# If in a triangle $\dpi{100} ABC$  ,$\dpi{100} a\cos ^{2}\left ( \frac{C}{2} \right )+c\cos ^{2}\left ( \frac{A}{2} \right )= \frac{3b}{2},$  then the sides  $\dpi{100} a,b \: and\: c$ Option 1) $are\: in \: G.P.$ Option 2) $are\: in \: H.P.$ Option 3) $satisf\! y\: \: a+b=c$ Option 4) $are\: in \: A.P.$

As we learnt in

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

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$a \cos ^{2}\frac{c}{2}+c \cos ^{2}\frac{a}{2}=\frac{3b}{2}$

$\Rightarrow 2a \cos ^{2}\frac{c}{2}+2c \cos ^{2}\frac{a}{2}=3b$

$\Rightarrow a[1+\cos c]+c[1+\cos a]=3b$

$\Rightarrow a+c+a\cos c+c\cos a =3b$

now by sine law , we have  $\frac{a}{\sin a}=\frac{b}{\sin b}=\frac{c}{\sin c}$

Thus , using it , we have

$\sin a+\sin c+\sin a\cos c+\sin c\cos a=3\sin b$

$\Rightarrow \sin a+\sin c+\sin (a+c)=3\sin b$

$\Rightarrow \sin a+\sin c+\sin (\pi-b)=3\sin b$

$\Rightarrow \sin a+\sin c+\sin b=3\sin b$

$\Rightarrow \sin a+\sin c=2\sin b$

using sine law, we have

$a+c=2b$

$a,b,c\ are\ in\ A.P$

Option 1)

$are\: in \: G.P.$

This is incorrect option

Option 2)

$are\: in \: H.P.$

This is incorrect option

Option 3)

$satisf\! y\: \: a+b=c$

This is incorrect option

Option 4)

$are\: in \: A.P.$

This is correct option

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