What is the largest number of members that a geometric progression can have, whose members are various natural numbers, greater than $210$ and less than $350$?

I'm not really sure what method to use solving this. sorry i know it doesn't look that hard but i couldn't do it


1 Answer 1


First note that we need to exclude the case $r=1$ which gives a sequence with infinitely many identical terms.

Also note that there is a seqence 216,252,294,343 with 4 terms.

Suppose there is a sequence with 5 terms. By reversing the sequence if necessary we can suppose $r>1$. Let $r=\frac{u}{v}$, where $u$ and $v$ are coprime positive integers.

Then the sequence can be written as $av^4,av^3u, ..., au^4$. We therefore require $u^4$ to be less than 350. Therefore $u\le 4$ and $v\le 3$. We also require $(\frac{u}{v})^4<\frac{350}{210}$ and therefore $\frac{u}{v}<1.2$. This is impossible and so we already have a sequence of maximum length.


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