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 $G.P.$
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 $G.P.$
Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given condition,
$a_{4}=a r^{3}=x$ .......$(1)$
$a_{10}=a r^{9}=y$ .......$(2)$
$a_{16}=a r^{15}=z$ .......$(3)$
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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