Puzzle...

Let the numbers be $a/r^2, a/r,a,ar,ar^2$

The sum is now given as $a(1/r^2+1/r+1+r+r^2)=6$

Sum of reciprocals is given by $(1/r^2+1/r+1+r+r^2)/a=3$

Dividing the 2 equations, we get $a^2=2$

The product is given as $a^5$ which is $4\sqrt{2}$

Solution 1:

Let the numbers be a/r², a/r, a, ar, ar²

The reciprocals are r²/a, r/a,1/a, 1/ar, 1/ar²

Sum of the numbers = a/r² (r^5-1)/(r-1)

Sum of reciprocals = r²/a (1-1/r^5)/(1-1/r)

= r²/a [(r^5-1)/r^5]/[(r-1)/r] = (r^5-1)/(r-1)* r²/a * 1/r^4

= a/r² (r^5-1)/(r-1) * 1/a² = 6/a² = 3

=> a² = 2

Product = a^5 = 4√2