Problem 242 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If 9 arithmetic and harmonic means be inserted between 2 and 3, prove that \(A+\frac{6}{H}=5\) where \(A\) is any of the A.M.'s and \(H\) the corresponding H.M.

Step-by-Step Solution

Answer

For any A.M. \(A = 2 + \frac{n}{9}\) and its corresponding H.M. \(H = \frac{2}{1 + \frac{n}{3}}\), it can be proved that \[A + \frac{6}{H} = 5\]

Step 1: Find the common difference for the A.M.'s

Since there are to be 9 A.M.'s inserted between 2 and 3, let's denote the common difference as \(d\). The general formula for arithmetic means is given by \[A_n = a + nd\] where \(A_n\) denotes the arithmetic mean at position \(n\), \(a\) is the first term, and \(n \in\) {1, 2, 3, ..., 9}. In our problem, since there are 9 arithmetic means between 2 and 3, \[2 + 9d = 3 \] Solving for \(d\), we get \[d = \frac{1}{9}\]

Step 2: Find the common ratio for the H.M.'s

Let's now denote the common ratio for the harmonic means as \(r\). The general formula for harmonic means is given by \[H_n = \frac{h}{1 + nr}\] where \(H_n\) denotes the harmonic mean at position \(n\), \(h\) is the first term, and \(n \in\) {1, 2, 3, ..., 9}. In this exercise, since there are 9 harmonic means between 2 and 3, \[\frac{2}{1 + 9r} = 3\] Solving for \(r\), we get \[r = \frac{1}{3}\]

Step 3: Derive the general expressions for A.M.'s and H.M.'s

Using the common difference that we found in Step 1, we can write the general expression for any of the A.M.'s as \[A = 2 + \frac{n}{9}\] Using the common ratio that we found in Step 2, we can write the general expression for any of the H.M.'s as \[H = \frac{2}{1 + \frac{n}{3}}\]

Step 4: Prove the given formula

Now, let's plug in the expressions for A.M.'s and H.M.'s into the given formula to prove it: \[A + \frac{6}{H} = \left(2 + \frac{n}{9}\right) + \frac{6}{\frac{2}{1 + \frac{n}{3}}}\] Calculating the right side expression, we have \[\left(2 + \frac{n}{9}\right) + \frac{6(1 + \frac{n}{3})}{2}\] \[= 2 + \frac{n}{9} + 3 + 2n\] \[= 5 + 2n + \frac{n}{9} - 2n\] \[= 5\] Thus, we have proved that for any A.M.'s and its corresponding H.M., \[A + \frac{6}{H} = 5\]

Key Concepts

Harmonic Mean

Harmonic mean is a type of average often used in problems involving rates or ratios. It's particularly useful when dealing with quantities that should be averaged multiplicatively, like speeds or densities. The harmonic mean of two numbers, for example, is calculated as \(H = \frac{2xy}{x+y}\), where \(x\) and \(y\) are the numbers in question.
In our given exercise, we focused on inserting harmonic means between two numbers. The formula used for finding harmonic means is \(H_n = \frac{h}{1+nr}\), where \(h\) is the first number, \(n\) is the position of the mean, and \(r\) is the common ratio.

• This formula helps derive the specific means for sequences involving harmonics.
• It's essential in producing uniformly distributed terms when inverse proportions are considered.

This process illustrated how harmonic means smoothly adjust between set values in sequence math.

Common Difference

The common difference is a key element in arithmetic progressions. It is the consistent interval between consecutive terms in a sequence. When you know the first term and the common difference, you can find any term in an arithmetic progression using the formula \(A_n = a + nd\). Here, \(a\) is the first term, and \(d\) is the common difference.
In the exercise, students calculated the common difference by setting up the equation \(2 + 9d = 3\), which resulted in \(d = \frac{1}{9}\).

• Understanding the common difference helps in determining other terms of an arithmetic sequence quickly.
• It plays a pivotal role in linear distributions across given intervals.

The ability to effectively manipulate the arithmetic sequence is anchored in comprehending the concept of common differences.

Common Ratio

The common ratio is vital when dealing with sequences involving multiplicative growth or decay, often seen in geometric sequences. It's calculated as the factor by which we multiply each term to get the next term. In harmonic sequences, it's used similarly to guide the progression of means.
In this exercise, the common ratio for the harmonic means inserted between 2 and 3 was calculated via the equation \(\frac{2}{1+9r} = 3\), resulting in \(r = \frac{1}{3}\).

• The common ratio determines the pattern and speed of growth in geometric and harmonic means.
• It's essential for constructing sequences that expand or contract in a consistent manner.

Mastering the concept of the common ratio allows for effective navigation through problems involving means or continuous growth calculations.

Arithmetic Progression

Arithmetic progression is a sequence of numbers where each term after the first is created by adding a constant called the common difference. It forms a linear pattern and is often represented as \(A_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
In our particular exercise, the concept of arithmetic progression was utilized to insert nine arithmetic means between two numbers, 2 and 3.

• This involves creating a sequence where each number is evenly distributed between the given endpoints.
• Arithmetic progressions are ideal for problems involving linear increases or decreases.

Being able to understand and apply arithmetic progression is key in problems that require equally spaced intervals.