Problem 203 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c\) be respectively the \(p\) th, \(q\) th and \(r\) th terms of an H.P., then prove that \(b c(q-r)+c a(r-p)+a b(p-q)=0 .\)

Step-by-Step Solution

Answer

The short answer to the question is: To prove that \(b c(q-r)+c a(r-p)+a b(p-q)=0\), we express the terms of the H.P. in terms of the first term and common difference: \(\frac{1}{a} - \frac{1}{a} = (p-1) \cdot \frac{1}{d}\) \(\frac{1}{b} - \frac{1}{a} = (q-1) \cdot \frac{1}{d}\) \(\frac{1}{c} - \frac{1}{a} = (r-1) \cdot \frac{1}{d}\) Substituting these relationships into the given expression and simplifying, we get: \[\frac{1}{d} \left[(q-r) \cdot b c + (r-p) \cdot c a + (p-q) \cdot a b \right] + \frac{1}{a} \left[b c(q-r) + ca(r-p) + ab(p-q) \right] = 0\] Since both expressions in the brackets have the same terms but with opposite signs, their sum becomes zero, proving that \(b c(q-r)+c a(r-p)+a b(p-q)=0\).

Step 1: Write the general form of H.P.

The general form of H.P. is given by: \[\frac{1}{a_n} = \frac{1}{a} + (n-1) \cdot \frac{1}{d}\] where \(a_n\) is the \(n\) th term, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term's position.

Step 2: Find the pth, qth and rth terms of H.P.

Using the general form, we have: For pth term (\(a\)): \[\frac{1}{a} = \frac{1}{a} + (p-1) \cdot \frac{1}{d}\] For qth term (\(b\)): \[\frac{1}{b} = \frac{1}{a} + (q-1) \cdot \frac{1}{d}\] For rth term (\(c\)): \[\frac{1}{c} = \frac{1}{a} + (r-1) \cdot \frac{1}{d}\]

Step 3: Simplify the equations and find the relationship between terms

Rewriting the equations from Step 2 in terms of the first term and common difference, we get: For term \(a\), \[\frac{1}{a} - \frac{1}{a} = (p-1) \cdot \frac{1}{d}\] For term \(b\), \[\frac{1}{b} - \frac{1}{a} = (q-1) \cdot \frac{1}{d}\] For term \(c\), \[\frac{1}{c} - \frac{1}{a} = (r-1) \cdot \frac{1}{d}\] Now, we have to prove that: \[b c(q-r)+c a(r-p)+a b(p-q)=0\]

Step 4: Substitute the relationships found in Step 3 into the given expression

Substitute the equations for \(a, b,\) and \(c\) in the given expression: \[b \left[ (q-1) \cdot \frac{1}{d} + \frac{1}{a} \right] c (q-r)+ c \left[ (r-1) \cdot \frac{1}{d} + \frac{1}{a} \right] a (r-p)+ a \left[ (p-1) \cdot \frac{1}{d} + \frac{1}{a} \right] b (p-q)\]

Step 5: Simplify the expression

Expand the terms and simplify the given expression: \[b c\left( (q-1) \cdot \frac{1}{d}\right)(q-r) + bc\frac{1}{a}(q-r) + c a\left( (r-1) \cdot \frac{1}{d}\right)(r-p) + ca\frac{1}{a}(r-p) + a b\left( (p-1) \cdot \frac{1}{d}\right)(p-q) + ab\frac{1}{a}(p-q)\] Now, rearrange the terms, and we have: \[(q-r) \cdot b c \cdot \frac{1}{d} + (r-p) \cdot c a \cdot \frac{1}{d} + (p-q) \cdot a b \cdot \frac{1}{d} + b c\frac{1}{a}(q-r) + ca\frac{1}{a}(r-p) + ab\frac{1}{a}(p-q)\] Factoring out \(\frac{1}{d}\) from the first three terms and \(\frac{1}{a}\) from the last three terms, we have: \[\frac{1}{d} \left[(q-r) \cdot b c + (r-p) \cdot c a + (p-q) \cdot a b \right] + \frac{1}{a} \left[b c(q-r) + ca(r-p) + ab(p-q) \right]\] Now, let's observe the two expressions inside the brackets. It can be seen that both of them have the same terms, but with opposite signs. Therefore, their sum becomes zero: \[\frac{1}{d} \left[ 0 \right] + \frac{1}{a} \left[ - b c(q-r) - c a(r-p) - a b(p-q) \right] = 0\] Hence, the given expression is proved: \[b c(q-r)+c a(r-p)+a b(p-q)=0\]

Key Concepts

Understanding Harmonic Progression

Harmonic Progression (HP) is a sequence of numbers where the reciprocals of the terms form an arithmetic progression (AP). In simpler terms, if you take the reciprocal of each term in an HP, you'll have an AP. This concept is fundamental in solving problems related to rates, time, and work in mathematical exercises.

To clarify, if we have an HP with terms denoted by \(a_n\), then the reciprocals \(1/a_n\) will form an AP with some common difference \(d\). If \(a, b, c\) are the \(p\)th, \(q\)th, and \(r\)th terms of an HP respectively, their reciprocals \(1/a, 1/b, 1/c\) are correspondingly the \(p\)th, \(q\)th, and \(r\)th terms of an AP.

Understanding this relationship allows us to derive algebraic expressions and establish proofs regarding the properties of harmonic sequences, like the one seen in the exercise. A pivotal step in solving such problems is always to transform the HP terms into an AP, which can be manipulated using the familiar rules of arithmetic sequences.

Using Algebraic Identities

Algebraic identities are equations that hold true for all values of the variables involved. Common identities like \(a^2 - b^2 = (a+b)(a-b)\) or \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) are tools that can greatly simplify complex algebraic expressions and proofs.

In our exercise, algebraic manipulation plays a crucial role in establishing the relationship needed to prove the given expression. The proof requires expanding and factoring expressions, which rely heavily on understanding how to apply these identities. For instance, when we simplify the expression, we recognize a pattern that resembles the null factor law, where products of sums and differences lead to a zero result. By reorganizing and factoring out common elements, we can simplify complex algebraic expressions to more manageable ones — eventually leading to the proof that the given expression equals zero. Recognizing these patterns is critical in dissecting and solving algebraic problems efficiently.

Leveraging Mathematical Induction

Mathematical induction is a powerful proof technique often used to verify the truth of an infinite number of cases, such as the properties of a sequence. While this specific exercise doesn't directly apply mathematical induction, understanding the principle is beneficial for tackling more complex sequences and series problems.

The process starts by proving the base case (the simplest case, usually for \(n=1\)), and then assuming that a statement is true for some arbitrary case \(n=k\). From there, you prove that if the statement holds for \(n=k\), it must also hold for \(n=k+1\). This 'domino effect' allows us to conclude the statement is true for all natural numbers \(n\).

In the realm of HP, induction can be employed to prove formulas or properties that hold for all terms of the sequence. Although not explicitly used in our exercise, it's a thought-provoking concept that solidifies the understanding of sequences and their behaviors in mathematics, thus improving overall problem-solving skills.