Consider a parabola y=x24 and the point F(0,1). Let A1(x1,y1),A2(x2,y2),A3(x3,y3),...,Ak(xk,yk) are 'n' points on parabola such as xk>0 and ∠OFAk=kπ2n,(k=1,2,3,...,n). Then the value of limn→∞1nn∑k=1FAk is

Consider a parabola y=x24 and the point F(0,1). Let A1(x1,y1),A2(x2,y2

1.	Consider a parabola y=x24 and the point F(0,1). Let A1(x1,y1),A2(x2,y2),A3(x3,y3),...,Ak(xk,yk) are 'n' points on parabola such as xk>0 and ∠OFAk=kπ2n,(k=1,2,3,...,n). Then the value of limn→∞1nn∑k=1FAk is
Answer» Consider a parabola $y = \frac{x^{2}}{4}$ and the point $F (0, 1)$ . Let $A_{1} (x_{1}, y_{1}), A_{2} (x_{2}, y_{2}), A_{3} (x_{3}, y_{3}), . . ., A_{k} (x_{k}, y_{k})$ are $' n'$ points on parabola such as $x_{k} > 0$ and $∠ O F A_{k} = \frac{k π}{2 n}, (k = 1, 2, 3, . . ., n)$ . Then the value of $lim n \to \infty \frac{1}{n} n \sum k = 1 F A_{k}$ is