Question

If $\log \left(\frac{5c}{a}\right),\log \left(\frac{3b}{5c}\right),\log \left(\frac{a}{3b}\right),$ are in A.P. where a, b, c are in G.P. then a, b, c are the lengths of sides of

• A. an isosceles triangle
• B. an equilateral triangle
• C. a scalene triangle
• D. none of these.

Answer 1

Answer: (D)

Since given nos. are in $A.P$.
$\therefore 2log\frac{3b}{5c}=log\frac{5c}{a}+log\frac{a}{3b}$
$\Rightarrow {\left(\frac{3b}{5c}\right)}^2=\frac{5c}{a}\cdot \frac{a}{3b}$
$\Rightarrow 3b=5c$
Also $b^2=ac$
$\therefore \frac{25c^2}{9}=ac$
$\Rightarrow 9a=25c$
$\therefore \frac{9a}{5}=5c=3b$
$\Rightarrow \frac{a}{5}=\frac{b}{3}=\frac{c}{9/5}$
$\Rightarrow b+c<a$
$\therefore a,b,c$ cannot be sides of a triangle