Question

The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP. [CBSE 2013]

Answer 1

Let a be the first term and d be the common difference of the AP. Then,

$$\begin{gathered} a_{19}=3a_6\left(\mathrm{Given}\right) \\ \Rightarrow a+18d=3\left(a+5d\right)\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+18d=3a+15d \\ \Rightarrow 3a-a=18d-15d \end{gathered}$$
$\Rightarrow 2a=3d.....\left(1\right)$

Also,
$$\begin{gathered} a_9=19\left(\mathrm{Given}\right) \\ \Rightarrow a+8d=19.....\left(2\right) \end{gathered}$$

From (1) and (2), we get
$$\begin{gathered} \frac{3d}{2}+8d=19 \\ \Rightarrow \frac{3d+16d}{2}=19 \\ \Rightarrow 19d=38 \\ \Rightarrow d=2 \end{gathered}$$

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

So,
$$\begin{gathered} a_2=a+d=3+2=5 \\ {a}_3=a+2d=3+2\times 2=7,... \end{gathered}$$

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