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NDA Math Properties of Triangles

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Consider a triangle ABC satisfying

2a\sin^2\left(\frac{C}{2}\right)+2c\sin^2\left(\frac{A}{2}\right)=2a+2c-3b

[NDA II 2015]

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Question : 12 of 16

Marks: +1, -0

The sides of the triangle are in

G.P
A.P
H.P
Neither in G.P nor in A.P nor in H.P

Solution:

2a\sin^{2}\left(\;\;\frac{C}{2}\right)+2c\sin^{2}\left(\;\;\frac{A}{2}\right)=2a+2c-3b

\Rightarrow 2a\times\;\;\frac{(s-a)(s-b)}{ab}+2c\times\;\;\frac{(s-b)(s-c)}{bc}=2a+2c-3b

\Rightarrow\;\;\frac{2}{b}(s-b)[s-a+s-c]=2a+2c-3b

\Rightarrow\;\;\frac{2}{b}(s-b)b=2a+2c-3b

[\because 2s-a-c=b]

\Rightarrow a+c=2b

So a, b, care in A.P.

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