What can be said about $a,b,c$ or $(b+c),(c+a),(a+b)$ if three real numbers $a,b,c$ are such that $a^2,b^2,c^2$ are terms of an AP?

Question

Question: Three real numbers $a,b,c$ are such that $a^2,b^2,c^2$ are terms of an arithmetic progression. Then

(A) $a,b,c$ are terms of a geometric progression

(B) $(b+c),(c+a),(a+b)$ are terms of an arithmetic progression

(C) $(b+c),(c+a),(a+b)$ are terms of a harmonic progression

(D) none of the foregoing statements is necessarily true

I cannot figure out how to solve this question. Initially I misread the question and took $a^2,b^2,c^2$ to be in arithmetic progression but then I realized that they are terms of an arithmetic progression and not necessarily in arithmetic progression. I cannot figure how to proceed. Please help.

Answer 1

I think the question means that the terms are in AP. I've solved it accordingly.

$\begin{align} 2b^2&=a^2+c^2\\ b^2-a^2&=c^2-b^2\\ \frac{b-a}{b+c}&=\frac{c-b}{b+a}\\ \frac{(b+c)-(a+c)}{(b+c)(a+c)}&=\frac{(a+c)-(a+b)}{(a+b)(a+c)}&&\text{...(divide both sides by $(a+c)$)}\\ \frac{1}{a+c}-\frac{1}{b+c}&=\frac{1}{a+b}-\frac{1}{a+c}\\ \frac{2}{a+c}&=\frac{1}{a+b}+\frac{1}{b+c} \end{align}$

Hence, $(b+c), \;(c+a), \;(a+b)$ are in HP.

Answer 2

To prove that nor a) neither b) are true, it suffices to consider $a,b,c>0,\; a^2=1, b^2=5,c^2=9.$

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The numbers $b+c, c+a, a+b$ from the above example are in harmonic progression. Let us see if it is so in general.

It suffices to prove that the harmonic mean of $b+c$ and $a+b$ equals $c+a,$ assuming that $(b+c)\neq0\neq (a+b).$

$$HM(b+c,a+b)=\frac{2}{\frac{1}{b+c}+\frac{1}{a+b}}$$ Therefore $$HM(b+c,a+b)=\frac{2(b+c)(a+b)}{a+2b+c}=\frac{2ab+2ac+a^2+c^2+2bc}{a+2b+c}$$

Since $a^2+c^2=2b^2,$ we have $$HM(b+c,a+b)=\frac{(a+2b+c)(c+a)}{a+2b+c}.$$

If $(b+c)\neq0\neq (a+b),$ then the statement c) is true.

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Without assuming $(b+c)\neq0\neq (a+b)$ or, e.g. $a,b,c>0$ (there is no constraint in the exercise!) holds d).
As a counterexample to c), consider $a=1, b=-1, c=1.$