The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Here, a = 1 and d = 1.
Sum to n terms of an AP is given by
S_n = n/2[ 2a + (n -1)d]
The sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Therefore
Sx-₁ = S₄₉ – Sx
⇒ (x -1/2)[2a + (x – 1 – 1)d]
⇒ 49/2[2a + (49 – 1)d] – x/2[2a + (x -1)d]
⇒ x -1/2[2(1) + (x -2)(1)]
⇒ 49/2[2(1) + 48(1)] – x/2[(1) + (x – 1)(1)]
⇒ (x – 1)[x] = 49[50] – x[x + 1]
⇒ x² – x = 2450 – x² – x
⇒ 2x² = 2450 x² = 1225 ⇒ x = 35
Hence, the value of x is 35.