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Chapter 8: Problem 598

If the harmonic mean and geometric mean of two numbers a and \(b\) are 4 and \(3 \sqrt{2}\) respectively then the interval \([a, b]=\) (A) \([3,6]\) (B) \([2,7]\) (C) \([4,5]\) (D) \([1,8]\)

Short Answer

Expert verified
The correct answer to the question is (A) \([3, 6]\), based on the calculations from the given harmonic mean and geometric mean values.

Step by step solution

01

Harmonic mean formula

The harmonic mean (HM) of two numbers a and b is given by the formula: \[ HM = \frac{2ab}{a+b} \]
02

Geometric mean formula

The geometric mean (GM) of two numbers a and b is given by the formula: \[ GM = \sqrt{ab} \] **Step 2: Use the given harmonic mean and geometric mean values to create equations.**
03

Harmonic mean equation

The given harmonic mean is 4, so using the formula, we have: \[ 4 = \frac{2ab}{a+b} \]
04

Geometric mean equation

The given geometric mean is \(3\sqrt{2}\), so using the formula, we have: \[ 3\sqrt{2} = \sqrt{ab} \] **Step 3: Solve the geometric mean equation for one variable and substitute it into the harmonic mean equation.**
05

Solving for b in geometric mean equation

From the geometric mean equation, we have: \[ ab = (3\sqrt{2})^2 = 18 \] So, we can obtain b in terms of a: \[b = \frac{18}{a}\]
06

Substituting b in harmonic mean equation

Substitute the value of b into the harmonic mean equation: \[ 4 = \frac{2a\left(\frac{18}{a}\right)}{a+\frac{18}{a}} \] **Step 4: Solve the equation for a.**
07

Solving the equation

Simplify and solve for a: \[ 4 = \frac{36}{a + \frac{18}{a}} \] Multiply both sides by a and simplify: \[ 4a^2 = 36(a + \frac{18}{a}) \\ 4a^2 = 36a + 648 \\ 0 = 9a^2 - 36a + 162 \\ \] Solve the quadratic equation: \[a = 3, 6\] **Step 5: Find the corresponding values of b.**
08

Find b for each value of a

Using the relation \(b = \frac{18}{a}\), we find the values of b: \[b_a = \frac{18}{3} = 6\] \[b_b = \frac{18}{6} = 3\] **Step 6: Identify the correct interval.**
09

Identify the interval [a, b]

With the values of a and b we obtained, the smallest value is 3, and the largest value is 6. Thus, the interval [a, b] is: \[[a, b] = [3, 6]\] The correct answer is (A) \([3, 6]\).

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Most popular questions from this chapter

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{A} . \mathrm{P}\). and geometric means of ac and \(\mathrm{ab}, \mathrm{ab}\) and \(b c\), ba nad \(c b\) are \(d\), e, f respectively then \(d^{2}, e^{2}, f^{2}\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G P.

\([1 /(2 \times 5)]+[1 /(5 \times 8)]+[1 /(8 \times 11)]+\ldots 100\) terms (A) \([(25) /(160)]\) (B) \((1 / 6)\) (C) \([(25) /(151)]\) (D) \([(25) /(152)]\)

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.

First term of a G. P. of \(2 \mathrm{n}\) terms is \(\mathrm{a}\), and the last term is 1 . The product of all the terms of the G. P. is (A) \((\mathrm{a} \ell)(\mathrm{n} / 2)\) (B) \((\mathrm{a} \ell)^{(\mathrm{n}-1)}\) (C) \((\mathrm{a} \ell)^{\mathrm{n}}\) (D) \((\mathrm{a} \ell)^{2 \mathrm{n}}\)

6 th term of the sequence \((7 / 3),(35 / 6),[(121) /(12)]\), \([(335) /(24)], \ldots .\) is (A) \([(2113) /(96)]\) (B) \([(2112) /(96)]\) (C) \([(865) /(48)]\) (D) \([(2111) /(96)]\)
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