# If in a triangle, ABC$p \cos ^{2}(\frac{C}{2})+r \cos ^{2}(\frac{A}{2})=\frac{3q}{2},$ then the sides p, q and r Option 1) are in AP Option 2) are in GP Option 3) are in HP Option 4) satisfy p+q=r

As we discussed in concept

Double Angle Formula -

- wherein

These are formulae for double angles.

$Pcos^{2}(\frac{C}{2})+rcos^{2}(\frac{A}{2})=\frac{3q}{2}$

=> $P[\frac{1+cosC}{2}]+r[\frac{1+cosA}{2}]=\frac{3q}{2}$

$\therefore (P+r)+(PcosC+rcosA)=3q$

$\therefore P+r+q=3q$                                                                          $[\because Since\:\:acosC+CcosA=b]$

$\therefore P+r=2q$                                                                                 $\therefore\:P,q,r\:in\:AP$

Option 1)

are in AP

This option is correct.

Option 2)

are in GP

This option is incorrect.

Option 3)

are in HP

This option is incorrect.

Option 4)

satisfy p+q=r

This option is incorrect.

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