If - - are natural numbers such that 100-199-100100-99101-98102-1199- then the slope of the line passing through - - and origin is-

RHS=∑_r=0^99(100 r)(100+r) =(100)^3 99×100× 1996 =(100)^3 (1650)199 LHS=(100)^α (199)β So, α=3, β=1650 Slope =tanθ=βα=550 ...

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Standard XII

Mathematics

Reflection of a Point about a Line

If α, β are n...

Question

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:

510

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550

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540

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530

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Solution

The correct option is B $550$
$R H S = 99 \sum r = 0 (100 - r) (100 + r)$
$= (100)^{3} - \frac{99 \times 100 \times 199}{6} = (100)^{3} - (1650) 199$

$L H S = (100)^{α} - (199) β$
So, $α = 3, β = 1650$
Slope $= tan θ = \frac{β}{α} = 550$

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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:

The correct option is B 550RHS=99∑r=0(100−r)(100+r) =(100)3−99×100×1996 =(100)3−(1650)199 LHS=(100)α−(199)β So, α=3,β=1650 Slope =tanθ=βα=550

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:

The correct option is B $550$
$R H S = 99 \sum r = 0 (100 - r) (100 + r)$
$= (100)^{3} - \frac{99 \times 100 \times 199}{6} = (100)^{3} - (1650) 199$

$L H S = (100)^{α} - (199) β$
So, $α = 3, β = 1650$
Slope $= tan θ = \frac{β}{α} = 550$