2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $$a^2=4b^2= bc$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Hint:

Eliminate $c$

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

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

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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.