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Place three numbers in between 15, 31, 104 in such way, that they would be successive members of geometric progression.

PS! I am studying for a test and need help. I would never ask anyone at to do my homework for me. Any help at all would be strongly appreciated. Thank you!

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Are you putting 3 numbers inbetween 15 & 31 and then 3 more numbers between 31 & 104? – Dan W Aug 24 '11 at 18:27
No. 3 new numbers between them. 6 numbers in total including the 3 numbers given. – BeatShot Aug 24 '11 at 18:34
DO you have the right numbers? – Aryabhata Aug 24 '11 at 18:35
$31$ does not belong to any geometric progression other than $\{31^k\}_k$ (since it is prime), so something seems wrong here. Perhaps instead you mean a generalized geometric progression? – JavaMan Aug 24 '11 at 18:43
Sadly, I don't have the answers. – BeatShot Aug 24 '11 at 18:45
up vote 4 down vote accepted

There is no solution. The terms of a geometric series are $ar^i$, where $r$ is the ratio and we will count from $i=0$ for convenience. This gives $a=15$. If there is one term between $15$ and $31$, it must be $\sqrt{15\cdot 31}\approx 21.56$ and the ratio would be about $1.4376$. Then the next terms would be about $44.565, 64.067, 92.103$ and we don't arrive at $104$. Other configurations of where to put the terms can be checked similarly.

Added: to avoid checking cases, from $15$ and $104$ the ratio must be $\sqrt[5]{\frac{104}{15}}\approx 1.47295$ and we don't hit $31$, though we come sort of close with about $32.544$

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What if the numbers needed to insert would be between 31 and 104? (15;31;x;x;x;104) – BeatShot Aug 24 '11 at 18:48
I have to agree. Perhaps there's a typo/mistake in my math book. – BeatShot Aug 24 '11 at 19:10
@BeatShot: if you insert all three between 31 and 104, the ratio would be $\sqrt[4]{\frac{104}{31}}\approx 1.3533$, but $\frac{31}{15}$ is almost $2$ – Ross Millikan Aug 24 '11 at 19:20

There is no geometric progression that contains all of $15$, $31$, and $104$, let alone also the hypothetical "extra" numbers.

For suppose that $15=kr^a$, $31=kr^b$, and $104=kr^c$ where $a$, $b$, $c$ are integers ($r$ need not be an integer, and $a$, $b$, $c$ need not be consecutive). Then $31=kr^ar^{b-a}=15r^{b-a}$. Similarly, $104=31r^{c-b}$. Without loss of generality, we may assume that $r>1$. So $b-a$ and $c-b$ are positive integers.

Let $b-a=m$ and $c-b=n$. Then $$\frac{31}{15}=r^m \qquad \text{and}\qquad \frac{104}{31}=r^n.$$

Take the $n$-th power of $31/15$, and the $m$-th power of $104/31$. Each is $r^{mn}$. It follows that

From this we conclude that $$31^{m+n}=15^n \cdot 104^m.$$ This is impossible, since $5$ divides the right-hand side, but $5$ does not divide the left-hand side.

Comment: Have we really answered the question? It asks us to place $3$ numbers between $15$, $31$, $104$ "in such a way that they would be successive members of a geometric progression." Who does "they" refer to? Certainly not all $6$ numbers, since already as we have seen, $15$, $31$, and $104$ cannot all be members of a (single) geometric progression of any kind.

But maybe "they" refers to the interpolated numbers! Then there are uncountably many solutions, and even several solutions where the interpolated numbers are all integers. For example, we can use $16$, $32$, $64$. The numbers $15$ and $31$ could be a heavy-handed hint pointing to this answer.

Or else we can use $16$, $24$, $36$. Or else $16$, $40$, $100$. Then there is $18$, $24$, $32$, or $18$, $30$, $50$, and so on.

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Thank you so much for this detailed answer. According to the exercise all 6 numbers (including the 3 numbers given) must be successive members of geometric progression. As pointed out before there are no solutions and it's impossible. Only conclusion I can make from this, is that there must be a mistake in the book. Thanks again. – BeatShot Aug 24 '11 at 20:46
@BeatShot: When you are asking a question on Stack Exchange, it is helpful if the exact language of the motivating question is used. It sounds reasonable to summarize, but that can introduce ambiguities. My answer does show that the $6$ can't be in any GP, but it is overkill. – André Nicolas Aug 24 '11 at 20:57

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