Question

What is the serial number of a boy in the middle in a row of 45 boys?

Answer 1

Hint: In this question we have to find out the serial number of boys in the middle of 45 boys, so first of all we have to find out how many boys in the middle. If total number of boys are an even number that is in the form of $2n$ then we have two middle terms that is \[{{\left( \dfrac{2n-1}{2} \right)}^{th}}\] and ${{\left( \dfrac{2n+1}{2} \right)}^{th}}$ terms. If number are odd number, suppose $2m+1$ then only one term if middle term that is \[{{\left( \dfrac{2m+1+1}{2} \right)}^{th}}\] term.
In the given question the total number of students is 45 which is an odd number, so we apply here an odd number case.

Complete step by step answer:
In the given question we have a total of 45 students, so mark them as first, second, third, fourth, and so on up to $ 45^{th}$.
As we know that if $2m+1=45$ is an odd number then we have only one middle term, and this is ${{\left( \dfrac{2m+1+1}{2} \right)}^{th}}$ term.
So, suppose the middle term is denoted by $M$, hence we can write
\[M=\dfrac{45+1}{2}\]
Here we put $2m+1=45$
Further on calculation we can write
$\begin{align}
  & M=\dfrac{46}{2} \\
 & \Rightarrow M=23 \\
\end{align}$
Hence ${{23}^{th}}$ term is middle term.
As we mark the serial numbers of boys as ${{1}^{th}},{{2}^{nd}},{{3}^{rd}}...$ and so on up to ${{45}^{th}}$, so we can say that serial number of a boy in the middle in a row of 45 boys is twenty third.

Note: It should be noted that if total number of terms is odd number then the middle term is also at odd place, and if total number of terms even then we have two middle terms, one at even place and one at odd place.