gathered}$$

Suppose a_m denote the m^th term of the AP.

$$\begin{gathered} \therefore a_m=S_m-S_{m-1} \\ =\left(4m^2-m\right)-\left(4m^2-9m+5\right) \\ =8m-5.....\left(1\right) \end{gathered}$$
Now,

$$\begin{gathered} a_n=107\left(\mathrm{Given}\right) \\ \Rightarrow 8n-5=107\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow 8n=107+5=112 \\ \Rightarrow n=14 \end{gathered}$$

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.