If 1 1 + 2 + 1 2 + 3 + + 1 99 + 100 = m and 11 2 + 12 3 + + 199 100 = n, then the point (m, n) lies on the line
- 11(x - 1) - 100y = 0
- 11x - 100y = 0
- 11(x - 1) - 100(y - 2) = 0
- 11(x - 2) - 100(y - 1) = 0
Answer
(2) 11x - 100y = 0
Solution
Rationalise: 1 r + r + 1 = r + 1 - r
m = 100 - 1 = 9 (telescoping)
1r(r + 1) = 1r - 1r + 1
n = 1 - 1100 = 99100
∵ 11(9) - 100 ( 99100 ) = 99 - 99 = 0 ⇒ (m, n) lies on 11x - 100y = 0
Answer: (2)