• # question_answer If a, b are positive quantities and if ${{a}_{1}}=\frac{a+b}{2},{{b}_{1}}=\sqrt{{{a}_{1}}b,}{{a}_{2}}=\frac{{{a}_{1}}+{{b}_{1}}}{2},{{b}_{2}}=\sqrt{{{a}_{2}}{{b}_{1}}}$... and so on, then 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

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)}$