Question

Let $a,b,c$ be three positive numbers in $G.P$. If $\log \left(\frac{5c}{a}\right),\log \left(\frac{3b}{5c}\right),\log \left(\frac{a}{3b}\right)$ are in $A.P.$, then which of the following options is CORRECT ?

• A. $a,b,c$ are the lengths of the sides of a right-angled triangle.
• B. $a,b,c$ are the lengths of the sides of a scalene triangle
• C. $a,b,c$ are the lengths of the sides of an isosceles triangle
• D. no triangle is possible having lengths of the sides as $a,b,c$

Answer 1

The correct option is D no triangle is possible having lengths of the sides as $a,b,c$

$\log \left(\frac{5c}{a}\right),\log \left(\frac{3b}{5c}\right),\log \left(\frac{a}{3b}\right)$ are in A.P.
$\Rightarrow \log \left(\frac{5c}{a}\right)+\log \left(\frac{a}{3b}\right)=2\log \left(\frac{3b}{5c}\right)\Rightarrow \frac{5c}{3b}=\frac{(3b{)}^2}{(5c{)}^2}\Rightarrow (5c{)}^3-(3b{)}^3=0\Rightarrow 5c=3b\cdots \left(1\right)$

Also, $b^2=ac$
$\Rightarrow b^2=a\cdot \frac{3b}{5}\left[\text{From}\right(1\left)\right]\Rightarrow a=\frac{5b}{3}$
Now, $b+c=b+\frac{3b}{5}=\frac{8b}{5}<\frac{5b}{3}$

Hence, no triangle can be formed as $b+c<a$