Find the sum of the first ten terms

Given: $a_5=-8.5$ and $a_n=-71$

$a_n=a_m+(n-m)d$

$-71=-8.5+(n-5)d$

$-71=-8.5+dn-5d$

$-62.5=dn-5d$

$d=\dfrac{-62.5}{n-5}$

Compute $a_1$ :

$a_5=a_1+4d$

$-8.5=a_1+4\left(\dfrac{-62.5}{n-5}\right)$

$a_1=\dfrac{250}{n-5}-8.5$

Compute $a_{10}$ :

$a_{10}=a_1+9d=\dfrac{250}{n-5}-8.5+9\left(\dfrac{-62.5}{n-5}\right)=\dfrac{250}{n-5}-8.5-\dfrac{562.5}{n-5}=\dfrac{-270-8.5}{n-5}$

Compute the sum of the first ten terms:

$s_{10}=\dfrac{10}{2}(a_1+a_{10})=5\left(\dfrac{250}{n-5}-8.5+\dfrac{-270-8.5}{n-5}\right)=5\left(\dfrac{250-8.5n+42.5-270-8.5n}{n-5}\right)=5\left(\dfrac{22.5-17n}{n-5}\right)=$ $\boxed{\dfrac{112.5-85n}{n-5}}$