Question & Answer

If ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in A.P. Show that $b+c,c+a,a+b$ are in H.P.

ANSWER Verified Verified
Hint: Apply the reverse process and prove that if $b+c,c+a,a+b$ are in H.P then ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ will be in A.P. Use the formula for harmonic mean of three numbers $x,y,z$ given by: $\dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z}$. Simplify the harmonic mean relation to get the proof.

Complete step-by-step answer:

An arithmetic progression is a sequence of numbers such that the difference between any two successive members is a constant while harmonic progression is a sequence of terms in which the reciprocal of the terms are in A.P.

Let us assume that $b+c,c+a,a+b$ are in A.P. So, we have to prove that ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ must be in A.P. We know that harmonic mean of three numbers $x,y,z$ is given by: $\dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z}$. Therefore, harmonic mean relation of $b+c,c+a,a+b$ will be,


Taking L.C.M in R.H.S we get,


  & \dfrac{2}{a+c}=\dfrac{b+c+a+b}{(a+b)(b+c)} \\

 & \dfrac{2}{a+c}=\dfrac{a+c+2b}{ab+ac+{{b}^{2}}+bc} \\


Cross-multiplying we get,


  & 2ab+2ac+2{{b}^{2}}+2bc=(a+c)(a+c+2b) \\

 & 2ab+2ac+2{{b}^{2}}+2bc={{a}^{2}}+{{c}^{2}}+2ac+2ab+2bc \\


Cancelling the common terms we get,


We can see that, this is the relation of arithmetic mean of three numbers which are ${{a}^{2}},{{b}^{2}},{{c}^{2}}$. Therefore, we can conclude that ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in A.P.

Note: You may note that if we will try to solve the question straight forward then we will encounter some problems like what to add or subtract in the next step to get the numbers in harmonic progression and you may get confused. Therefore, it is better to simplify the harmonic mean relation and to prove ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in A.P.