Question

Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term. [CBSE 2012]

Answer 1

Let d be the common difference of the AP.

Here, a = 10 and n = 14
Now,
$$\begin{gathered} S_{14}=1505\left(\mathrm{Given}\right) \\ \Rightarrow \frac{14}{2}\left[2\times 10+\left(14-1\right)\times d\right]=1505\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ \Rightarrow 7\left(20+13d\right)=1505 \\ \Rightarrow 20+13d=215 \end{gathered}$$
$$\begin{gathered} \Rightarrow 13d=215-20=195 \\ \Rightarrow d=15 \end{gathered}$$

∴ 25th term of the AP, a₂₅
$$\begin{gathered} =10+\left(25-1\right)\times 15\left[a_n=a+\left(n-1\right)d\right] \\ =10+360 \\ =370 \end{gathered}$$

Hence, the required term is 370.