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

Question

If α, β are natural numbers such that 100α-199 β=(100)(100)+(99)(101)+(98)(102) + ldots .+(1)(199), then the slope of the line passing through (α, β) and origin is : (A) 540 (B) 550 (C) 530 (D) 510

Tardigrade Admin · Accepted Answer

S=(100)(100)+(99)(101)+(98)(102) ldots ldots ldots(2)(198)+(1)(199) S= displaystyle∑x=099(100-x)(100+x)=∑ 1002-x2 =1003-(99 × 100 × 199/6) α=3 β=1650 text slope =(1650/3)=550

Q. If $α, β$ are natural numbers such that $10 0^{α} - 199 β = (100) (100) + (99) (101) + (98) (102)$ $+ \dots . + (1) (199)$ , then the slope of the line passing through $(α, β)$ and origin is :

540

550

530

510

$S = (100) (100) + (99) (101) + (98) (102) \dots\dots\dots (2) (198) + (1) (199)$
$S = x = 0 \sum 99 (100 - x) (100 + x) = \sum 10 0^{2} - x^{2}$
$= 10 0^{3} - \frac{99 \times 100 \times 199}{6}$
$α = 3$
$β = 1650$
$slope = \frac{1650}{3} = 550$

Q. 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 :

540

550

530

510

S=(100)(100)+(99)(101)+(98)(102)………(2)(198)+(1)(199) S=x=0∑99​(100−x)(100+x)=∑1002−x2 =1003−699×100×199​ α=3 β=1650 slope =31650​=550

Q. If $α, β$ are natural numbers such that $10 0^{α} - 199 β = (100) (100) + (99) (101) + (98) (102)$ $+ \dots . + (1) (199)$ , then the slope of the line passing through $(α, β)$ and origin is :

$S = (100) (100) + (99) (101) + (98) (102) \dots\dots\dots (2) (198) + (1) (199)$
$S = x = 0 \sum 99 (100 - x) (100 + x) = \sum 10 0^{2} - x^{2}$
$= 10 0^{3} - \frac{99 \times 100 \times 199}{6}$
$α = 3$
$β = 1650$
$slope = \frac{1650}{3} = 550$