Inserting numbers to create a geometric progression Place three numbers in between 15, 31, 104 in such way, that they would be successive members of geometric progression. 
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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 $32.544$
