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Chapter 8: Problem 670

If \(\mathrm{a}, 4, \mathrm{~b}\) are in \(\mathrm{A} . \mathrm{P}\). and \(\mathrm{a}, 2, \mathrm{~b}\) are in G. P. then \((1 / \mathrm{a}), 1\), \((1 / \mathrm{b})\) are in (A) G. P. (B) A. P. (C) H. P. (D) A. G. P.

Short Answer

Expert verified
(A) A.P.

Step by step solution

01

Use the A.P. condition on a, 4, b

Since a, 4, and b are in A.P., the average of a and b is equal to 4. We can write this as: \[(a + b)/2 = 4\]
02

Use the G.P. condition on a, 2, b

Since a, 2, and b are in G.P., the square of the middle term (2) is equal to the product of a and b. We can write this as: \[2^2 = a \cdot b\] Which simplifies to, \[4 = a \cdot b\]
03

Solve for a and b

We now have two equations: 1. \((a + b)/2 = 4\) 2. \(4 = a \cdot b\) Solve equation 1 for a: \[a = 8 - b\] Substitute the value of a in equation 2: \[4 = (8 - b) \cdot b\] This is a quadratic equation, so we can rewrite it as: \[b^2 - 8b + 4 = 0\] Solve for b: \(b = 4 \pm 2\sqrt{3}\) But as b is in G.P. with a, its value must be positive and |a| is larger in G.P.: \(b = 4 + 2\sqrt{3}\) Then we find the value of \(a = 8 - (4 + 2\sqrt{3})\) Which simplifies to, \(a = 4 - 2\sqrt{3}\)
04

Examine the sequence (1/a, 1, 1/b)

Now let's examine the sequence: \[\left(\frac{1}{a}, 1, \frac{1}{b}\right)\] Generating the new sequence: \[\left(\frac{1}{4 - 2\sqrt{3}}, 1, \frac{1}{4 + 2\sqrt{3}}\right)\] Now we can examine this sequence. If they are in A.P., G.P., or H.P., we can express the sequence relationship as one of the following: - A.P.: \((1 / a) + (1 / b) = 2\) - G.P.: \[(1 / a) \cdot (1 / b) = 1^2\] - H.P.: \[((1 / a) + (1 / b)) / 2 = 1 / 1\] For A.P., we have: \[\frac{1}{4 - 2\sqrt{3}} + \frac{1}{4 + 2\sqrt{3}} = 2\] This simplifies to: \[\frac{(4 + 2\sqrt{3}) + (4 - 2\sqrt{3})}{16 - 12} = 2\] Which is true. So, \((1 / a), 1, \)and \( (1 / b)\) are in A.P. Therefore, the correct answer is: (A) A.P.

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