Q79 / 120 · VITEEE 2019 · Mathematics
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$, be in A.P. with a common difference $\mathrm{d}$. Then $\mathrm{e}^{1 / \mathrm{c}}, \mathrm{e}^{\mathrm{b} / \mathrm{ac}}, \mathrm{e}^{1 / \mathrm{a}}$ are in:
- AGP. with common ratio $\mathrm{e}^{\mathrm{d}}$
- BG.P with common ratio $\mathrm{e}^{1 / d}$
- Correct answerCG.P. with common ratio $\mathrm{e}^{\mathrm{d} /\left(\mathrm{b}^{2}-\mathrm{d}^{2}\right)}$
- DA.P.
Solution
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. $\Rightarrow 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$
Now,
$$
\mathrm{e}^{1 / \mathrm{c}} \times \mathrm{e}^{1 / \mathrm{a}}=\mathrm{e}^{(\mathrm{a}+\mathrm{c}) / \mathrm{ac}}=\mathrm{e}^{2 \mathrm{~b} / \mathrm{ac}}=\left(\mathrm{e}^{\mathrm{b} / \mathrm{ac}}\right)^{2}
$$
$\therefore \mathrm{e}^{1 / \mathrm{c}}, \mathrm{e}^{\mathrm{b} / \mathrm{ac}}, \mathrm{e}^{1 / \mathrm{a}}$ in G.P. with common ratio
$$
\begin{array}{l}
=\frac{\mathrm{e}^{\mathrm{b} / \mathrm{ac}}}{\mathrm{e}^{1 / \mathrm{c}}}=\mathrm{e}^{(\mathrm{b}-\mathrm{a}) / \mathrm{ac}}=\mathrm{e}^{\mathrm{d} /(\mathrm{b}-\mathrm{d})(\mathrm{b}+\mathrm{d})} \\
=\mathrm{e}^{\mathrm{d} /\left(\mathrm{b}^{2}-\mathrm{d}^{2}\right)}
\end{array}
$$
$[\because \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. with common difference $d \therefore b-a=c-b=d$ ]
Hence, the correct answer is option: 'C'.