Question

If $a\ne b$ and $a:b$ is a duplicate ratio of $\left( a+c \right):\left( b+c \right)$, prove that c is mean proportional between a and b.

Answer 1

Hint: We start solving this question by first going through the definition of duplicate ratio. Then we find the duplicate ratio of $\left( a+c \right):\left( b+c \right)$ and equate it to $a:b$. Then we solve it to obtain a simplified relation between a, b and c. Then we go through the definition of mean proportional and compare the obtained relation with it and prove that c is mean proportional of a and b.

Complete step-by-step answer:
First let us go through the definition of duplicate ratio.
A duplicate ratio can be said as the product of the ratio two times that is square of the ratio.
For any ratio $\dfrac{p}{q}$, its duplicate ratio is $\dfrac{{{p}^{2}}}{{{q}^{2}}}$.
We are given that $a\ne b$ and $a:b$ is duplicate ratio of $\left( a+c \right):\left( b+c \right)$. As duplicate ratio of any ratio is its square, duplicate ratio of $\left( a+c \right):\left( b+c \right)$ is,
$\Rightarrow \dfrac{{{\left( a+c \right)}^{2}}}{{{\left( b+c \right)}^{2}}}$
As $a:b$ is a duplicate ratio of $\left( a+c \right):\left( b+c \right)$, we can equate $a:b$ with the above obtained value. Then we get,
$\Rightarrow \dfrac{{{\left( a+c \right)}^{2}}}{{{\left( b+c \right)}^{2}}}=\dfrac{a}{b}$
Now, let us consider the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using that we can transform the above equation as,
$\begin{align}
  & \Rightarrow \dfrac{{{a}^{2}}+2ac+{{c}^{2}}}{{{b}^{2}}+2bc+{{c}^{2}}}=\dfrac{a}{b} \\
 & \Rightarrow b\left( {{a}^{2}}+2ac+{{c}^{2}} \right)=a\left( {{b}^{2}}+2bc+{{c}^{2}} \right) \\
 & \Rightarrow {{a}^{2}}b+2abc+b{{c}^{2}}=a{{b}^{2}}+2abc+a{{c}^{2}} \\
 & \Rightarrow {{a}^{2}}b+b{{c}^{2}}=a{{b}^{2}}+a{{c}^{2}} \\
 & \Rightarrow a{{c}^{2}}-b{{c}^{2}}={{a}^{2}}b-a{{b}^{2}} \\
 & \Rightarrow {{c}^{2}}\left( a-b \right)=ab\left( a-b \right) \\
\end{align}$
As we are given that a is not equal to b, we can cancel the term (a-b) on the both sides. Then we get,
$\Rightarrow {{c}^{2}}=ab$
Now, let us go through the definition of mean proportional.
Any number z is said to be mean proportional of x and y if ${{z}^{2}}=xy$.
So, by comparing the above attained relation between a, b, c with the definition of mean proportional above we can see that they are similar.
So, we can say that c is mean proportional of a and b.
Hence Proved.

Note: The common mistake one does while solving this problem one might confuse the term mean proportional of a and b as mean of a and b and try to prove the problem and get stuck in the end. So, one needs to remember that the mean proportional of a and b is $\sqrt{ab}$ which is equal to c. So, we get ${{c}^{2}}=ab$.