1)

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


A) GP. with common ratio $e^{d}$

B) GP. with common ratio $e^{1/d}$

C) GP. with common ratio $e^{d/(b^{2}-d^{2})}$

D) A.P

Answer:

Option C

Explanation:

a,b,c are in A.P. 2b = a+c Now,

$e^{1/c}\times e^{1/a} = e^{(a+c)/ac}$ = $e^{2b/ac}$ = $\left(e^{b/ac}\right)^{2}$

.'. $e^{1/c}, e^{b/ac},e^{1/a}$ in G.P with common ratio

$\frac{e^{b/ac}}{e^{1/c}}$ = $e^{b-a/ac}$ = $e^{d/(b-d)(b+d)}$

= $e^{d/(b^{2}-d^{2})}$

'.' a, b, c are in A.P. with common difference d .'. b- a = c-b = d