In a triangle ABC,A,B,C are in G.P with common ratio 2 then 1b+1c−1a is =
1
-2
3
0
We have B=2A,C=4A⇒7A=π⇒A=π/7
→1b+1c−1a=12R1sin2A+1sin4A−1sinA=2sinAsin4A+2sinAsin2A−2sin2Asin4A4RsinAsin2Asin4A=14Rcos2A−cos2A+cosA−cosAsinAsin2Asin4A=0