$a=2 b+1$
$2 b=a-1$
$R=\{(3,1),(5,2), \ldots,(99,49)\}$
Let $(2 m+1, m),(2 \lambda-1, \lambda)$ are such ordered pairs.
According to the condition
$m =2 \lambda-1 \Rightarrow m=$ odd number
$\Rightarrow 1^{\text {st }}$ element of ordered pair $( a , b )$
$a=2(2 \lambda-1)+1=4 \lambda-1$
Hence $a \in\{3,7, \ldots, 99\}$
$\Rightarrow \lambda \in\{1,2, \ldots, 25\}$
$\Rightarrow$ set of sequence
$\left\{(4 \lambda-1,2 \lambda-1),(2 \lambda-1, \lambda-1),\left(\lambda-1, \frac{\lambda-2}{2}\right), \ldots \ldots\right\}$
$2^{\text {nd }}$ element of each ordered pair $=\frac{\lambda-2^{ r -2}}{2^{ r -2}}$
For maximum number of ordered pairs in such sequence
$\frac{\lambda-2^{r-2}}{2^{r-2}}=1 \text { or } 2 ; 1 \leq \lambda \leq 25$
$\lambda=2^{r-1} \text { or } \lambda=3.2^{r-2}$
$\text { Case-I }: \lambda=2 r-1$
$\lambda=2,2^2, 2^3, 2^4$
$r=2,3,4,5$
Hence maximum value of $r$ is $5$ when $\lambda=16$
Case-II : $\lambda =3.2^{ r -2}$
$\lambda=3, 6, 12,$
$r=24 3, 4,$
Hence maximum value of $r$ is $5$ when $\lambda=24$