Question

If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term. [CBSE 2012]

Answer 1

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

$4\times a_4=18\times a_{18}$ (Given)
$$\begin{gathered} \Rightarrow 4\left(a+3d\right)=18\left(a+17d\right)\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2\left(a+3d\right)=9\left(a+17d\right) \\ \Rightarrow 2a+6d=9a+153d \\ \Rightarrow 7a=-147d \end{gathered}$$
$$\begin{gathered} \Rightarrow a=-21d \\ \Rightarrow a+21d=0 \\ \Rightarrow a+\left(22-1\right)d=0 \\ \Rightarrow a_{22}=0 \end{gathered}$$

Hence, the 22nd term of the AP is 0.