If the angles A, B, and C of a triangle ABC are in AP and the sides a, b and c opposite to these angles are in GP; then a2, b2 and c2 are related as
b2=a2c2
b2=a2+c2
2b2=a2+c2
none
A
2b2=a2+c2
B
none
C
b2=a2+c2
D
b2=a2c2
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Solution
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Since, A,B and C are in AP
2B−A+C
A+B+C=180o
2B+B=180B=60o
a,b,c are in GP then
b2=ac
cosB=c2+a2−b22ac
⇒cos60o=c2+a2−b22b2
12=c2+a2−b22b2
c2+a2−b2=b2
2b2=c2+a2
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