Question

If $a,b(a<b)$ 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. $a_{\infty }=\frac{\sqrt{b^2+a^2}}{{\cos }^{-1}(\frac{a}{b})}$
• B. $a_{\infty }=\frac{\sqrt{a^2+b^2}}{{\cos }^{-1}(\frac{b}{a})}$
• C. $a_{\infty }=\frac{\sqrt{a^2+b^2}}{{\cos }^{-1}(\frac{a}{b})}$
• D. $b_{\infty }=\frac{\sqrt{b^2-a^2}}{{\cos }^{-1}(\frac{a}{b})}$

Answer 1

The correct option is D $b_{\infty }=\frac{\sqrt{b^2-a^2}}{{\cos }^{-1}(\frac{a}{b})}$

Let $a=b\cos \theta$
Given $a_1=\frac{a+b}{2}$
$a_1=\frac{b\cos \theta +b}{2}=b{\cos }^2\frac{\theta }{2}$
$b_1=\sqrt{a_{1}b}=\sqrt{b^2{\cos }^2\frac{\theta }{2}}=b\cos \frac{\theta }{2}$
Given $a_2=\frac{a_1+b_1}{2}=\frac{b{\cos }^2\frac{\theta }{2}+b\cos \frac{\theta }{2}}{2}=\frac{b\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+1)}{2}=b\cos \frac{\theta }{2}{\cos }^2\frac{\theta }{4}$
$b_2=\sqrt{a_2{b}_1}$

$=\sqrt{b^2{\cos }^2\frac{\theta }{2}{\cos }^2\frac{\theta }{4}}$
$b_2=b\cos \frac{\theta }{2}\cos \frac{\theta }{4}$
Similarly, $b_3=b\cos \frac{\theta }{2}\cos \frac{\theta }{4}\cos \frac{\theta }{8}$
So,$b_n=b\cos \frac{\theta }{2}\cos \frac{\theta }{4}\cos \frac{\theta }{8}......\cos \frac{\theta }{2^n}$
Now, $b_{\infty }=\underset{n\to \infty }{lim}{b}_n$

we can reduce $b_n=\frac{b\cos \frac{\theta }{2}\cos \frac{\theta }{4}\cos \frac{\theta }{8}......2\sin \frac{\theta }{2^n}\cos \frac{\theta }{2^n}}{2\sin \frac{\theta }{2^n}}=\frac{b\cos \frac{\theta }{2}\cos \frac{\theta }{4}\cos \frac{\theta }{8}......\cos \frac{\theta }{2^{n-1}}}{2\sin \frac{\theta }{2^n}}$
and thus reducing so on, we get
$=\underset{n\to \infty }{lim}\frac{\left(\frac{\theta }{2^n}\right)b\sin \theta }{\theta \sin (\frac{\theta }{2^n})}=\frac{b\sin \theta }{\theta }=\frac{\sqrt{1-\frac{a^2}{b^2}}}{{\cos }^{-1}\left(\frac{a}{b}\right)}=\frac{\sqrt{b^2-a^2}}{{\cos }^{-1}\left(\frac{a}{b}\right)}$