CBSE 10th Maths Paper

2016 Q27. Sum of an AP

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 a1 is the first term and d is the common difference = \\frac{n}{2}) (2a1 + (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)
x2 – x = 49x + 49 × 50- x2 – 50x
x2 - x = -x2 - x + 49 × 50
2x2 = 49 × 50
x2 = 49 × 25
x = 7 × 5 = 35