Question
Let $$a, b, c$$ be in A.P. with a common difference $$d.$$ Then $${e^{\frac{1}{c}}},{e^{\frac{b}{{ac}}}},{e^{\frac{1}{a}}}$$ are in :
A.
G.P. with common ratio $$e^d$$
B.
G.P. with common ratio $${e^{\frac{1}{d}}}$$
C.
G.P. with common ratio $${e^{\frac{d}{{\left( {{b^2} - {d^2}} \right)}}}}$$
D.
A.P.
Answer :
G.P. with common ratio $${e^{\frac{d}{{\left( {{b^2} - {d^2}} \right)}}}}$$
Solution :
$$a, b, c$$ are in A.P.
⇒ $$2b = a + c$$
Now,
$$\eqalign{
& {e^{\frac{1}{c}}} \times {e^{\frac{1}{a}}} = {e^{\frac{{\left( {a + c} \right)}}{{ac}}}} = {e^{\frac{{2b}}{{ac}}}} = {\left( {{e^{\frac{b}{{ac}}}}} \right)^2} \cr
& \therefore {e^{\frac{1}{c}}},{e^{\frac{b}{{ac}}}},{e^{\frac{1}{a}}}{\text{ in G}}{\text{.P}}{\text{. with common ratio}} \cr
& = \frac{{{e^{\frac{b}{{ac}}}}}}{{{e^{\frac{1}{c}}}}} = {e^{\frac{{\left( {b - a} \right)}}{{ac}}}} = {e^{\frac{d}{{\left( {b - d} \right)\left( {b + d} \right)}}}} \cr
& = {e^{\frac{d}{{\left( {{b^2} - {d^2}} \right)}}}} \cr
& \left[ {\because a,b,c{\text{ are in A}}{\text{.P}}{\text{. with common difference }}d\,\,\therefore b - a = c - b = d} \right] \cr} $$