If a,b are positive quantitis and if `a_(1)=(a+b)/(2), b_(1)=sqrt(a_(1)b) , a_(2)=(a_(1)+b_(1))/(2), b_(2)=sqrt(a_(2)b_(1))` and so on then

Options

A. `a_(oo)=sqrt(b^(2)-a^(2))/(cos^(-1)(a/b))`
B. `b_(oo)=sqrt(b^(2)-a^(2))/(cos^(-1)(a/b))`
C. `b_(oo)=sqrt(a^(2)+b^(2))/(cos^(-1)(b/a))`
D. none of these

Answer

To solve the problem step by step, we will analyze the given sequences \( a_n \) and \( b_n \) defined as follows:

1. **Definitions**:
- \( a_1 = \frac{a + b}{2} \)
- \( b_1 = \sqrt{a_1 b} \)
- \( a_2 = \frac{a_1 + b_1}{2} \)
- \( b_2 = \sqrt{a_2 b_1} \)
- And so on...

2. **Assumption**:
- Let \( a = b \cos \theta \) for some angle \( \theta \).

3. **Calculate \( a_1 \)**:
\[ a_1 = \frac{a + b}{2} = \frac{b \cos \theta + b}{2} = \frac{b(1 + \cos \theta)}{2} \]

4. **Using the identity for cosine**:
We know that:
\[ \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) - 1 \implies \cos^2 \left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{2} \]
Therefore:
\[ a_1 = b \cos^2 \left(\frac{\theta}{2}\right) \]

5. **Calculate \( b_1 \)**:
\[ b_1 = \sqrt{a_1 b} = \sqrt{b \cos^2 \left(\frac{\theta}{2}\right) \cdot b} = b \cos \left(\frac{\theta}{2}\right) \]

6. **Calculate \( a_2 \)**:
\[ a_2 = \frac{a_1 + b_1}{2} = \frac{b \cos^2 \left(\frac{\theta}{2}\right) + b \cos \left(\frac{\theta}{2}\right)}{2} = \frac{b \left(\cos^2 \left(\frac{\theta}{2}\right) + \cos \left(\frac{\theta}{2}\right)\right)}{2} \]
Factoring out \( b \):
\[ a_2 = b \cdot \frac{\cos^2 \left(\frac{\theta}{2}\right) + \cos \left(\frac{\theta}{2}\right)}{2} \]

7. **Using the identity again**:
\[ 1 + \cos \left(\frac{\theta}{2}\right) = 2 \cos^2 \left(\frac{\theta}{4}\right) \implies a_2 = b \cdot \frac{2 \cos^2 \left(\frac{\theta}{4}\right)}{2} = b \cos^2 \left(\frac{\theta}{4}\right) \]

8. **Calculate \( b_2 \)**:
\[ b_2 = \sqrt{a_2 b_1} = \sqrt{b \cos^2 \left(\frac{\theta}{4}\right) \cdot b \cos \left(\frac{\theta}{2}\right)} = b \cos \left(\frac{\theta}{4}\right) \cos \left(\frac{\theta}{2}\right) \]

9. **Continuing this pattern**:
- We can see that:
\[ a_n = b \cos^2 \left(\frac{\theta}{2^{n-1}}\right) \]
\[ b_n = b \prod_{k=1}^{n-1} \cos \left(\frac{\theta}{2^k}\right) \]

10. **Finding the limit as \( n \to \infty \)**:
- As \( n \to \infty \), \( \cos \left(\frac{\theta}{2^n}\right) \to 1 \).
- Therefore:
\[ b_n \to b \cdot \prod_{k=1}^{\infty} \cos \left(\frac{\theta}{2^k}\right) \]

11. **Using the sine identity**:
\[ \prod_{k=1}^{\infty} \cos \left(\frac{\theta}{2^k}\right) = \frac{\sin \theta}{\theta} \]

12. **Final result**:
\[ b_\infty = b \cdot \frac{\sin \theta}{\theta} \]

13. **Expressing \( \theta \)**:
- Since \( \theta = \cos^{-1} \left(\frac{a}{b}\right) \), we can express \( b_\infty \) in terms of \( a \) and \( b \):
\[ b_\infty = b \cdot \frac{\sqrt{b^2 - a^2}}{b} = \sqrt{b^2 - a^2} \]

Thus, the final answer is:
\[ b_\infty = \frac{\sqrt{b^2 - a^2}}{\cos^{-1} \left(\frac{a}{b}\right)} \]