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If the 4th, 10th and 16th terms of a GP ...

If the 4th, 10th and 16th terms of a GP are x, y, z respectively, prove that x, y, z are in GP.

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Let a be the first term and r be the common ratio of the given GP. Then, `T_(4)=x, T_(10)=y` and `T_(16)=z`.
`rArr ar^(3)=x, ar^(9)=y` and `ar^(15)=z`.
`:. y^(2)=(ar^(9))^(2)=a^(2)r^(18)` and `xz=(ar^(3))(ar^(15))=a^(2)r^(18)`.
Consequently, we have `y^(2)=xz`.
Hence, x, y, z are in GP.
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