Two consecutive number p,p + 1 are removed from natural numbers. 1,2,3, 4……….., `n-1,` n such that arithmetic mean of these number reduces by 1, then (n-p) is equal to.

Two consecutive numbers k,k+1 are removed from natural numbers 1,2,3,....n such that arithmetic mean reduces by 1 ,then (n-k) is equal to "

Two consecutive number from n natural numbers 1,2,3,……, n are removed. Arithmetic mean of the remaining numbers is (105)/(4). The value of n is:

Two consecutive number from n natural numbers 1,2,3,……, n are removed. Arithmetic mean of the remaining numbers is (105)/(4). The G.M. of the removed numbers is :

Two consecutive number from n natural numbers 1,2,3,……, n are removed. Arithmetic mean of the remaining numbers is (105)/(4). Let removed numbers are x _(1), x _(2) then x _(1)+ x _(2) + n=

Two consecutive numbers from 1,2,3,..., n are removed,then arithmetic mean of the remaining numbers is (105)/(4) then (n)/(10) must be equal to

Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic mean of the remaining numbers is 105/4 The sum of all numbers

Two consecutive numbers from 1,2,3,"……n" are removed. The arithmetic mean of the remaining numbers is (105)/4) . The value of n lies in

If the number of ways of selecting 3 numbers out of 1, 2, 3, ……., 2n+1 such that they are in arithmetic progression is 441, then the sum of the divisors of n is equal to

Two consecutive numbers from 1, 2, 3,...., n are removed. The arithmetic mean of the remaining numbers is (105)/(4) . Find n and those removed numbers.