Solve the following
Question:

If the 4th, 10th and 16th terms of a G.P. are xy and z respectively. Prove that xyz are in G.P.

Solution:

$a_{4}=x$

$\Rightarrow a r^{3}=x$

Also, $a_{10}=y$

$\Rightarrow a r^{9}=y$

And, $a_{16}=z$

$\Rightarrow a r^{15}=z$

$\because \quad \frac{y}{x}=\frac{a r^{9}}{a r^{3}}=r^{6}$

and $\frac{z}{y}=\frac{a r^{15}}{a r^{9}}=r^{6}$

$\therefore \frac{y}{x}=\frac{z}{y}$

Therefore, $x, y$ and $z$ are in G.P.

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