# If $a,b,c$ be in Arithmetic Progression,

If $a,b,c$ be in Arithmetic Progression, $b,c,a$ in Harmonic Progression, prove that $c,a,b$ are in Geometric Progression.

My Attempt: $a,b,c$ are in AP so $$b=\dfrac {a+c}{2}$$

$b,c,a$ are in HP so $$c=\dfrac {2ab}{a+b}$$

Multiplying these relations: $$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$ $$=\dfrac {2a^2b+2abc}{2(a+b)}$$ $$=\dfrac {2a^2b+2abc}{2a+2b}$$

Since $a,b,c$ are in arithmetic progression, we get \begin{align*} &c-b=b-a\\[4pt] \implies\;&a = 2b - c\\[4pt] \end{align*} Since $b,c,a$ are in harmonic progression, we get \begin{align*} &\frac{1}{a}-\frac{1}{c} = \frac{1}{c}-\frac{1}{b}\\[4pt] \implies\;&a = \frac{bc}{2b-c}\\[4pt] \implies\;&a = \frac{bc}{a}\\[4pt] \implies\;&\frac{a}{c}=\frac{b}{a}\\[4pt] \end{align*} hence, $c,a,b$ are in geometric progression, as was to be shown.

Since $a,b,c$ are in AP we have $a=b-x$ and $c=b+x$ for some $x$ and since $b,c,a$ are in HP we have $$(b+x)(2b-x) = 2(b-x)b$$

Solwing this we get: $x=0$ or $x=3b$. So in the later case we get $a=-2b$ and $c=4b$ and so $$a^2=4b^2= bc$$

Hint:

Eliminate $c$

$$2ab=(a+b)c=(a+b)(2b-a)$$

Simplify to find $$0=a^2+ab-2b^2=(a+2b)(a-b)$$

• what to do after that? – pi-π Jan 29 '18 at 14:14
• @blue_eyed_..., what can we conclude if the product of two terms is $0$ – lab bhattacharjee Jan 29 '18 at 14:16
• Either of one or both terms are 0. – pi-π Jan 29 '18 at 14:39

you did everthing right. Now just multiply both sides with the denominator of RHS and simplify.