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 x^th 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 = S_x-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 – S_x = {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.