Understanding the Problem: Mean of Squares vs Square of Mean

This question involves calculating two related quantities for the first \(n\) natural numbers: the mean of their squares and the square of their mean. We are given a ratio of these two quantities and need to find the value of \(n\).

Let's define the terms clearly:

• Natural Numbers: These are the positive integers starting from 1, i.e., 1, 2, 3, ..., n.
• Mean of first n natural numbers: The sum of the first n natural numbers divided by n.
• Square of mean: The square of the mean calculated above. This is denoted by \(y\).
• Squares of first n natural numbers: The squares of the natural numbers, i.e., \(1^2, 2^2, 3^2, ..., n^2\).
• Mean of squares: The sum of the squares of the first n natural numbers divided by n. This is denoted by \(x\).

Formulas for Sums

We need the formulas for the sum of the first n natural numbers and the sum of the squares of the first n natural numbers.

• Sum of first n natural numbers: \(S_n = 1 + 2 + \dots + n = \frac{n(n+1)}{2}\)
• Sum of squares of first n natural numbers: \(S_{n^2} = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}\)

Calculating x (Mean of Squares)

The mean of squares, \(x\), is the sum of the squares of the first n natural numbers divided by n.

\[ x = \frac{S_{n^2}}{n} = \frac{\frac{n(n+1)(2n+1)}{6}}{n} \]

Assuming \(n \neq 0\), we can cancel \(n\) from the numerator and denominator:

\[ x = \frac{(n+1)(2n+1)}{6} \]

Calculating y (Square of Mean)

First, let's find the mean of the first n natural numbers.

Mean of first n natural numbers \( = \frac{S_n}{n} = \frac{\frac{n(n+1)}{2}}{n} \)

Assuming \(n \neq 0\), we can cancel \(n\):

Mean \( = \frac{n+1}{2} \)

Now, we find the square of this mean, which is \(y\).

\[ y = \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)^2}{4} \]

Setting up the Given Equation

The problem states that the ratio of x to y is \(\frac{55}{42}\). So, we have:

\[ \frac{x}{y} = \frac{55}{42} \]

Substitute the expressions we found for \(x\) and \(y\):

\[ \frac{\frac{(n+1)(2n+1)}{6}}{\frac{(n+1)^2}{4}} = \frac{55}{42} \]

Solving the Equation for n

Now, we simplify the complex fraction on the left side. Dividing by a fraction is the same as multiplying by its reciprocal.

\[ \frac{(n+1)(2n+1)}{6} \times \frac{4}{(n+1)^2} = \frac{55}{42} \]

Assuming \(n+1 \neq 0\) (which is true for natural numbers \(n \geq 1\)), we can cancel one factor of \((n+1)\) from the numerator and denominator:

\[ \frac{(2n+1)}{6} \times \frac{4}{(n+1)} = \frac{55}{42} \]

Multiply the fractions on the left:

\[ \frac{4(2n+1)}{6(n+1)} = \frac{55}{42} \]

Simplify the fraction \(\frac{4}{6}\) to \(\frac{2}{3}\):

\[ \frac{2(2n+1)}{3(n+1)} = \frac{55}{42} \]

Now, we can cross-multiply:

\[ 42 \times [2(2n+1)] = 55 \times [3(n+1)] \]

\[ 84(2n+1) = 165(n+1) \]

Distribute the numbers on both sides:

\[ 84 \times 2n + 84 \times 1 = 165 \times n + 165 \times 1 \]

\[ 168n + 84 = 165n + 165 \]

Gather the terms with \(n\) on one side and the constant terms on the other side:

\[ 168n - 165n = 165 - 84 \]

\[ 3n = 81 \]

Solve for \(n\):

\[ n = \frac{81}{3} \]

\[ n = 27 \]

Since 27 is a natural number, this is a valid solution.

Conclusion on the Value of n

Based on the calculations, the value of \(n\) that satisfies the given condition \(\frac{\text{x}}{\text{y}} = \frac{55}{42}\) is 27.

Revision Table: Key Concepts and Formulas

Additional Information: Mean vs. Mean of Squares

It's important to understand the difference between the "square of the mean" and the "mean of squares". In general, for a set of numbers, the mean of their squares is not equal to the square of their mean. This problem highlights that difference and provides a specific condition where their ratio is fixed.

For example, consider the first 2 natural numbers (n=2):

• Numbers: 1, 2
• Mean: \(\frac{1+2}{2} = \frac{3}{2}\)
• Square of Mean (y): \((\frac{3}{2})^2 = \frac{9}{4}\)
• Squares: \(1^2=1, 2^2=4\)
• Sum of Squares: \(1+4 = 5\)
• Mean of Squares (x): \(\frac{5}{2}\)

Here, \(\frac{x}{y} = \frac{5/2}{9/4} = \frac{5}{2} \times \frac{4}{9} = \frac{20}{18} = \frac{10}{9}\). This ratio is different from \(\frac{55}{42}\), which is why n=2 is not the answer.

The problem essentially asks for what value of n this ratio \(\frac{\frac{(n+1)(2n+1)}{6}}{\frac{(n+1)^2}{4}}\) simplifies to \(\frac{55}{42}\).