A) \[{{a}_{\infty }}=\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{{{\cos }^{-1}}\left( \frac{a}{b} \right)}\]
B) \[{{b}_{\infty }}=\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{{{\cos }^{-1}}\left( \frac{a}{b} \right)}\]
C) \[{{b}_{\infty }}=\frac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{{{\cos }^{-1}}\left( \frac{b}{a} \right)}\]
D) None of these
Correct Answer: B
Solution :
Given that a and b are positive quantities and\[{{a}_{1}}=\frac{a+b}{2},{{b}_{1}}=\sqrt{{{a}_{1}}b,}{{a}_{2}}=\frac{{{a}_{1}}+{{b}_{1}}}{2},{{b}_{2}}=\sqrt{{{a}_{2}}{{b}_{1}}},...\] Let, \[a=b\cos \theta \] Then, \[{{a}_{1}}=\frac{a+b}{2}=\frac{b(1+\cos \theta )}{2}=b{{\cos }^{2}}\frac{\theta }{2}\] \[\Rightarrow \] \[{{b}_{1}}=\sqrt{{{a}_{1}}b}=b\cos \frac{\theta }{2}\] Now, \[{{a}_{2}}=\frac{{{a}_{1}}+{{b}_{2}}}{2}=b\cos \frac{\theta }{2}\cdot {{\cos }^{2}}\frac{\theta }{4}\] \[\therefore \] \[{{b}_{2}}=b\cos \frac{\theta }{2}\cdot \cos \frac{\theta }{{{2}^{2}}}\] Similarly, \[{{b}_{3}}=b\cos \frac{\theta }{2}\cdot \cos \frac{\theta }{{{2}^{2}}}\cdot \cos \frac{\theta }{{{2}^{3}}}\]and so on. Now, \[{{b}_{n}}=b\cos \frac{\theta }{2}\cdot \cos \frac{\theta }{{{2}^{2}}}...\cos \frac{\theta }{{{2}^{n}}}\] \[\therefore \]\[{{b}_{\infty }}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{b\sin \theta }{{{2}^{n}}\sin \left( \frac{\theta }{{{2}^{n}}} \right)}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{b\sin \theta }{\theta \left( \frac{\sin (\theta /{{2}^{n}})}{\theta /{{2}^{n}}} \right)}\] \[=b\frac{\sin \theta }{\theta }=\frac{b\sqrt{1-{{\cos }^{2}}}\theta }{{{\cos }^{-1}}(a/b)}\]\[\therefore \]\[{{b}_{\infty }}=\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{{{\cos }^{-1}}(a/b)}\]
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