Question

If the mean of squares of first n natural numbers is 105,then find the median of first n natural numbers?

Answer 1

Here is your answer....

SORRY!!!

Answer 2

The median of the first 17 natural nos. will be 9.

Step-by-step explanation:

Given:

The mean of  the squares of the first n natural numbers $(1^{2} +2^{2} +3^{2} .........n^{2} )$ is 105.

To Find:

The median of  the first n natural numbers $(1 +2 +3 .........n )$.

Formula Used:

The sum of the  squares of first n natural numbers $(1^{2} +2^{2} +3^{2} .........n^{2} )$

$=\frac{n(n+1)(2n+1)}{6}$ .

Mean= The sum of  the squares of first n natural numbers $(1^{2} +2^{2} +3^{2} .........n^{2} )$/ total  of  the natural numbers.

The Median means middle term.

Solution:

As given- The mean of squares of first n natural numbers is 105.

The sum of squares of first n natural numbers $=\frac{n(n+1)(2n+1)}{6}$

Total  of natural numbers =n

Mean= The sum of squares of first n natural numbers $(1^{2} +2^{2} +3^{2} .........n^{2} )$/ total  of natural numbers    

    $105=\frac{[\frac{n(n+1)(2n+1)}{6}]}{n}$

     $105=\frac{n(n+1)(2n+1)}{6\ n}$

     $105=\frac{(n+1)(2n+1)}{6}$

     $2n^{2} +3n-629=0$

     $2n^{2} +37n-34n-629=0\\n(2n+37)-17(2n+37)=0\\(n-17)(2n+37)=0$

    $(n-17)=0\\n=17$

    $(2n+17)=0\\n=-\frac{17}{2}$( n can not be negative}

Hence  n value is 17.

Therefore, the natural numbers are 1,2,3,…..16, and 17.

The Median will be 9.

Thus. the median of the first 17 natural nos. will be 9.