Question

The 8th term of an AP ia zero. Prove that its 38th term is triple its 18th term. [CBSE 2010]

Answer 1

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

$$\begin{gathered} a_8=0\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+\left(8-1\right)d=0 \\ \Rightarrow a+7d=0 \\ \Rightarrow a=-7d.....\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} \frac{a_{38}}{a_{18}}=\frac{a+\left(38-1\right)d}{a+\left(18-1\right)d} \\ \Rightarrow \frac{a_{38}}{a_{18}}=\frac{-7d+37d}{-7d+17d}\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow \frac{a_{38}}{a_{18}}=\frac{30d}{10d}=3 \\ \Rightarrow a_{38}=3\times a_{18} \end{gathered}$$

Hence, the 38th term of the AP is triple its 18th term.