Question

An AP 8, 10, 12, ... has 60 terms. Find its last term. Hence, find the sum of its last 10 terms. [CBSE 2015]

Answer 1

The given AP is 8, 10, 12, ... .

Here, a = 8, d = 10 − 8 = 2 and n = 60

Since there are 60 terms in the AP, so the last term of the AP is a₆₀.

$$\begin{gathered} l=a_{60}=8+\left(60-1\right)\times 2\left[a_n=a+\left(n-1\right)d\right] \\ =8+118 \\ =126 \end{gathered}$$

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of the AP

$$\begin{gathered} =S_{60}-S_{50} \\ =\frac{60}{2}\left[2\times 8+\left(60-1\right)\times 2\right]-\frac{50}{2}\left[2\times 8+\left(50-1\right)\times 2\right]\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ =30\times \left(16+118\right)-25\times \left(16+98\right) \\ =30\times 134-25\times 114 \end{gathered}$$
$$\begin{gathered} =4020-2850 \\ =1170 \end{gathered}$$

Hence, the required sum is 1170.