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 know 17 is a solution but there are apparently multiple solutions to this.
An arithmetic sequence is a sequence of numbers such that the difference between any two consecutive terms is constant. Let's call this constant "d". Then, the nth term of the sequence can be found by a_n = a_1 + (n-1)d.
Since a_1 + a_3 + a_5 = -12, we have: a_1 + (a_1 + 2d) + (a_1 + 4d) = -12 3a_1 + 6d = -12 3a_1 = -12 - 6d a_1 = -4 - 2d
Also, we have a_1 * a_3 * a_5 = 80, so: (-4 - 2d) * ( -4 - 2d + 2d) * (-4 - 2d + 4d) = 80 Expanding this expression and simplifying, we get: d^2 - 8d + 15 = 0
This is a quadratic equation, and we can use the quadratic formula to solve for d: d = (-b ± √(b^2 - 4ac)) / 2a where a = 1, b = -8, and c = 15.'
Substituting these values, we get: d = (8 ± √(8^2 - 4 * 1 * 15)) / 2 * 8 d = 3 d = 1
The possible values of d are then 3 and 1.
If d = 3, then a = -10, so a_{10} = 17.
If d = 1, then a = 2, so a_{10} = 11.
The possible values of a_{10} are 17 and 11.
+2668 Let the starting term be s and the common difference be d.
Then \(a_3 = s + 2d\) and \(a_5 = s + 4d\).
We know that \(3s + 6d = -12\), meaning that \(s + 2d = -4 = a_3\)
This means that \(a_1 + a_5 = -8\) and their product is \(80 \div -4 = -20\).
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So, we have the system \(a_1 + a_5 = -8\) and \(a_1 \times a_5 = -20\).
Solving these gives us \((a_1, a_5) = (-10, 2), (2, -10)\).
In the first case, the starting term is -10 and the common difference is 3, so the 10th term is \(-10 + 3 \times 9 = \color{brown}\boxed{17}\).
In the second case, the starting term is 2 and the common difference is -3, so the 10th term is \(2 - (3 \times 9) = \color{brown}\boxed{-25}\)
