the houses of a row are numbered consecutively from 1 to 49 , show that there is a number m such that the summer of the numbers of the houses preceding the house marked m is equal to the sum of the numbers of the house following it . find m
Answers
Let us assume that the required house no. is m
ie Sum of the numbers of houses preceding m is equal
to sum of the numbers of houses following it.
Sum of first n natural numbers is given by n*(n+1)/2.
So,Sum of numbers of houses preceeding x is sum of
first x-1 numbers = (m-1)*x/2.
Now, Sum of numbers of houses following it is
equal to sum of first n houses minus sum of firstm houses = n*(n+1)/2 - m*(m +1)/2.
A/Q,
They both are equal.
So, We have
(m-1)*m /2 = n*(n+1)/2 - m*(m +1)/2
=> (m-1)*m /2+(x+1)*m /2 = n*(n+1)/2
=>m *(m -1+m +1)/2 = n*(n+1)/2
=>m *2m/2 = n*(n+1)/2
Putting n=49, we get
=>m ^2=49*50/2
=>m ^2=49*25
=>m ^2=7^2*5^2
=>m = sqrt(7^2*5^2)
Therefore, m =7*5=35
Hence 35 is the number of the house such that
Sum of the numbers of houses preceding it is equal to sum of the numbers of
houses following it.