Question

STATEMENT 1: In a $\Delta ABC$, if the sides $a,b,c$ are in $A.P.$, then $a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=3b$

STATEMENT 2: In a $\Delta ABC,$ if $\angle A,\angle B,\angle C$ are in $A.P$., then $\angle B=\frac{\pi }{3}$

• A. Statement-1 is true, Statement-2 is true, Statement-2 is correct explanation of Statement- 1
• B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true

Answer 1

The correct option is D Statement-1 is false, Statement-2 is true

Statement 1:

Given, $a+c=2b$

Now, $a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=a\left(\frac{1+\cos C}{2}\right)+c\left(\frac{1+\cos A}{2}\right)$

$=\frac{a+c}{2}+\frac{a\cos C+c\cos A}{2}$

$=b+\frac{b}{2}=\frac{3b}{2}$

So, statement 1 is false.

Statement 2:

Let $A=x-d,B=x,C=x+d(\because A,B,C$ are in AP$)$

Also $A+B+C=\pi \Rightarrow x-d+x+x+d=3x=\pi$

$\Rightarrow \angle B=x=\frac{\pi }{3}$

Hence, statement 2 is true.