In my time zone, (UTC +8), it's the night of the 5th day to 2015, so here comes the first question...

Given four terms $a_1$ , $a_2$ , $a_3$ and $a_4$ .

Let $a_1$ , $a_2$ , $a_3$ be a arithmetic progression with a product of 20. (i.e. $a_1a_2a_3=20$ )

Let $a_2$ , $a_3$ , $a_4$ be a geometric progression with a sum of 15. (i.e. $a_2+a_3+a_4=15$ )

A closed form is pretty hard to get (even wolfram alpha spits at me...) so something else has to be asked...

Given that $a_1$ , $a_2$ , $a_3$ and $a_4$ are in the set of real numbers, and that $a_1<a_2<a_3<a_4$ , find the H.M. of $a_1$ , $a_2$ , $a_3$ and $a_4$ .

Round the answer off to three significant decimal digits.

A closed form is appreciated (although most likely complicated).

The answer is 0.717.

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