Question

Very-Short and Short-Answer Questions

Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40. [CBSE 2011]

Answer 1

Let a be the first term and d be the common difference of the AP. Then,
$$\begin{gathered} a_4=9 \\ \Rightarrow a+\left(4-1\right)d=9\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+3d=9.....\left(1\right) \end{gathered}$$

Now,
$a_6+a_{13}=40$ (Given)
$$\begin{gathered} \Rightarrow \left(a+5d\right)+\left(a+12d\right)=40 \\ \Rightarrow 2a+17d=40.....\left(2\right) \end{gathered}$$

From (1) and (2), we get
$$\begin{gathered} 2\left(9-3d\right)+17d=40 \\ \Rightarrow 18-6d+17d=40 \\ \Rightarrow 11d=40-18=22 \\ \Rightarrow d=2 \end{gathered}$$

Putting d = 2 in (1), we get
$$\begin{gathered} a+3\times 2=9 \\ \Rightarrow a=9-6=3 \end{gathered}$$

Hence, the AP is 3, 5, 7, 9, 11, ... .