Question:
If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z 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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