Question

Solve the following questions:
a) If \[a \ne b{\text{ }}and{\text{ }}a{\text{ }}:{\text{ }}b\]is equal to the duplicate ratio of \[a{\text{ }} + {\text{ }}c{\text{ }}:{\text{ }}b{\text{ }} + {\text{ }}c\;\]prove that 'c' is the mean proportional between a and b.
b) Find the two numbers such that their mean proportional is 24 and the third proportional is 1,536.

Answer 1

Hint: The duplicate ratio of the ratio \[x{\text{ }}:{\text{ }}y\]is the ratio\[{x^2}\;:{\text{ }}{y^2}\].
Also, if ‘b’ is a mean proportional between a and c then b 2 = ac.
In the given question \[a{\text{ }}:{\text{ }}b\] is equal to the duplicate ratio of \[a{\text{ }} + {\text{ }}c{\text{ }}:{\text{ }}b{\text{ }} + {\text{ }}c\;\]
 $ \Rightarrow a:b = {\left( {a + c} \right)^2}:{\left( {b + c} \right)^2} $
Use this equation and simplify it to prove: $ {{\text{c}}^2} = ab $
For the second part of the question, assume the two numbers to be x and y.
It is given 24 is the mean proportional between them $ \Rightarrow {24^2} = xy $
Also, its given that 1536 is third proportional foe x and y $ \Rightarrow x:y::y:1536 = \dfrac{x}{y} = \dfrac{y}{{1536}} $
Now, just solve the two equations for x and y.

Complete step-by-step answer:
a) First let us understand what is duplicate ratio and mean proportional:
 Duplicate ratio: Duplicate ratio is defined as the ratio of two equal ratios where one ratio is equal to the square of another ratio.
For example:
The duplicate ratio of the ratio \[x{\text{ }}:{\text{ }}y\]is the ratio\[{x^2}\;:{\text{ }}{y^2}\].
Mean proportional: The term ‘b’ is said to be mean proportional between a and c if and only if b 2 = ac.
So, here in the above question it’s given:
 If \[a \ne b{\text{ }}and{\text{ }}a{\text{ }}:{\text{ }}b\]is equal to the duplicate ratio of \[a{\text{ }} + {\text{ }}c{\text{ }}:{\text{ }}b{\text{ }} + {\text{ }}c\;\]
 $ \Rightarrow a:b = {\left( {a + c} \right)^2}:{\left( {b + c} \right)^2} $ ----------------[1]
And we have to prove that: c is the mean proportional between a and b.
 $ {\text{i}}{\text{.e}}{\text{. }}{{\text{c}}^2} = ab $
Now, from [1]:
 $ \dfrac{a}{b} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{{{\left( {b + c} \right)}^2}}} $
$
\Rightarrow \dfrac{a}{b}{\text{ }} = {\text{ }}\dfrac{{{a^2}\; + {\text{ }}{c^2}\; + {\text{ }}2ac}}{{{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2bc}} $
$\Rightarrow a\left( {{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2bc} \right) = b\left( {{a^2}\; + {\text{ }}{c^2}\; + {\text{ }}2ac} \right) $

$ { \Rightarrow {\text{ }}a{b^2}\; + {\text{ }}a{c^2}\; + {\text{ }}2abc{\text{ }} = {\text{ }}{a^2}b{\text{ }} + {\text{ }}b{c^2}\; + {\text{ }}2abc} $
$ { \Rightarrow {\text{ }}a{c^{2\;}}-{\text{ }}b{c^{2{\text{ }}\;}} = {\text{ }}\;{a^2}b{\text{ }}-{\text{ }}a{b^2}} \\ $
$\Rightarrow c^2 {(a-b)} = ab {(a-b)}$
$ \Rightarrow {c^2} = ab $
Hence, proved that c is the mean proportional between a and b.
b) Let x and y be the two numbers which are required to be found.
Given, mean proportional between x and y = 24

Also, 1536 is the third proportional then:
For a continued proportion: $ a:b::b:c \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} $ where, c is the third proportion.
Here,
\[\begin{array}{*{20}{l}}
  {x{\text{ }}:{\text{ }}y{\text{ }} = {\text{ }}y{\text{ }}:{\text{ }}1536} \\
  { \Rightarrow \dfrac{x}{y} = \dfrac{y}{{1536}}} \\
  \begin{gathered}
  From{\text{ }}\left( 1 \right),{\text{ }} \\
  {y^2}\;{\text{ }} = \;{\text{ }}1536{\text{ }} \times {\text{ }}576/y \\
\end{gathered} \\
  { \Rightarrow {y^3}\; = {\text{ }}1576{\text{ }} \times {\text{ }}576} \\
  { \Rightarrow {y^3}\; = {\text{ }}24{\text{ }} \times {\text{ }}24{\text{ }} \times {\text{ }}24{\text{ }} \times \;{\text{ }}64} \\
  { \Rightarrow y{\text{ }} = {\text{ }}24 \times 4} \\
  { \Rightarrow y{\text{ }} = {\text{ }}96}
\end{array}\]
Again from (1), we get
\[x{\text{ }} = {\text{ }}\dfrac{{576}}{{96}}{\text{ }} = \;{\text{ }}6\]
Hence, the required numbers are 6 and 96.

Note: The given question was based upon the concept of continued proportionality.
Continued proportion : Three numbers $ a,b{\text{ and c}} $ are said to be continued proportions if a, b and c are in proportion.
Thus, if $ a,b{\text{ and c}} $ are in continued-proportion, then
a, b, b, c are in proportion, which means
\[a{\text{ }}:{\text{ }}b\;:{\text{ }}:\;b{\text{ }}:{\text{ }}c\]
 $ \Rightarrow \dfrac{a}{b} = \dfrac{b}{c} $
\[\begin{gathered}
   \Rightarrow {\text{Product of extremes = Product of means}} \\
   \Rightarrow a \times c{\text{ }} = {\text{ }}b \times b \\
   \Rightarrow a \times c{\text{ }} = {\text{ }}b{\;^2} \\
\end{gathered} \]
Continued-proportion is also known as mean proportional, because here \[b{\;^2}\; = {\text{ }}ac\]and b is the mean proportional between a and c.