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Chapter 10: Problem 95

If \(a, b, c\) are positive numbers in G.P. and log \(\left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)\) and \(\log \left(\frac{a}{3 b}\right)\) are in A.P. then \(a, b, c\) (A) form the sides of an equilateral triangle (B) form the sides of an isosceles triangle (C) form the sides of a right angled triangle (D) can not form the sides of a triangle

Short Answer

Expert verified
(D) can not form the sides of a triangle.

Step by step solution

01

Conditions for Geometric Progression

Since \(a, b, c\) are in Geometric Progression, we know that \( \frac{b}{a} = \frac{c}{b} \). This implies that \( b^2 = ac \).
02

Set up the Arithmetic Progression

Given that \( \log\left(\frac{5c}{a}\right), \log\left(\frac{3b}{5c}\right), \log\left(\frac{a}{3b}\right) \) are in Arithmetic Progression (A.P.), the difference between consecutive terms should be constant. Thus, \( \log\left(\frac{3b}{5c}\right) - \log\left(\frac{5c}{a}\right) = \log\left(\frac{a}{3b}\right) - \log\left(\frac{3b}{5c}\right) \).
03

Simplify Logarithmic Expressions

This simplifies to: \[\log\left(\frac{3b}{5c}\right) = \frac{1}{2}\left(\log\left(\frac{5c}{a}\right) + \log\left(\frac{a}{3b}\right)\right)\]Combining logs we get: \[\log\left(\frac{3b}{5c}\right) = \frac{1}{2} \log\left(\frac{5c}{3b}\right)^{\frac{1}{2}} = 0\]Thus: \(\frac{3b}{5c} = 1\rightarrow 3b = 5c\)
04

Solve the System of Equations

Using the results from Step 1 and Step 3, we solve the system of equations:1. \(b^2 = ac\)2. \(3b = 5c\)Substituting \(c = \frac{3}{5}b\) into equation 1 gives \(b^2 = a\left(\frac{3}{5}b\right) \rightarrow b = \frac{3a}{5}\).Thus, \(c = 3b/5 = 9a/25\).
05

Check Triangular Inequalities

The sides \(a, b, c\) must satisfy the triangle inequalities:1. \(a + b > c\)2. \(a + c > b\)3. \(b + c > a\)Substituting \(b = \frac{3a}{5}, c = \frac{9a}{25}\), we check each inequality:1. \(a + \frac{3a}{5} > \frac{9a}{25}\) simplifies to \(\frac{8a}{5} > \frac{9a}{25}\), which holds true.2. \(a + \frac{9a}{25} > \frac{3a}{5}\) simplifies to \(\frac{34a}{25} > \frac{3a}{5}\), which is false.3. \(\frac{3a}{5} + \frac{9a}{25} > a\) simplifies to \(\frac{24a}{25} < a\), which holds true.Since one condition is false, they cannot form the sides of a triangle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Expressions in Arithmetic Progression
Logarithmic expressions are a way to represent numbers using powers or exponents. In this context, we are looking at three specific logarithmic expressions: \( \log\left(\frac{5c}{a}\right) \), \( \log\left(\frac{3b}{5c}\right) \), and \( \log\left(\frac{a}{3b}\right) \). These expressions are in arithmetic progression (A.P.), which means that the difference between the consecutive terms remains constant.
  • A key property of logarithms used here is that \( \log(x/y) = \log(x) - \log(y) \).
  • For the expressions to be in A.P., the difference \( \log\left(\frac{3b}{5c}\right) - \log\left(\frac{5c}{a}\right) \) must equal \( \log\left(\frac{a}{3b}\right) - \log\left(\frac{3b}{5c}\right) \).
  • This simplifies by combining logs and using their properties to determine relationships between \( a, b, \) and \( c \).
Understanding how logarithmic differences relate in this context is crucial to solving the problem. This reveals hidden equations that lead to finding the specific proportional relationships of \( a, b, \) and \( c \).
Arithmetic Progression
Arithmetic progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant. This property becomes useful when evaluating sequences in various mathematical problems, including our exercise.
  • In an A.P., if you know two terms, you can always find the third using the formula: \( x_{n+1} = x_n + d \), where \( d \) is the common difference.
  • To verify if numbers form an A.P., calculate the difference between consecutive numbers. If these differences are the same, the numbers are in A.P.
Let's consider our problem. Using A.P. conditions on logs, once simplified, shows us relationships like \( 3b = 5c \), providing insights on the progression and connections among \( a, b, \) and \( c \). These relationships help determine not only the value structure but also potential characteristics when formed together, such as side lengths of geometric shapes.
Triangle Inequalities
Triangle inequalities state that for any triangle with sides \( a \), \( b \), and \( c \), certain conditions must be met. These conditions are essential in establishing whether a particular set of three numbers can be sides of a triangle.
  • The inequality \( a + b > c \) ensures that two sides combined will always be greater than the third.
  • Similarly, \( a + c > b \) and \( b + c > a \) confirm the need for the two other combinations of sides to add up to more than the third side.
When you substitute the values derived from the logarithmic analysis into these inequalities, some do not hold true, showing that \( a, b, \) and \( c \) cannot form a triangle. Specifically, \( a + c < b \) in our derived scenario, hence violating the triangle inequality condition, which is critical proof in solving these types of geometric questions.

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Most popular questions from this chapter

If \(a_{1}, a_{2}, \ldots, a_{n}\) are in A.P. with common difference \(d \neq 0\), then sum of the series \(\sin d\left[\sec a_{1} \sec a_{2}+\sec a_{2}\right.\) \(\left.\sec a_{3}+\ldots+\sec a_{n-1} \sec a_{n}\right]\) is (A) \(\tan a_{n}-\tan a_{1}\) (B) \(\cot a_{n}-\cot a_{1}\) (C) \(\sec a_{n}-\sec a_{1}\) (D) \(\operatorname{cosec} a_{n}-\operatorname{cosec} a_{1}\)

The consecutive numbers of a three digit number form a G.P. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order and if we increase the second digit of the required number by 2, the resulting number forms an A.P. The number is (A) 139 (B) 193 (C) 931 (D) None of these

The sum of of first ten terms of an A.P. is equal to 155 and the sum of first two terms of a G.P. is 9 . If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P, then (A) first term of the G.P. is \(\frac{2}{3}, 3\) (B) first term of the A.P. is \(\frac{2}{3}, 3\) (C) Common ratio of the G.P. is \(\frac{25}{2}, 2\) (D) Common difference of the A.P is \(\frac{2}{3}, 3\)

The sum to \(n\) terms of the series \(\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots\) is (A) \(n-\frac{\left(3^{n}-2^{n}\right)}{2^{n}}\) (B) \(n-\frac{2\left(3^{n}-2^{n}\right)}{3^{n}}\) (C) \(2^{n}-1\) (D) \(3^{n}-1\)

In a sequence of \(4 n+1\) terms, the first \(2 n+1\) terms are in A.P. having common difference 2 and the last \(2 n+1\) terms are in G.P. having common ratio \(\frac{1}{2}\), If the middle term of the A.P. is equal to the middle term of the G.P. then the middle term of the sequence is (A) \(\frac{n \cdot 2^{n+1}}{2^{n}+1}\) (B) \(\frac{n \cdot 2^{n+1}}{2^{n}-1}\) (C) \(\frac{n \cdot 2^{n}}{2^{n}-1}\) (D) None of these
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