Question

If $\alpha ,\beta$ are natural numbers such that ${100}^{\alpha }–199\beta =\left(100\right)\left(100\right)+\left(99\right)\left(101\right)+\left(98\right)\left(102\right)+\dots .+\left(1\right)\left(199\right)$, then the slope of the line passing through $(\alpha ,\beta )$ and origin is

• A. $510$
• B. $550$
• C. $540$
• D. $530$

Answer 1

The correct option is B

$550$

Explanation for the correct option:

Find the slope of the line:

$\begin{array}{rcl}{100}^{\alpha }–199\beta & =& \left(100\right)\left(100\right)+\left(99\right)\left(101\right)+\left(98\right)\left(102\right)+\dots .+\left(1\right)\left(199\right)\\ & =& \sum _{r=0}^{99}\left(100-r\right)\left(100+r\right)\\ & =& \sum _{r=0}^{99}\left({100}^2-r^2\right)\\ & =& {100}^3-\frac{99\times 100\times 199}{6}\\ & =& {100}^3-199\times 1650\end{array}$

$\Rightarrow$$\alpha =3,\beta =199$

$\begin{array}{rcl}\tan \theta & =& \frac{b}{a}\\ & =& \frac{1650}{3}\\ & =& 550\end{array}$

$\therefore$The required slope of the line is $\begin{array}{rcl}& & \mathbf{550}\end{array}$

Hence, Option ‘B’ is Correct.