A point on the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$  at a distance equal to the arithmetic mean of the length of the semi-major axis and semi-minor axis from the centre is 

Lengths of semi-major axis and semi-minor axis of the ellipse  $\frac{x^2}{16}+\frac{y^2}{9}=1$ are 4 and 3 respectively. So that the mean of these length is  $\frac{7}{2}$. Let the co-ordinates of any point on the ellipse be  $P(4\cos \theta ,4\sin \theta )$ . If the distance of p from the centre O(0,0) of the ellipse is  $\frac{7}{2}$,  then  $16{\cos }^2\theta +9{\sin }^2\theta =\frac{49}{4}\Rightarrow 28{\cos }^2\theta =13$ $\Rightarrow \cos \theta =\pm \sqrt{\frac{13}{28}}=\pm \frac{\sqrt{91}}{14}and\sin \theta =\pm \frac{\sqrt{105}}{14}$ 
 
So, the co-ordinates the required point is  $(\pm \frac{4\sqrt{91}}{14},\pm \frac{3\sqrt{105}}{14})i.e.(\pm \frac{2\sqrt{91}}{7},\pm \frac{3\sqrt{105}}{14})$