Question

If a, b, c are positive numbers in G.P. and $\log \left(\frac{5c}{a}\right)\log \left(\frac{3b}{5c}\right)$ and $\log \left(\frac{a}{3b}\right)$ are in A.P., then a, b, c forms thesides of a triangle which is

• A. equilateral
• B. right angled
• C. isosceles
• D. none of these

Answer 1

Answer: (D)

none of these

We have $b^2=ac$ .......... (1)
and $2\log \left(\frac{3b}{5c}\right)=\log \left(\frac{5c}{a}\right)+\log \left(\frac{a}{3b}\right)$
$\Rightarrow 2\log \left(\frac{3b}{5c}\right)=\log \left(\frac{5c}{a}\cdot \frac{a}{3b}\right)=-\log \left(\frac{3b}{5c}\right)$
$\Rightarrow 3\log \left(\frac{3b}{5c}\right)=0\Rightarrow b=\frac{5}{3}c$ .......... (ii)
from (i) & (ii) . we have $b+c=\frac{5c}{3}+c=\frac{8c}{3}=\frac{24}{25}a<a$
Hence a, b, c cannot be the sides of a triangle.