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.
Question:
If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ terms of a G.P. are $x, y$ and $z$, respectively. Prove that $x, y, z$ are in G.P.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
$a_{4}=a r^{3}=x \ldots(1)$
$a_{10}=a r^{9}=y \ldots(2)$
$a_{16}=a r^{15}=z \ldots$ (3)
Dividing $(2)$ by $(1)$, we obtain
$\frac{y}{x}=\frac{a r^{9}}{a r^{3}} \Rightarrow \frac{y}{x}=r^{6}$
Dividing (3) by (2), we obtain
$\frac{z}{y}=\frac{a r^{15}}{a r^{9}} \Rightarrow \frac{z}{y}=r^{6}$
$\therefore \frac{y}{x}=\frac{z}{y}$
Thus, $x, y, z$ are in G. P.
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