hey guys help please.
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The AP condition gives 2b=a+c and the HP condition gives
2b2=1a2+1c2⟺8(a+c)2=a2+c2a2c2⟺(a−c)2(a2+4ac+c2)=0
Now either (a−c)2=0⟹a=b=c
or a2+4ac+c2=0⟹(a+c)2+2ac=0⟹2b2+ac=0⟹a,b,−c2 are in GP.
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