Question

If log $\left(\frac{5c}{a}\right),log\left(\frac{3b}{5c}\right)andlog\left(\frac{a}{3b}\right)$ are in an AP (where a, b, and c are in a GP), then a, b, and c, are the lengths of sides of

• A. An isosceles traingle
• B. An equilateral triangle
• C. A scalene triangle
• D. None of these

Answer 1

The correct option is D None of these

2 log$\left(\frac{3b}{5c}\right)$ = log $\left(\frac{5c}{a}\right)$ + log$\left(\frac{a}{3b}\right)$
log${\left(\frac{3b}{5c}\right)}^2$ = log $(\frac{5c}{a}\times \frac{a}{3b})$
log${\left(\frac{3b}{5c}\right)}^2$ = log$\left(\frac{5c}{3b}\right)$
${\left(\frac{3b}{5c}\right)}^2$ = $\left(\frac{5c}{3b}\right)$
$b=\frac{5}{3}c$
$b^2=ac(wherea,b,andcareinaGP)$
$a=\frac{25}{9}c$
$a>b+c$
Therefore, a, b and c do not form a triangle.