Solution:
Given,
Row houses have numbers ranging from 1,2,3,4,5,……49.
As a result, we can observe that the houses in a row are in the form of AP.
So,
The first term, a = 1
The common difference, d=1
Let’s say the total number of xth houses can be as follows;
Sum of the nth term of AP = n/2[2a+(n-1)d]
Sum of the number of houses beyond the x house = Sx-1
= (x-1)/2[2.1+(x-1-1)1]
= (x-1)/2 [2+x-2]
= x(x-1)/2 ………………………………………(i)
Based of the given condition, we can write,
S49 – Sx = {49/2[2.1+(49-1)1]}–{x/2[2.1+(x-1)1]}
= 25(49) – x(x + 1)/2 ………………………………….(ii)
Equations I and (ii) are equal under the stated condition;
Therefore,
x(x-1)/2 = 25(49) – x(x-1)/2
x = ±35
We all know that the number of houses cannot be negative. As a result, x has a value of 35.