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Question: If \(a\ne b\) and \(a:b\) is a duplicate ratio of \(\left( a+c \right):\left( b+c \right)\), prove t...

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

Explanation

Solution

We start solving this question by first going through the definition of duplicate ratio. Then we find the duplicate ratio of (a+c):(b+c)\left( a+c \right):\left( b+c \right) and equate it to a:ba: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 pq\dfrac{p}{q}, its duplicate ratio is p2q2\dfrac{{{p}^{2}}}{{{q}^{2}}}.
We are given that a≠ba\ne b and a:ba:b is duplicate ratio of (a+c):(b+c)\left( a+c \right):\left( b+c \right). As duplicate ratio of any ratio is its square, duplicate ratio of (a+c):(b+c)\left( a+c \right):\left( b+c \right) is,
⇒(a+c)2(b+c)2\Rightarrow \dfrac{{{\left( a+c \right)}^{2}}}{{{\left( b+c \right)}^{2}}}
As a:ba:b is a duplicate ratio of (a+c):(b+c)\left( a+c \right):\left( b+c \right), we can equate a:ba:b with the above obtained value. Then we get,
⇒(a+c)2(b+c)2=ab\Rightarrow \dfrac{{{\left( a+c \right)}^{2}}}{{{\left( b+c \right)}^{2}}}=\dfrac{a}{b}
Now, let us consider the formula (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
Using that we can transform the above equation as,
⇒a2+2ac+c2b2+2bc+c2=ab ⇒b(a2+2ac+c2)=a(b2+2bc+c2) ⇒a2b+2abc+bc2=ab2+2abc+ac2 ⇒a2b+bc2=ab2+ac2 ⇒ac2−bc2=a2b−ab2 ⇒c2(a−b)=ab(a−b) \begin{aligned} & \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{aligned}
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,
⇒c2=ab\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 z2=xy{{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 ab\sqrt{ab} which is equal to c. So, we get c2=ab{{c}^{2}}=ab.