Problem 19
Question
If \(a, b, c\), and \(d\) are four unequal positive numbers which are in A.P., then a. \(\frac{1}{a}+\frac{1}{d}>\frac{1}{b}+\frac{1}{c}\) b. \(\frac{1}{a}+\frac{1}{d}<\frac{1}{b}+\frac{1}{c}\) c. \(\frac{1}{b}+\frac{\mathrm{I}}{c}>\frac{4}{a+d}\) d. \(\frac{1}{a}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c}\)
Step-by-Step Solution
Verified Answer
Option a: \(\frac{1}{a} + \frac{1}{d} > \frac{1}{b} + \frac{1}{c}\).
1Step 1: Understand the Given Condition
The numbers \(a, b, c, d\) are in Arithmetic Progression (A.P.). In an A.P., the difference between consecutive terms is constant. If \(a, b, c, d\) are in A.P., then \(b = a + n\), \(c = a + 2n\), and \(d = a + 3n\) where \(n\) is the common difference.
2Step 2: Express in Reciprocal Terms
Calculate \(\frac{1}{a} + \frac{1}{d}\) and \(\frac{1}{b} + \frac{1}{c}\) using inverse expressions:\[\frac{1}{a} + \frac{1}{d} = \frac{1}{a} + \frac{1}{a + 3n} = \frac{a + 3n + a}{a(a + 3n)} = \frac{2a + 3n}{a^2 + 3an}\]\[\frac{1}{b} + \frac{1}{c} = \frac{1}{a + n} + \frac{1}{a + 2n} = \frac{a + 2n + a + n}{(a+n)(a+2n)} = \frac{2a + 3n}{a^2 + 3an + 2n^2}\]
3Step 3: Compare the Two Expressions
Compare the fractions \(\frac{2a + 3n}{a^2 + 3an}\) and \(\frac{2a + 3n}{a^2 + 3an + 2n^2}\). The denominator \(a^2 + 3an\) is smaller than \(a^2 + 3an + 2n^2\), hence:\[\frac{2a + 3n}{a^2 + 3an} > \frac{2a + 3n}{a^2 + 3an + 2n^2}\]
4Step 4: Conclude with the Correct Choice
Since \(\frac{1}{a} + \frac{1}{d} > \frac{1}{b} + \frac{1}{c}\), the correct answer is option a.
Key Concepts
InequalitiesReciprocal ExpressionsMathematical Proofs
Inequalities
Understanding inequalities is essential in comparing two values when there isn't a direct equivalence. An inequality shows either a "greater than" (>) or "less than" (<) relationship between two expressions. For example, if we say "x > y," it means that x is greater than y. In mathematics, inequalities are crucial for expressing ranges, limits, and optimizing solutions. They are often used in statistical predictions, economic modeling, and scientific computations.
When dealing with inequalities, it's important to remember that multiplying or dividing both sides by a negative number reverses the inequality sign. This rule is vital in solving inequality-based equations and must be carefully applied.
When dealing with inequalities, it's important to remember that multiplying or dividing both sides by a negative number reverses the inequality sign. This rule is vital in solving inequality-based equations and must be carefully applied.
- For example: if -2x > 4, dividing both sides by -2 gives x < -2.
Reciprocal Expressions
Reciprocal expressions flip the roles of numerators and denominators in fractions. If a number is represented as a fraction with numerator and denominator, its reciprocal inverts these parts. For instance, the reciprocal of a number x is 1/x. This concept is key in various mathematical operations like division and equations set in fractional forms.
In our exercise, reciprocal expressions are used to compare terms in an Arithmetic Progression. This process involves examining the expressions:
In our exercise, reciprocal expressions are used to compare terms in an Arithmetic Progression. This process involves examining the expressions:
- \( \frac{1}{a} + \frac{1}{d} \)
- \( \frac{1}{b} + \frac{1}{c} \)
Mathematical Proofs
Mathematical proofs transform our understanding from assumptions to verified facts using logical reasoning. They form the backbone of mathematical rigor, affirming what is true through step-by-step deductions. In general terms, proofs may demonstrate properties of numbers, validate equations, or verify complex theories.
In our problem, proving the inequality involves several stages:
In our problem, proving the inequality involves several stages:
- Setting expressions from an A.P. context: identify terms and relate them using the common difference.
- Ensuring correct reciprocal conversion of terms to uncover insights about their equation relationships.
- Applying logical analysis to determine the inequality's validity based on comparative denominators.
Other exercises in this chapter
Problem 18
Let \(a_{1}, a_{2}, a_{3}, \ldots\) be terms of an A.P. If \(\frac{a_{1}+a_{2}+\cdots+a_{p}}{a_{1}+a_{2}+\cdots+a_{9}}\) \(=\frac{p^{2}}{q^{2}}, p \neq q\), the
View solution Problem 18
If \(x^{2}+9 y^{2}+25 z^{2}=x y z\left(\frac{15}{x}+\frac{5}{y}+\frac{3}{z}\right)\), then a. \(x, y\), and \(z\) are in H.P. b. \(\frac{1}{x}, \frac{1}{y}, \fr
View solution Problem 20
Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P. is a. \(2-\sqrt{3} \quad\)
View solution Problem 20
The next term of the G.P. \(x, x^{2}+2\), and \(x^{3}+10\) is a. \(\frac{729}{16}\) b. 6 c. 0 d. 54
View solution