Answer :
Let [tex]a_n[/tex] denote the [tex]n[/tex]th term in the sequence. Then
[tex]a_n=a_{n-1}+d[/tex]
[tex]a_n=(a_{n-2}+d)+d=a_{n-2}+2d[/tex]
[tex]a_n=(a_{n-3}+d)+2d=a_{n-3}+3d[/tex]
and so on down to
[tex]a_n=a_1+(n-1)d[/tex]
where [tex]d[/tex] is the common difference between terms in the sequence.
This means we have
[tex]a_6=a_1+5d[/tex]
[tex]a_9=a_1+8d[/tex]
Given that [tex]a_6+a_9=20[/tex] and [tex]a_6a_9=64[/tex], we can substitute the relationships above to solve for [tex]a_1[/tex]:
[tex]\begin{cases}(a_1+5d)+(a_1+8d)=20\\(a_1+5d)(a_1+8d)=64\end{cases}[/tex]
[tex]\implies\begin{cases}2a_1+13d=20\\{a_1}^2+13a_1d+40d^2=64\end{cases}[/tex]
From the first equation, we get
[tex]d=\dfrac{20-2a_1}{13}[/tex]
Substitute this into the second equation and solve for [tex]a_1[/tex]:
[tex]{a_1}^2+a_1(20-2a_1)+\dfrac{40}{169}(20-2a_1)^2=64[/tex]
[tex]\implies{a_1}^2-20a_1-576=(a_1-36)(a_1+16)=0[/tex]
from which it follows that the first term is [tex]a_1=-16[/tex] (because we're told it's negative).
Then the common difference is [tex]d=\frac{20-2(-16)}{13}=4[/tex], and so the 10th term in the sequence is
[tex]a_{10}=a_1+9d=-16+9\cdot4=\boxed{20}[/tex]