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

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$If α, β are natural numbers such that 100^{α} - 199 β = (100) (100) + (99) (101) + (98) (102) + \dots + (1) (199), then the slope of the line$ $passing through (α, β) and origin is$

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510

530

550

540

answer is C.

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Detailed Solution

$\begin{array}{l} 100.100 + 99 . (101) + 98 . (102) + \dots . . + 1.199 \\ \sum_{r = 0}^{99} (100 - r) (100 + r) \\ \sum_{r = 0}^{99} (100^{2} - r^{2}) \\ = 100^{2} . 100 - \sum r^{2} \\ = 100^{3} - \frac{99.100.199}{6} \\ = 100^{3} - 199.33.50 = 100^{3} - 199.1650 \\ α = 3, β = 1650 \\ Slope = \frac{β}{α} = \frac{1650}{3} = 550 \end{array}$