+10 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 $a_{10}$.
Let the common difference of the arithmetic sequence be d. Then [a_1 + a_3 + a_5 = 3a_2 = 3 \left( a_1 + d \right) = -12]and [a_1 a_3 a_5 = a_2^3 = \left( a_1 + d \right)^3 = a_1^3 + 3a_1^2 d + 3a_1 d^2 + d^3 = 80.]We can solve the first equation for d to get d=−4. Substituting this into the second equation, we get [a_1^3 - 48a_1 - 64 = 80]or [a_1^3 - 48a_1 - 144 = 0.]This factors as [(a_1 - 8)(a_1 + 18)(a_1 + 2) = 0,]so a1=8, a1=−18, or a1=−2. The corresponding values of a10 are -16,-10, and 4.
Since a1+a3+a5=−12, we can write
$a_1 + a_3 + a_5 = 3 \cdot \frac{a_1 + a_3 + a_5}{3} = 3 \cdot \frac{a_1 + (a_1 + d) + (a_1 + 2d)}{3} = 3a_1 + d^2 = -12$.
Then d2=−36, so d=±6. If d=6, then a1+a3+a5=0, which is a contradiction, so d=−6. Then
a_{10} = a_1 + 9d = a_1 - 54 = -12 - 54 = -66.
