#### Question: 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 proceeding the house numbered x is equal to sum of the numbers of houses following x.

#### Video Explanation

#### Explanatory Answer

**Objective** : To find x

**Condition**: Sum of positive integers from 1 to (x - 1) = sum of positive integers from (x + 1) to 49

Both parts are in AP with common difference of 1.

Sum of first n terms of an AP, where a_{1} is the first term and d is the common difference = \\frac{n}{2}) (2a_{1} + (n - 1)d)

∴ Sum of positive integers from 1 to (x – 1) = \\frac{x-1}{2}) (2 + x - 2)

= \\frac{x (x-1)}{2}) .....(1)

Number of terms from (x + 1) to 49 = 49 - x

Sum of positive integers from x + 1 to 49

= \\frac{49-x}{2}) (2 (x + 1) + (49 - x - 1))

= \\frac{49-x}{2})(2x + 2 + 48 - x)

= \\frac{49-x}{2}) (x + 50) ......(2)

Equate (1) and (2) because sum from 1 to (x - 1) is equal to sum from (x + 1) to 49.

\\frac{x (x-1)}{2}) = \\frac{49 - x}{2}) (x + 50)

x^{2} – x = 49x + 49 × 50- x^{2} – 50x

x^{2} - x = -x^{2} - x + 49 × 50

2x^{2} = 49 × 50

x^{2} = 49 × 25

x = 7 × 5 = 35