Question

In an acute angle triangle $ABC,$ if $\frac{\sin 3B}{\sin B}={\left(\frac{a^2-c^2}{2ac}\right)}^2,$ then $a^2,b^2,c^2$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $A.G.P.$

Answer 1

The correct option is A $A.P.$

Given: $3-4{\sin }^{2}B=\frac{(a^2-c^2{)}^2}{4a^2{c}^2}\Rightarrow 4{\cos }^{2}B-1=\frac{(a^2-c^2{)}^2}{4a^2{c}^2}$
$\Rightarrow 4{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2-1=\frac{(a^2-c^2{)}^2}{4a^2{c}^2}$
$\Rightarrow 4{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2=1+\frac{(a^2-c^2{)}^2}{4a^2{c}^2}=\frac{(a^2+c^2{)}^2}{4a^2{c}^2}$
$\Rightarrow 4(a^2-b^2+c^2{)}^2=(a^2+c^2{)}^2$
$\Rightarrow a^2+c^2=2b^2$(only)
Thus, we get $a^2,b^2,c^2$ are in $A.P.$