Question

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

• A. 540
• B. 550
• C. 530
• D. 510

Answer 1

Answer: (B)

$S=(100)(100)+(99)(101)+(98)(102) \ldots \ldots \ldots(2)(198)+(1)(199) $
$S=\displaystyle\sum_{x=0}^{99}(100-x)(100+x)=\sum 100^2-x^2$
$=100^3-\frac{99 \times 100 \times 199}{6} $
$\alpha=3 $
$\beta=1650$
$\text { slope }=\frac{1650}{3}=550$