Answer
$v_0-\frac{g}{2}(t_2+t_1)$
Work Step by Step
The average velocity over [t1, t2] is
$\frac{s(t_2)-s(t_1)}{t_2-t_1}$ = $\frac{(s_0+v_0t_2-\frac{1}{2}gt_2^{2})-(s_0+v_0t_1-\frac{1}{2}gt_1^{2})}{t_2-t_1}$ = $\frac{v_0(t_2-t_1)+\frac{g}{2}(t_2^{2}-t_1^{2})}{t_2-t_1}$ = $v_0-\frac{g}{2}(t_2+t_1)$
the average of the instantaneous velocities at the beginning and end of [t1, t2] is
$\frac{s'(t_1)+s'(t_2)}{2}$ = $\frac{1}{2}[(v_0-gt_1)+(v_0-gt_2)]$ = $v_0-\frac{g}{2}(t_2+t_1)$
