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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.
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Let $a$ be the first term and $r$ be the common ratio.
$T_4=x \Rightarrow a r^3=x$ $T \quad \ldots(i)$
$T_{10}=y \Rightarrow a r^9=y \quad \ldots(ii)$
$T_{16}=z=a r^{15}=z \quad \ldots(iii)$
Now, $x, y, z$ will be in G.P.
If $a r^3, a r^9, a r^{15}$ are in G.P.
i.e. $\frac{a r^9}{a r^3}=\frac{a r^{15}}{a r^9} \Rightarrow r^6=r^6$, which is true.
Hence, $x, y, z$ are in G.P.
$T_4=x \Rightarrow a r^3=x$ $T \quad \ldots(i)$
$T_{10}=y \Rightarrow a r^9=y \quad \ldots(ii)$
$T_{16}=z=a r^{15}=z \quad \ldots(iii)$
Now, $x, y, z$ will be in G.P.
If $a r^3, a r^9, a r^{15}$ are in G.P.
i.e. $\frac{a r^9}{a r^3}=\frac{a r^{15}}{a r^9} \Rightarrow r^6=r^6$, which is true.
Hence, $x, y, z$ are in G.P.
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