Problem 224 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

A G.P. and a H.P. have the same \(p\) th, \(q\) th and \(r\) th terms as \(a, b, c\) respectively. Show that \(a(b-c) \log a+b(c-a) \log b+c(a-b) \log c=0 .\)

Step-by-Step Solution

Answer

In summary, we derived expressions for the terms of the G.P. and H.P. in terms of \(a, b, c\), solved for the common ratios of both progressions, and used these expressions to prove that \(a(b-c) \log a+b(c-a) \log b+c(a-b) \log c=0\).

Step 1: Derive expressions for the terms of the G.P. and H.P.

To determine the terms of the G.P. in terms of \(a, b, c\), we should use the formula for the nth term of a G.P.: The nth term of a G.P with first term \(A\) and common ratio \(R\), is given by \(A R^{n-1}\). For \(p\)th term, we get: \( a=AR^{p-1} \) (1) For \(q\)th term, we get: \( b=AR^{q-1} \) (2) For \(r\)th term, we get: \( c=AR^{r-1} \) (3) A harmonic progression is nothing but the arithmetic progression of the reciprocals of the terms. For \(p\)th term, we get: \( \frac{1}{a}=A'R^{p-1} \) (4) For \(q\)th term, we get: \( \frac{1}{b}=A'R^{q-1} \) (5) For \(r\)th term, we get: \( \frac{1}{c}=A'R^{r-1} \) (6)

Step 2: Solve for the common ratio of the G.P. and the H.P.

In order to solve for the common ratio of the G.P. (\(R\)) and the H.P. (\(A'\)), we will need to use the expressions derived for the terms of both progressions in steps 1. Divide equations (2) by (1), we get: \( \frac{b}{a}=R^{q-p} \) (7) Divide equations (3) by (2), we get: \( \frac{c}{b}=R^{r-q} \) (8) Similarly, divide equations (5) by (4), we get \( \frac{b}{a}=A'^{q-p} \) (9) Divide equations (6) by (5), we get \( \frac{c}{b}=A'^{r-q} \) (10)

Step 3: Prove the desired result

Now let's substitute the expressions derived in steps 2 into the equation that we want to prove: \( a(b-c) \log a+b(c-a) \log b+c(a-b) \log c = 0 \) Using equations (7) and (8), substitute for \(\log a\), \(\log b\), and \(\log c\). We get: \( a(b-c) (q-p) \log R + b(c-a) (r-q) \log R + c(a-b) (r-p) \log R = 0 \) Factor out \((\log R)\): \( (a(b-c)(q-p)+b(c-a)(r-q)+c(a-b)(r-p))\log R = 0 \) Since log R cannot be equal to 0, the term inside the parenthesis must be 0. \( a(b-c)(q-p)+b(c-a)(r-q)+c(a-b)(r-p) = 0 \) Using equations (9) and (10). This equation can also be written as: \( a(b-c)(r-q)A'^{p-r} + b(c-a)(r-p)A'^{q-r} + c(a-b)(q-p)A'^{r-p} = 0 \) This equation is equal to our original equation. Thus, we have shown that \(a(b-c)\log a+b(c-a)\log b+c(a-b)\log c=0.\)

Key Concepts

Geometric Progression

A geometric progression (G.P.) is a sequence where each term is found by multiplying the previous term by a constant number, known as the common ratio, denoted as \( R \). In a G.P., the sequence is constructed as follows: if the first term is \( A \), then the sequence follows as \( A, AR, AR^2, AR^3, \ldots \).
The formula for the \( n \)th term of a G.P. is:

• \( T_n = A \cdot R^{n-1} \)

This formula helps to find any term when the first term and the common ratio are known.
In the given problem, we have specific terms of a G.P.—\( a, b, \) and \( c \) for the \( p \)th, \( q \)th, and \( r \)th terms, respectively. This helps derive equations such as \( a = AR^{p-1}, \) \( b = AR^{q-1}, \) and \( c = AR^{r-1}. \) These equations are crucial for solving and relating G.P. to H.P. concepts further in the exercise.

Harmonic Progression

A harmonic progression (H.P.) is a sequence of numbers derived from the reciprocals of an arithmetic progression (A.P.). In simple terms, if the sequence \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \ldots \) forms an A.P., then \( a, b, c, \ldots \) form an H.P.
To comprehend the nth term of an H.P., consider that if \( A' \) is the first term and \( R \) is the common difference of the A.P. of the reciprocals, the formula for reciprocals of the \( n \)th term is:

• \( \frac{1}{T_n} = A' + (n-1)R \)

From the problem's context, the terms \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are analogous to \( A'R^{p-1}, A'R^{q-1}, A'R^{r-1} \). By cross-relating these with G.P., students can solve for common terms and build inter-relations between these groups of sequences.

Logarithmic Identities

Logarithmic identities are mathematical rules that simplify logarithmic expressions. A key aspect of solving the initial exercise is leveraging these identities.
We frequently use logarithmic identities like:

• Product Property: \( \log(ab) = \log a + \log b \)
• Quotient Property: \( \log\left(\frac{a}{b}\right) = \log a - \log b \)
• Power Rule: \( \log(a^b) = b \cdot \log a \)

Such properties are the pillar in transforming the logarithmic equations in the problem.
The primary equation to prove involves terms like \( a(b-c) \log a \) etc., requiring substitution and simplification based on logarithmic properties. Recognizing and interpreting these identity rules is fundamental for solving complex equations in mathematical sequences.