The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X. (CBSE 2016)

The AP in the above problem is

1, 2, 3, - - - , 49

With first term, a = 1

Common difference, d = 1

nth term of AP = a + (n - 1)d

a_n = 1 + (n - 1)1

a_n = n …[1]

Suppose there exist a mth term such that, (m<49)

Sum of first m - 1 terms of AP = Sum of terms following the mth term

Sum of first m - 1 terms of AP = Sum of whole AP - Sum of first m terms of AP

As we know sum of first n terms of an AP is, if last term a_n is given

(m - 1)(1 + m - 1) = 49(1 + 49) - m(1 + m) [using 1]

(m - 1)m = 2450 - m(1 + m)

m² - m = 2450 - m + m²

2m² = 2450

m² = 1225

m = 35 or m = - 35 [not possible as no of terms can't be negative.]

and a_m = m = 35 [using 1]

So, sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35.