If α, β are natural numbers such that
100 α - 199β = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through (α, β) and origin is :

Solution

Maths sir · 4 steps

Each pair of factors sits equally either side of 100, so every product is a difference of squares.

1. Spot the pattern. The r-th term is (100-r)(100+r) for r = 0, 1, 2, …, 99. The first term has r = 0 and the last, (1)(199), has r = 99. So there are 100 terms, and each one equals 10000 - r².

   Σᵣ₌₀⁹⁹(10000 - r²) = 100 × 10000 - (99 × 100 × 199)/6

2. Keep 199 visible. Do not multiply out the second term completely. Write it as 199 × (99 × 100)/6 = 199 × 1650. The right side becomes 100³ - 199 × 1650.
3. Match both sides. Compare with 100^α - 199β. This gives α = 3 and β = 1650, and both are natural numbers as required.
4. So. The line through the origin and (3, 1650) has slope 1650/3 = 550, which is option (b).

The idea to keep. When the two factors move in opposite directions with a constant sum, write them as (a-r)(a+r) = a² - r². Then use Σ r², and keep the factors in the form the question is asking for.

Answer from the NTA final answer key. Solution by PrepWiser.