Need help to prove an identity about arithmetic and geometric progression

Question

Suppose $a=a_5,b=a_{17},c=a_{37}$ of an arithmetic sequence are also geometric sequence .

How can prove $$a^{b-c}\times b^{c-a}\times c^{a-b}=1$$ ?
I tried to find geometric common ratio, that is $q=\frac{37-17}{17-5}=\frac{20}{12}=\frac{5}{3}$ and plug into formula ...but I am lost in calculations. I tried also to write $$a,b,c\\\frac{k}{q},k,kq$$ but I get stuck on this . I do appreciate your help ?(or any other idea)

Answer 1

Using condition of geometric progression we have: $ac=b^2$, so that $$\log(a)+\log(c) = 2\log(b)$$

Then lets assume:

$$a^{b-c}\times b^{c-a}\times c^{a-b}=k\\ (b-c)\log(a) + (c-a) \log(b) +(a-b)\log(c) = \log(k)$$

Let the arithmetic progression have common difference $d$ between its terms. Then we can say $$a-b=-12d;\ b-c = -20;\ c-a = 32d$$

Using this, our equation becomes:

$$-20d\log(a) +32d \log(b) -12d\log(c) = \log(k)\\ 8d \log\left(\frac{b}{a}\right) = \log(k)$$

If $k$ has to be $1$, we must have $d=0$, or identically $\frac{b}{a} = 1$.

Answer 2

HINT:

We have $\dfrac ba=\dfrac cb=k$(say) $\implies a=bk,c=ak^2$

Replace the values of $b,c$ in the base

As $b=A+(17-1)D,a=a+(5-1)D$ etc. where $A,D$ are the first term & the common difference respectively.

$b-a=?$

The values of $b-c$ etc. in the exponent.

Answer 3

Hint:

We have $\dfrac ba=\dfrac cb=k$(say) $\implies a=bk,c=ak^2$

$$\left(\dfrac cb\right)^a\left(\dfrac ac\right)^b\left(\dfrac ba\right)^c=\dfrac{k^ak^c}{(k^2)^b}=k^{a+c-2b}$$

So, we need $a+c-2b=0\iff a,b,c$ in arithmetic progression as well.