<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\Ans{Let, $$x=\sqrt{N}$$
	$$x^2-N=0$$
	$$f(x)=x^2-N$$
	$$f'(x)=2x$$
	By Newton-Raphson,
	$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
	$$x_{n+1}=x_n-\frac{{x_n}^2-N}{2x_n}$$
	$$x_{n+1}=x_n-\frac{x_n}{2}-\frac{N}{2x_n}$$
	$$x_{n+1}=\frac{x_n}{2}+\frac{N}{2x_n}$$
	Suppose, $$x_n=\frac{A+B}{2}$$
	So, $$x_{n+1}=\frac{A+B}{2*2}+\frac{N}{A+B}=\frac{S}{4}+\frac{N}{S}$$
	$$\therefore S=A+B$$}</pre></body></html>