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 (Hint: `S_[x-1]=S_[49]-S_x`)

The houses of a row are numbered consecutively from 1 to 49. Show that

1.	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 (Hint: `S_[x-1]=S_[49]-S_x`)
Answer» We are given an AP, namely 1, 2, 3,…… (x-1), x, (x +1),…, 49 Such that 1+2+3+…+ (x-1) = (x+1) + (x+2) +….+49. Thus, we have `S_(x-1) = S_(49)-S_(x) " "…(i)` Using the formula, `S_(n) = (n)/(2) (a+1)` in (i), we have `((x-1))/(2) * {1+(x-1)} = (49)/(2) * (1+49) - (x)/(2) * (1+x)` `rArr (x(x-1))/(2) + (x (x +1))/(2) = 1225` `rArr 2x^(2) = 2450 rArr x^(2) = 1225 rArr x = sqrt(1225) = 35` Hence, x = 35.