4. 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 :Sx – 1 = S49 – Sx ]

Given, Row houses are numbers from 1,2,3,4,5…….49. Thus we can see the houses numbered in a row are in the form of AP. So, First term, a = 1 Common difference, d=1 Let us say the number of xth houses can be represented as; Sum of nth term of AP = n/2[2a+(n-1)d] Sum of number of houses beyond x house = Sx-1 = (x-1)/2[2.1+(x-1-1)1] = (x-1)/2 [2+x-2] = x(x-1)/2 ………………………………………(i) By the given condition, we can write, S49 – Sx = {49/2[2.1+(49-1)1]}–{x/2[2.1+(x-1)1]} = 25(49) – x(x + 1)/2 ………………………………….(ii) As per the given condition, eq.(i) and eq(ii) are equal to each other; Therefore, x(x-1)/2 = 25(49) – x(x-1)/2 x = ±35 As we know, the number of houses cannot be a negative number. Hence, the value of x is 35