I recently got the question:
Is there a geometric sequence that contains each of the numbers 1, 2 and 3?
I have tried my best to make a start with the $x_n=ar^{(n-1)}$ theorem but I'm still puzzled!
On stack exchange I have also seen this very similar question but I don't see how it really works...
Any help would be greatly appreciated! :)
Assume that $\{a_n\}_{n\in\mathbb{N}}$ is a geometric progression, $\frac{a_{n+1}}{a_n}=r$, and $$ a_m = 1,\qquad a_{m+i}=2,\qquad a_{m+j}=3. $$ That implies $3=r^j$ and $2=r^{i}$, hence: $$ 3^i = r^{ij}$$ $$ 2^j = r^{ij}$$ but that is equivalent to: $$ 3^i = 2^j $$ but that is only possible when $i=j=0$ but it is not possible because if $i=j=0$ then both $ a_{m+i}=2,a_{m+j}=3 $ become $a_m$ but $a_m$ cannot be 1, 2 and 3 simultaneously
