If $a_3=15,S_{10}=125$, then d and $a_{10}=$

$a_3=15and{S}_{10}=125$

$a+2d=15\_\_\_\_\left(1\right)$

$\frac{10}{2}[2a+9d]=125\Rightarrow 2a+9d=25\_\_\_\_\left(2\right)$

Solving (1) & (2) we get $d=-1,a=17$

Now $a_{10}=a+9d=17+9(-1)=17-9=8$