Question

The average of first n whole numbers is X if the first whole number is removed from them, then the average of remaining whole numbers is​

Answer 1

Answer:

X + 1/2

Step-by-step explanation:

Hi,

Whole Numbers are

{ 0, 1, 2, 3, .....}

Given that the average of first n whole numbers is

X,

First n whole numbers will be

{0, 1, 2, 3, ......., n-1} which are total n in number,

Average of first n whole numbers

= X

= (Sum of n whole numbers)/n

= (0 + 1 + 2 +......+ n-1)/n

= (1 + 2 .....+ n-1)/n

= (n -1)*n/2n

= (n - 1)/2

Hence, X = (n - 1)/2--------------(1)

If the first whole number is removed,

then the remaining whole numbers will be

{1, 2, 3,......,n - 1}

Average of remaining whole numbers

= ( 1 + 2 +.....+ n - 1)/(n-1)

= (n - 1)*n/2(n - 1)

= n/2,

From (1), we can rewrite X = n/2 - 1/2

n/2 = X + 1/2,

Hence, average of remaining numbers

= X + 1/2,

Hope, it helps !

Answer 2

Answer:

X + 1/2

Step-by-step explanation:

The average of first 'N' whole numbers is. X. If the first whole number is removed from them.then what is the average

Average of first N whole numbers = (n+1)/2

(n+1)/2 = X

=> n + 1 = 2X

=> n = 2X-1

Sum of n number = nX

now first number is reduced so value reduce by 1 & number also decrease by

new Sum = nX - 1

Numbers = n -1

new average = (nX - 1) / n-1

= ((2X-1)X - 1) / (2X -1 - 1)

= (2X² - X - 1)/(2X -2)

= (2X² -2X + X -1)/ (2(X-1))

= (2X(X-1) + 1(X-1))/ (2(X-1))

= (2X +1)(X-1) / (2(X-1))

= (2X + 1 )/2

= X + 1/2