Let S=(100)(100)+(99)(101)+(98)(102)+...+(2)(198)+(1)(199)
S=x=0∑99(100−x)(100+x)=x=0∑991002−x=0∑99x2
S=1003−699×100×199=1003−199×1650
Given that, S=100α−199β
∴α=3, β=1650
The slope of line passing through (\alpha ,\beta )&(0,0)=3−01650−0=550
JEE Main 2021 — Mathematics Algebra
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:
Held on 18 Mar 2021 · Verified 6 Jul 2026.
540
550
530
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