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

Answer: (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 $d\therefore b-a=c-b=d$]