Let $A=\{1,2,3, \ldots ., 100\}$ and $R$ be a relation on $A$ such that $R=\{(a, b): a=2 b+1\}$. Let

Question

Let $A=\{1,2,3, \ldots ., 100\}$ and $R$ be a relation on $A$ such that $R=\{(a, b): a=2 b+1\}$. Let $\left(a_{1}, a_{2}ight),\left(a_{2}, a_{3}ight),\left(a_{3}, a_{4}ight), \ldots .$, $\left(a_{k}, a_{k+1}ight)$ be a sequence of $k$ elements of $R$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer $k$, for which such a sequence exists, is equal to:\

Professor PC Thomas Classes · Answer

(b): Let the smallest value is $a_{k+1}$ in $A=\{1,2,3 \ldots, 100\}$.\ Since, $a_{k}=2 a_{k+1}+1=(2 	imes 1)+1=3$, $$ \begin{aligned} 	herefore \quad a_{k-1} & =2 a_{k}+1=(2 	imes 3)+1=7 \ a_{k-2} & =2 a_{k-1}+1=(2 	imes 7)+1=15 \ a_{k-3} & =2 a_{k-2}+1=(2 	imes 15)+1=31 \ a_{k-4} & =2 a_{k-3}+1=(2 	imes 31)+1=63 \ a_{k-5} & =2 a_{k-4}+1=(2 	imes 63)+1=127 
otin A \end{aligned} $$ ∴ The sequence is $\{63,31),(31,15),(15,7)$, $(7,3),(3,1)\}$\ $	herefore \quad k=5$\