Question

The houses numbered consecutively 1to49. Show that there exists a value X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X

Answer 1

use a n = s n - s n-1

Answer 2

Consider the numbers as:
1, 2, 3, .....,x-1, x, x+1,.....,49
Let first term of AP is a and common difference is d.
Thus sum of first n terms,
$s = \frac{n}{2} (2a + (n - 1)d)$

Sum of first x-1 terms,
$s1 = \frac{x - 1}{2} (2 + (x - 2)) = \frac{x(x - 1)}{2}$
Sum of last x-49 terms,
$s2 = \frac{49 - x}{2} (2 \times (x + 1) + (48 - x))$
Given s1 = s2,
$x(x - 1) = (49 - x)(x + 50)$
I.e. x = 35.
Since 1 < x < 49, proved