The question is to find out the longest geometric progression (the common ratio $r \neq 1$) that can be obtained from the set $(100,101,102,...,1000)$
The common ratio must be greater than $1$ and for the common ratio $2$ the number of terms is $4$ which starts from $100$ I have no idea on how to proceed with this.Any thoughts?Thanks.
$128, 192, 288, 432, 648, 972$ is a geometric progression with common ratio $r = 3/2$. I'll prove this is the unique best possibility, up to reversing the order.
In general, the common ratio will be a reduced fraction $p/q$ (we can assume $p > q$). Suppose the progression has length $k+1$, then the initial term will be $aq^{k}$ and the final term will be $ap^{k}$, for some integer $a$.
Then $ap^{k} \leq 1000$. Since $p \geq q + 1$ and $a \geq 1$, this means $(q + 1)^{k} \leq 1000$. We also have $(p/q)^{k} \leq 10$, meaning $((q + 1)/q)^{k} \leq 10$. If $k = 6$, the first equation requires $q \leq 2$ while the second requires $q \geq 3$, meaning there are no solutions.
If $k = 5$, the first equation requires $q \leq 2$ and the second requires $q \geq 2$, so we must have $q = 2$, and $p^5 \leq 1000$ gives $p = 3$. Then the first term is $32a$ (meaning $a \geq 4$) and the last term is $243a$ (meaning $a \leq 4$), so $a = 4$ giving the above sequence.
This is related to a question asked in Canadian Mathematical Olympiad, 1972, p. 10 and I am only expanding the answer of Alex Zorn for the sake of better understanding and my self satisfaction.
