Question

The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms. [CBSE 2012, 14]

Answer 1

Suppose there are n terms in the AP.

Here, a = 5, l = 45 and S_n = 400

$$\begin{gathered} S_n=400 \\ \Rightarrow \frac{n}{2}\left(5+45\right)=400\left[S_n=\frac{n}{2}\left(a+l\right)\right] \\ \Rightarrow \frac{n}{2}\times 50=400 \\ \Rightarrow n=\frac{400\times 2}{50}=16 \end{gathered}$$

Thus, there are 16 terms in the AP.

Let d be the common difference of the AP.

$$\begin{gathered} \therefore a_{16}=45 \\ \Rightarrow 5+\left(16-1\right)\times d=45\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 15d=45-5=40 \\ \Rightarrow d=\frac{40}{15}=\frac{8}{3} \end{gathered}$$

Hence, the common difference of the AP is $\frac{8}{3}$.