Question

In an AP, the first term is −4, the last term is 29 and the sum of all its terms is 150. Find its common difference. [CBSE 2011]

Answer 1

Suppose there are n terms in the AP.

Here, a = −4, l = 29 and S_n = 150

$$\begin{gathered} S_n=150 \\ \Rightarrow \frac{n}{2}\left(-4+29\right)=150\left[S_n=\frac{n}{2}\left(a+l\right)\right] \\ \Rightarrow n=\frac{150\times 2}{25}=12 \end{gathered}$$

Thus, the AP contains 12 terms.

Let d be the common difference of the AP.

$$\begin{gathered} \therefore a_{12}=29 \\ \Rightarrow -4+\left(12-1\right)\times d=29\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 11d=29+4=33 \\ \Rightarrow d=3 \end{gathered}$$

Hence, the common difference of the AP is 3.