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Question

If the angles A,B,C of a triangle are in AP and the sides a,b,c opposite to these angles are in GP, then a2,b2,c2 are in

A
AP
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B
GP
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C
Neither AP nor GP
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D
HP
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Solution

The correct option is C AP
Since, A,B and C are in AP.
Therefore, B=60o and as a,b,c are in GP, therefore b2=ac
Consider, cosB=a2+c2b22ac
12=a2+c2b22b2
b2=a2+c2b2
2b2=a2+c2

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