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 ]
Mathematics
Class 12
2229
Emily
Solution:
Given, Row houses are numbers from 1,2,3,4,5…….49. The houses numbered in a row and all are in the form of AP. So, Therefore, First term, a = 1 and the common difference, d = 1
Now, Let us say the number of xth houses can be represented as the sum of preceding the numbers of x = sum of following numbers of x, i.e.
Sum of ( 1, 2, 3, …., x – 1) = sum of [(x + 1), (x + 2 ), . . . . . 48, 49]
That is 1 + 2 + 3 + …… + ( x – 1) = ( x + 1) + ( x + 2) . . . . . . + 49
⇒ (x – 1)/2 [1+ x – 1] = (49 – x)/2 [x + 1 + 49]
⇒ (x – 1) x = (49 – x) (x + 50)
⇒ x2 – x = 49x + 2450 – x2 – 50x
⇒ x2 – x = 2450 – x² – x
⇒ 2x2 = 2450
⇒ x2 = 1225
x = √1225
x = 35
Therefore, the value of x is 35.