Question

Let $a, b, c, be$ in $A.P.$ with a common difference $d.$ Then $e^{1/e}, e^{b/bc}, e^{1/a}$ are in :

• A. $G.P.$ with common ratio $e^d$
• B. $G.P $with common ratio $e^{1/d}$
• C. $G.P.$ with common ratio $e^{d/\left(b^2-d^2\right)}$
• D. $A.P.$

Answer 1

The Correct Option is C

$a, b, c$ are in $A.P. \Rightarrow 2b=a+c$ Now, $e^{1/c}\times e^{1/a}=e^{\left(a+c\right)/ac}=e^{2b/ac}=\left(e^{b/ac}\right)^{2}$ $\therefore e^{1/c}, e^{b/ac}, e^{1/a}$ in $G.P.$ with common ratio $=\frac{e^{b/ac}}{e^{1/c}}=e^{\left(b-a\right)/ac}=e^{d/\left(b-d\right)\left(b+d\right)}$ $=e^{d/\left(b^2-d^2\right)}$ [$\because a, b, c$ are in $A.P.$ with common difference $d \therefore b - a = c - b = d$]