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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 $a_{10}$.

Guest May 5, 2023

gb1falcon · Answer 1 · 2023-05-05T04:23:03

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.

Guest · Answer 2 · 2023-05-08T03:30:18

The possible values of a_{10} are -5 and 2.