Question

The students of a school decided to beautify the school on an annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored in the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. What is the maximum distance she travelled carrying a flag.

Answer 1

Hint:Here there are 27 flags, so by taking the midpoint, we get the 14th flag as the middle flag and arrange 13 flags on one side and 13 flags on the other side. The distance covered for carrying each flag should be taken as, distance covered for carrying one flag and in return. i.e. we will get the arithmetic series 4 + 8 + 12 +… till 13 terms. Next we have to apply the formula:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ where ${{S}_{n}}$ is the sum of the sequence, $a$ is the first term, $n$ is the number of terms and $d$ is the common difference.

Complete step-by-step answer:
We are given that the students decided to beautify the school on an annual day by fixing colourful flags on the straight passage of the school.
Here, the total number of flags = 27
Distance between each flag to be fixed = 2 m
Since, the flags are stored in the position of the middle most flag, first we have to find the middle most flag. i.e we will get the mid value or the median as:
$\begin{align}
  & Median=\dfrac{First\text{ }term+Last\text{ }term}{2} \\
 & Median=\dfrac{1+27}{2} \\
 & Median=\dfrac{28}{2} \\
 & Median=14 \\
\end{align}$
Therefore, we can say that the middle most flag is the 14th flag.
Next, we have to arrange 13 flags on one side, 13 flags on the other side and one flag in the middle.
Now, we have to calculate the distance travelled for carrying each flag, since it is given that Ruchi can carry only one flag at a time.
First, we have to place 13 flags on one side.
Now, the distance travelled to carry the first flag and in return = 2m + 2m =4m
The distance travelled to carry the second flag and in return = 4 m + 4 m = 8 m
The distance travelled to carry the third flag and in return = 6 m + 6 m = 12 m
Similarly the distance travelled to carry the thirteenth flag and in return = 4 m + 8 m + 12 m + …. + 13 terms.
Therefore, it will form an arithmetic series of the form:
4 + 8 + 12 + … + 13th term
Now, the distance travelled by Ruchi carrying 13 flags = 4 + 8 + 12 + … 13thterm.
We know that the sum of the terms in AP is calculated by the formula:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\text{ }.....\text{ (1)}$
Here, ${{S}_{n}}$ is the sum of the sequence, $a$ is the first term, $n$ is the number of terms and $d$ is the common difference. i.e.
$\begin{align}
  & n=13 \\
 & a={{a}_{1}}=4 \\
 & d={{a}_{2}}-{{a}_{1}} \\
 & d=8-4 \\
 & d=4 \\
\end{align}$
By substituting all these values in equation (1) we obtain:
$\begin{align}
  & {{S}_{n}}=\dfrac{13}{2}\left[ 2\times 4+(13-1)4 \right]\text{ } \\
 & {{S}_{n}}=\dfrac{13}{2}\left[ 8+12\times 4 \right]\text{ } \\
 & {{S}_{n}}=\dfrac{13}{2}\left[ 8+48 \right] \\
 & {{S}_{n}}=\dfrac{13}{2}\times 56 \\
\end{align}$
By cancellation we get:
$\begin{align}
  & {{S}_{n}}=28\times 13 \\
 & {{S}_{n}}=364 \\
\end{align}$
So, we can say that the distance travelled to carry 13 flags = $364m$
Therefore, we can say that the distance travelled to carry 27 flags = $2\times 364=728m$
Here, the maximum distance Ruchi travelled carrying a flag, D is:
$\begin{align}
  & D=2+2+2+...+13times \\
 & D=2\times 13 \\
 & D=26m \\
\end{align}$
Hence, the required maximum distance Ruchi travelled carrying a flag is $26m$

Note: Here, first we are finding the sum of 13 terms. So, to calculate total distance travelled carrying 27 flags you have to multiply it with 2, otherwise you will not get the total distance. Maximum distance carrying a flag is not the total distance. Maximum distance is 2+ 2 +… till 13 terms. So don’t get confused between total distance and maximum distance.