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
Question
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
Solution
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.
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