Question

The 4th, 7th and 10 th terms of a GP are a, b, c, respectively. Prove that `b^(2)=ac`.

Answer 1

Let A be the first term and r be the common ratio of the given GP. Then
`a=Ar^((4-1))=Ar^(3), b=Ar^((7-1))=Ar^(6)` and `c=Ar^((10-1))=Ar^(9)`.
`:. ac=(Ar^(3))xx(Ar^(9))=A^(2) r^(12)=(Ar^(6))^(2)=b^(2)`.
Hence, `b^(2)=ac`.