To solve this problem, we need to find the value of 'n' when given the ratio of x/y.

Given:
x/y = 55/42

Let's break down the problem into steps to find the value of 'n':

Step 1: Understanding the problem
We are given two variables, x and y, which represent the mean of squares and the square of the mean of the first 'n' natural numbers, respectively. We need to find the value of 'n' when x/y is equal to 55/42.

Step 2: Understanding the formulas
The mean of squares of the first 'n' natural numbers can be calculated using the formula:
x = (1^2 + 2^2 + 3^2 + ... + n^2) / n

The square of the mean of the first 'n' natural numbers can be calculated using the formula:
y = [(1 + 2 + 3 + ... + n) / n]^2

Step 3: Simplifying the expressions
Let's simplify the expressions for x and y:

x = (1^2 + 2^2 + 3^2 + ... + n^2) / n
= (n(n+1)(2n+1)) / 6n
= (n+1)(2n+1) / 6

y = [(1 + 2 + 3 + ... + n) / n]^2
= [(n(n+1))/2n]^2
= (n+1)^2 / 4

Step 4: Substituting the expressions into the given ratio
Now, we substitute the expressions for x and y into the given ratio:

x/y = 55/42
[(n+1)(2n+1) / 6] / [(n+1)^2 / 4] = 55/42

Step 5: Cross-multiplying and simplifying the equation
To simplify the equation, we cross-multiply and get rid of the fractions:

[(n+1)(2n+1)] * 4 = [(n+1)^2] * 55/42

Simplifying further:

[(n+1)(2n+1)] * 4 * 42 = [(n+1)^2] * 55

Step 6: Expanding and simplifying the equation
Expanding the equation:

8(n^2) + 12n + 4 = 55n^2 + 110n + 55

Simplifying further:

47n^2 + 98n + 51 = 0

Step 7: Solving the quadratic equation
To solve the quadratic equation, we can factorize it or use the quadratic formula. In this case, the equation cannot be easily factorized, so let's use the quadratic formula:

n = [-b ± √(b^2 - 4ac)] / 2a

Plugging in the values:

n = [-98 ± √(98^2 - 4 * 47 * 51)] / (2 * 47)

n = [-98 ± √(9604 - 9596)] / 94

n = [-98 ± √8] / 94

n = [-98 ± 2√2] / 94

Step