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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 $a_{10}$.

 May 5, 2023
 #1
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Say the common ratio is r. Then a14r=a5, a12r=a3, a1+a3+a5=3a16rand a1a3a5=a1(a12r)(a14r)

This is a system of equations. The first one reveals a1=2r4. The second one then becomes

(2r4)4(42r)=80

242(r2)(r+2)=80

(r2)(r+2)=5

r=3

So a1=2, a2=5, a3=8, a5=14, a10=29

So the answer you are looking for is 29.

 May 5, 2023
 #2
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The possible values of a_{10} are -5 and 2.

 May 8, 2023

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