Question

If three consecutive terms in an AP are \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\), then \(\frac{b-c}{a-b}=\) (a) \(\frac{\mathrm{a}}{\mathrm{c}}\) (b) \(\frac{\mathrm{b}}{\mathrm{a}}\) (c) a (d) \(\frac{c}{a}\)

Short answer

Answer: (d) \(\frac{c}{a}\)

Step-by-step solution

Identify the relationship among consecutive terms in an AP

In an arithmetic progression, the difference between any two consecutive terms is constant. In this case, we are given the terms \(\frac{1}{a}\), \(\frac{1}{b}\), and \(\frac{1}{c}\). The common difference between the terms is the same, so we can set up the following equations: \(\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}\)

Cross-multiply and simplify the equation

To simplify the equation, cross-multiply the terms: \((b - a)c = (c - b)a\) Next, expand the terms to obtain: \(bc - ac = ac - ab\)

Rearrange the equation to isolate the desired expression

We need to find the value of \(\frac{b-c}{a-b}\). From the equation obtained in Step 2, we can rearrange it as follows: \(bc - ac + ab = ac\) Now, factor out the common terms: \(c(b-a) = a(c-b)\) Divide both sides by \((b - a)(c - b)\): \(\frac{c}{a} = \frac{b - c}{a - b}\) Therefore, the correct answer is (d) \(\frac{c}{a}\).

Key concepts

Consecutive Terms in an Arithmetic Progression (AP)

When dealing with arithmetic progressions (AP), a foundational concept is understanding consecutive terms and their properties. Consecutive terms in an AP are a series of numbers in which the difference between any two subsequent terms is always constant. This common difference is denoted by 'd'.

For example, if we have three consecutive terms in an AP, such as \(\frac{1}{a}\), \(\frac{1}{b}\), and \(\frac{1}{c}\), we can represent them mathematically as follows: \(a_n = \frac{1}{a}\), \(a_{n+1} = \frac{1}{b}\), and \(a_{n+2} = \frac{1}{c}\). To maintain the property of an AP, the difference between \(a_{n+1}\) and \(a_n\), which is \(d\), should be equal to the difference between \(a_{n+2}\) and \(a_{n+1}\).

This is expressed as:

• \(a_{n+1} - a_n = d\)
• \(a_{n+2} - a_{n+1} = d\)

When considering fractions like in our example, finding this common difference is key to establishing the relationship between the consecutive terms and solving further parts of the problem.

Understanding the Term Relationship in AP

The term relationship in an AP is established by the common difference. If you are given any term in an AP, you can find subsequent or preceding terms by adding or subtracting the common difference, respectively. In the case of our exercise, we express this relationship algebraically.

We start with the provided terms \(\frac{1}{a}\), \(\frac{1}{b}\), and \(\frac{1}{c}\) and use the property that the difference between consecutive terms is constant. As such, we set up the fundamental AP equations:

• \(\frac{1}{b} - \frac{1}{a} = d\)
• \(\frac{1}{c} - \frac{1}{b} = d\)

These equations underline the inherent relationship between consecutive terms in an AP. It's important to understand that any term in an AP (except for the first one) is the arithmetic mean of its immediate neighbors. So when consecutive terms are fractions, this relationship helps us relate the terms to each other and greatly simplifies the process of solving for unknown variables within the progression.

Simplifying Equations in an AP

Now let's discuss simplifying equations in an AP, which involves manipulating algebraic expressions to find a desired outcome, like in our given exercise. Simplification often involves cross-multiplication when dealing with fractions, which is the process of multiplying the numerator of one fraction by the denominator of the other.

As seen in the solution steps:

• We start by cross-multiplying to eliminate the fractions: \( (b - a)c = (c - b)a \).
• Expanding the terms gives us \( bc - ac = ac - ab \).
• Rearrange and factor to find common terms: \(c(b - a) = a(c - b)\).
• Finally, we divide both sides by the product of differences, resulting in \( \frac{c}{a} = \frac{b - c}{a - b} \).

This process exemplifies the importance of understanding how to move between steps in your simplification, being careful to preserve the equality at each stage. Mastering this skill is integral to solving problems involving APs and other algebraic contexts.