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 $a_{10}$.
I am unsure, is this some type of system of equations question?
Yes, it is.
Let the starting term be \(a\) and let the common difference be \(d\).
The \(n_{\text {th}}\) term of this sequence is \(a + d(n-1)\). This means that the starting (1st) term is \(a\), and the second term is \(a + d\)
Thus, we can form the following system: \(a+(a+2d)+(a+4d) = -12\) and \(a(a+2d)(a+4d) = 80\)
Now, we just have to solve for the 10th term, or \(a + 9d\).
Can you take it from here?