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 the values of x.

Clearly, No of houses are in AP and

Given,

First term, a = 1

common difference d = 1

According to given statement: S_{1 to x- 1} = S_{(x+1) to 49}
⇒ S_{x+1 to 49} = S₄₉ – S_x and S_{1 to x- 1} = S_{x- 1}

Since the sum of n terms is S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Applying given condition:

S_{1 to x- 1} = S_{(x+1) to 49}
⇒ S_{x- 1} = S₄₉ – S_x
⇒ S_{x- 1} = S₄₉ – S_x


⇒ (x - 1)(x) = 49(50) - x(x+1)
⇒ x² – x = 49(50) – x² -x
⇒ 2x² = 49 × 50
⇒ x² = 49 × 25
⇒ x = 35 and x = -35
Here since the x is a number between 1 to 49, ∴ x = 35.

Hence, The value of x is 35.