If , are natural numbers such that 100-199 = (100)(100)+(99)(101)+(98)(102)++(1)(199), the…

Question

If $\alpha, \beta$ are natural numbers such that $100\alpha-199\beta = (100)(100)+(99)(101)+(98)(102)+\cdots+(1)(199)$, then the slope of the line passing through $(\alpha,\beta)$ and the origin is:

Accepted Answer

RHS = $$\sum\limits_{r = 0}^{99} {(100 - r)(100 + r)} $$

$$ = {(100)^3} - {{99 \times 100 \times 199} \over 6} = {(100)^3} - (1650)199$$

LHS = (100) (\alpha\] - (199) (\beta\]

So, $$\alpha$$ = 3, $$\beta$$ = 1650

Slope = tan$$\theta$$ = $${\beta  \over \alpha }$$

$$ \Rightarrow $$ tan$$\theta$$ = 550

Solutions · 1