Let $a_1, a_2, a_3,\dots$ be an arithmetic sequence. If $a_1 + a_3 + a_5 = -12$ and $a_1a_3a_5 = 80$, find all possible values of $a_{10}$.
Say the common ratio is r. Then a1−4r=a5, a1−2r=a3, a1+a3+a5=3a1−6rand a1a3a5=a1(a1−2r)(a1−4r)
This is a system of equations. The first one reveals a1=2r−4. The second one then becomes
(2r−4)⋅−4⋅(−4−2r)=80
2⋅−4⋅−2⋅(r−2)(r+2)=80
(r−2)(r+2)=5
r=3
So a1=2, a2=5, a3=8, a5=14, a10=29
So the answer you are looking for is 29.