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$a,b$ and $c$ are in geometric progression with $a^p=b^q=c^r$. What is the relation between $p,q$ and $r$?

I have three options:

  1. $p,q,r$ are in geometric progression
  2. $p,q,r$ are in arithmetic progression
  3. $1/p,1/q,1/r$ are in arithmetic progression

Any help with how to find out the solution will be appreciated.

2 Answers 2

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Let $$a^p=b^q=c^r=k\\ \Rightarrow a=k^{\frac 1p}, b=k^{\frac 1q}, c=k^{\frac 1r}$$ For $a,b,c$ to be in a GP, $$ \frac ba=\frac cb\\ k^{\frac 1q-\frac 1p}=k^{\frac 1r-\frac 1q}\\ \frac 1q-\frac 1p=\frac 1r-\frac 1q$$ i.e. $\frac 1p, \frac 1q, \frac 1r$ are in AP.

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if $a,b,c$ in a GP, then we have $$c:b=b:a$$ with $$a=t^{1/p},b=t^{1/q},c=t^{1/r}$$ we get $$t^{1/r-1/q}=t^{1/q-1/p}$$ comparing the exponents we have $$\frac{1}{r}+\frac{1}{p}=\frac{1}{2q}$$ if $$p,q,r$$ in a AP then we have $$r+p=2q$$ for the last question we have $$\frac{2}{q}=\frac{1}{r}+\frac{1}{p}$$

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    should we use the condition $$a^p=b^q=c^r$$ again? – 2017-02-02