Question

If bc, ca and ab are in GP then 1/a,1/b,1/c are in

no scam
​

Answer 1

Answer:

we know b²=ac for gp----1

hence take bc as a,ca as b and ab as c

now taking the reciprocal of 1 we get

1/b²=1/a²×1/c²

1/b²=1/ac

by cross multiplication we get ac=b²

hence putting proper values

(ca)²=(bc)(ab)

hence,(ca)²=ba+b²+ca+cb

by solving we get b²=(ca)