If the mean of the squares of first n natural numbers is 105, then the first n natural numbers is(a) 8 (b) 9(c) 10 (d) 11

The formula for the mean of the squares of the first n natural numbers is (n+1)(2n+1)/6.

So, we have (n+1)(2n+1)/6 = 105.

Solving this equation for n will give us the answer.

First, multiply both sides by 6 to get rid of the denominator on the left side:

(n+1)(2n+1) = 630.

Then, expand the left side:

2n^2 + 3n + 1 = 630.

Rearrange the equation to set it equal to zero:

2n^2 + 3n - 629 = 0.

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 2, b = 3, and c = -629.

We can solve this equation for n using the quadratic formula, n = [-b ± sqrt(b^2 - 4ac)] / 2a.

Substituting the values of a, b, and c into the formula gives us:

n = [-3 ± sqrt((3)^2 - 42(-629))] / 2*2.

Solving this gives us two possible values for n, but since n must be a natural number, we discard the negative solution.

So, the only possible value for n is 10.

Therefore, the answer is (c) 10.