Reciprocal or its reciprocal?

Great! Actually, we could use the intermediate value theorem here.

(Taking positive $r$ and positive $a$ ,) The necessary conditon is $a = \sqrt{ \frac{3}{r^{11} } }$ . When $r = 1$ , we have $\sqrt{ \frac{3}{r^{11} } } r ( 1 + r + r^2 + \ldots + r^9 ) = \sqrt{3}$ .
Since the term $\frac{ r^ { 10 } } { \sqrt{ r^ { 11 } } }$ dominates, we know that as $r \rightarrow \infty$ , the value of the expression tends to infinity.
In particular, for $r = 2$ , $\sqrt{ \frac{3}{r^{11} } } r ( 1 + r + r^2 + \ldots + r^9 ) > 18$ .

Hence, there is some solution between $r = 1$ and $r = 2$ .

We could apply a similar argument to show that there is some solution between $r = 1$ and $r= 0.5$ .

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Wait, that would mean that ar/a=1/1=1 but ar^2/ar=2, this means that r=2, and 1. You can try to make them in a different order like 1222122222 but r=2/2, 1/2, and 2.Since the only possible orders in which we can see different possible r's are when the 1's are between 2's like 2222211222 or 2221222122 in which case we can see that r=1, r=1/2, and r=2.In the other case, we have the one arrangement of 1122222222 in which case the 1's are in front of the 2's, making sure that r can not be equal to 1/2, we can still see that r can equal to 2, or 1.Also, we can be sure that there can be no more possible r's as each possible r can be in to form of one term divided by the term right below it, the possible cases are 2/1, 2/2, 1/1 and 1/2 which is 1/2, 1, and 2, and now we can say that the set 1,1,2,2,2,2,2,2,2,2 can not satisfy this geometric sequence.