Question

Given that $\frac{a+b{e}^y}{a-b{e}^y}=\frac{b+c{e}^y}{b-c{e}^y}=\frac{c+d{e}^y}{c-d{e}^y}$, then $a,b,c,d$ are in .....................

• A. G.P
• B. A.P
• C. H.P
• D. Both b and c

Answer 1

The correct option is A G.P

Given $\frac{a+b{e}^y}{a-b{e}^y}=\frac{b+c{e}^y}{b-c{e}^y}=\frac{c+d{e}^y}{c-d{e}^y}$

$\to \frac{a+b{e}^y}{a-b{e}^y}=\frac{b+c{e}^y}{b-c{e}^y}$

Applying componendo and dividendo rule gives

$\to \frac{2a}{2b{e}^y}=\frac{2b}{2c{e}^y}$

$\Rightarrow \frac{a}{b}=\frac{b}{c}$ ------(1)

Similarly, Applying componendo and dividendo rule for $\frac{b+c{e}^y}{b-c{e}^y}=\frac{c+d{e}^y}{c-d{e}^y}$, we get

$\frac{2b}{2c{e}^y}=\frac{2c}{2d{e}^y}$

$\Rightarrow \frac{b}{c}=\frac{c}{d}$ ------(2)

From (1) and (2), we get

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$

i.e ratio of consecutive terms is same

$\therefore$ $a,b,c,d$ are in GP