# CBSE 10th Maths Paper

2016 Q27. Sum of an AP

#### Video Explanation

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