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 $a_{10}$.
+214 Say the common ratio is r. Then \(a_1-4r=a_5, \space a_1-2r=a_3, \space a_1+a_3+a_5=3a_1-6r\)and \(a_1a_3a_5=a_1(a_1-2r)(a_1-4r)\)
This is a system of equations. The first one reveals \(a_1=2r-4\). The second one then becomes
\((2r-4)\cdot-4\cdot(-4-2r)=80\)
\(2\cdot-4\cdot-2\cdot(r-2)(r+2)=80\)
\((r-2)(r+2)=5\)
\(r=3\)
So \(a_1=2, \space a_2=5, \space a_3=8, \space a_5=14, \space a_{10}=29\)
So the answer you are looking for is 29.
