Question
Find two numbers whose mean proportional is $18$ and the third proportional is $486.$
✓
Answer
Let $a$ and $b$ be the two numbers, whose mean proportional is $18 .$
'therefore $"ab" =18^2\Rightarrow \sim " a b "=324 \Rightarrow \sim " b "=324 / " a^{\prime 2}$......(i)
Now, third proportional is 486
$\therefore a: b:: b: 486$
$\Rightarrow b^2=486 a $
$\Rightarrow\left(\frac{324}{a}\right)^2=486 a $
$\Rightarrow \frac{(324)^2}{a^2}=486 a $
$\Rightarrow a^3=\frac{324 \times 324}{486} $
$\Rightarrow a^3=216 $
$\Rightarrow a=6 $
$b=\frac{324}{a}=\frac{324}{6}=54$
Therefore, numbers are $6$ and $54 .$
'therefore $"ab" =18^2\Rightarrow \sim " a b "=324 \Rightarrow \sim " b "=324 / " a^{\prime 2}$......(i)
Now, third proportional is 486
$\therefore a: b:: b: 486$
$\Rightarrow b^2=486 a $
$\Rightarrow\left(\frac{324}{a}\right)^2=486 a $
$\Rightarrow \frac{(324)^2}{a^2}=486 a $
$\Rightarrow a^3=\frac{324 \times 324}{486} $
$\Rightarrow a^3=216 $
$\Rightarrow a=6 $
$b=\frac{324}{a}=\frac{324}{6}=54$
Therefore, numbers are $6$ and $54 .$
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