Let $\mathrm{A}=\{1,2,3, \ldots, 10\}$ and R be a relation on A such that $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=2 \mathrm{~b}+1\}$. Let $\left(\mathrm{a}_1, \mathrm{a}_2\right)$, $\left(a_2, a_3\right),\left(a_3, a_4\right), \ldots .,\left(a_k, a_{k+1}\right)$ be a sequence of $k$…

Question

Let $\mathrm{A}=\{1,2,3, \ldots, 10\}$ and R be a relation on A such that $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=2 \mathrm{~b}+1\}$. Let $\left(\mathrm{a}_1, \mathrm{a}_2\right)$, $\left(a_2, a_3\right),\left(a_3, a_4\right), \ldots .,\left(a_k, a_{k+1}\right)$ 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 :

Aicharya · Accepted Answer

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

Let $\mathrm{A}=\{1,2,3, \ldots, 10\}$ and R be a relation on A such that $\mathrm{R}=\{(\mathrm{a},…

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