$a, b, c$ form a geometric sequence and $\log_c a, \log_ b c, \log_a b$ form an arithmetic sequence.

Question

The common difference of the arithmetic sequence can be expressed as $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

So far, I rearranged the sequences to be

1: $a$, $ar$, $ar^2$

2: $\log_c a, \log_c a + d, \log_c a + 2d$

I plugged in a, ar and ar^2 to the corresponding a, b, and c in the log sequence.

Then, I don't know what to do!

Answer 1

Solution

We have to constrain that, $a,b,c$ are all positive numbers which do not equal $1$. According to the assumption conditions, we have $$b^2=ac,\tag1$$and$$2\log_b c=\log_a b+\log_c a.\tag2$$ From $(1)$, $b=\sqrt{ac}$. Put it into $(2)$ and apply the base-changing formula. Thus, $$\frac{4\ln c}{\ln a +\ln c}=\frac{\ln a+\ln c}{2\ln a}+\frac{\ln a}{\ln c}.\tag3$$Denote $t=\dfrac{\ln a}{\ln c}.$ Then,we can obtain the equation$$2t^3+3t^2-6t+1=(t-1)(2t^2+5t-1)=0.\tag4$$ By solving it,we have the roots $$t_1=1,t_2=\frac{-5-\sqrt{33}}{4},t_3=\frac{-5+\sqrt{33}}{4}.$$ On the other hand,let the common difference be $d$,then $$d=\log_c-\log_c a=\frac{\ln c}{\ln b}-\frac{\ln a}{\ln c}=\frac{2-t^2-t}{t+1}\tag5.$$

• If $t=t_1=1$, then $d=0$. In this case, $a=b=c$,which is trivial;
• If $t=t_2$ or $t_3$, then $t^2=\dfrac{1-5t}{2}$.Thus, $d=\dfrac{2-t^2-t}{t+1}=\dfrac{3(t+1)}{2(t+1)}=\dfrac{3}{2}$.

Summing up the above, it follows that $d=0$ or $d=\dfrac{3}{2}.$ We are done.

Answer 2

We have $a>0$, $b>0,$ $c>0$, $a\neq1$, $b\neq1$ and $c\neq1$.

Also, $b^2=ac$ and $$2\log_bc=\log_ca+\log_ab$$ or $$2\log_{\sqrt{ac}}c=\log_ca+\log_a\sqrt{ac}$$ or

$$4\log_{ac}c=\log_ca+\frac{1}{2}\log_aac$$ or $$\frac{8}{1+\log_ca}=2\log_ca+1+\frac{1}{\log_ca},$$ which gives $$\log_ca\in\left\{1,\frac{-5-\sqrt{33}}{4},\frac{-5+\sqrt{33}}{4}\right\}.$$ Now, can you choose the answer?