Question
Find two numbers whose mean proportional is 16 and the third proportional is 128.
✓
Answer
Let $x$ and $y$ be two numbers
Their mean proportion $=16$
and third proportion $=128$
$
\begin{aligned}
& \therefore \sqrt{x y}=16 \\
& \Rightarrow xy =256 \\
& \Rightarrow x =\frac{256}{y}
\end{aligned}
$
$
\text { and } \frac{y^2}{x}=128
$
$
\Rightarrow x =\frac{y^2}{128}
$From (i) and (ii)
$
\begin{aligned}
& \frac{256}{y}=\frac{y^2}{128} \\
& \Rightarrow y ^3=256 \times 128 \\
&= 32768 \\
& \Rightarrow y ^3=(32)^3 \\
& \Rightarrow y =32 \\
& \therefore x =\frac{256}{y} \\
& \frac{256}{32} \\
&= 8
\end{aligned}
$
$\therefore$ Numbers are 8,32 .
Their mean proportion $=16$
and third proportion $=128$
$
\begin{aligned}
& \therefore \sqrt{x y}=16 \\
& \Rightarrow xy =256 \\
& \Rightarrow x =\frac{256}{y}
\end{aligned}
$
$
\text { and } \frac{y^2}{x}=128
$
$
\Rightarrow x =\frac{y^2}{128}
$From (i) and (ii)
$
\begin{aligned}
& \frac{256}{y}=\frac{y^2}{128} \\
& \Rightarrow y ^3=256 \times 128 \\
&= 32768 \\
& \Rightarrow y ^3=(32)^3 \\
& \Rightarrow y =32 \\
& \therefore x =\frac{256}{y} \\
& \frac{256}{32} \\
&= 8
\end{aligned}
$
$\therefore$ Numbers are 8,32 .
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