Areas of circles follow the sequence 1,4,9,16,…………100${}^2$

Areas of Black & Red annuals – 1,3,5,7,9……………199 – General term 2n – 1

Red annuals – 3, 7, 11, …………….. 199 Sum = $\left(\frac{50}{2}\right)$*(3 + 199)

Largest circle’s area = 100${}^2$

Ratio = $\frac{101}{200}$

Note : Ignore pi as it is a ratio of areas

Shortcut: Reverse gear approach –

We know that the denominator is 100${}^2$ and num is an integer, hence the denominator cannot be 101 as it is in no-way a factor of 10000. Hence option d) has been eliminated.

We need to find $\frac{R}{(R+B)}$. R + B = 10000. Each of the red annual is greater in area by 2 units than the subsequent black annual and there are 50 such cases. Hence R = 5050 and B = 4950. $\frac{R}{(R+B)}$. = $\frac{5050}{10000}$ = $\frac{101}{200}$