Question

the average of first n whole numbers is x.if the first whole number is removed then the average of remaining whole numbers?

Answer 1

Answer:

X + ½ = (2X+1)/2

Step-by-step explanation:

The sum of first n whole numbers = n*(n+1)/2

Average of first n numbers = Sum/n = (n + 1)/2 = X

First whole number = 1.  

Sum of above whole numbers without first = n*(n+1)/2 – 1 = (n^2 + n – 2)/2 = (n – 1)(n + 2)/2

Average of above numbers = (n -1)(n + 2)/(2 * (n – 1)) = (n + 2)/2

= (n + 1)/2 + 1 / 2  

Since (n+1)/2 = X

Average of n – 1 numbers = X + ½ = (2X+1)/2

Answer 2

The average of first n whole numbers is x.if the first whole number is removed then to find the average of remaining whole numbers,we must keep in our mind that

Whole numbers: 0,1,2,3,4,5,...

So, given that First n whole numbers average is x.


$Average= \frac{sum \: of \: all \:n \: numbers}{total \: numbers}$
$x = \frac{sum \: of \: first \: n \: whole \: numbers}{n} \\ \\ sum \: of \: first \: n \: whole \: numbers = xn$
Now ATQ if 1st whole number is removed,i.e. 0 is removed

➡️➡️1 st whole number is 0

➡️it does not change the sum

➡️but total number are now n-1

So,new average is

$= \frac{xn}{n - 1}$
Hope it helps you.