Question

Find the mean of the first five multiples of 3.

Answer 1

Hint: : For solving this problem we need to generalise the equation of mean for first \['n'\] multiples of 3 and then we substitute \[n=5\] to get the required mean. We find the mean of data using the formula
\[\text{mean}=\dfrac{\text{sum of all the numbers}}{\text{number of numbers}}\].

Complete step-by-step solution:
Let us assume that there are \['n'\] multiples of 3.
So, we can say that the number of multiples is \['n'\]
Now, let us find the sum of first \['n'\] multiples of 3 as follows
\[\Rightarrow {{S}_{n}}=3\left( 1 \right)+3\left( 2 \right)+3\left( 3 \right)+.......+3\left( n \right)\]
By converting the above equation to summation we will get
\[\begin{align}
  & \Rightarrow {{S}_{n}}=\sum{3n} \\
 & \Rightarrow {{S}_{n}}=3\sum{n} \\
\end{align}\]
We know that the formula of first \['n'\] natural numbers as \[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
By substituting this result in above equation we will get
\[\begin{align}
  & \Rightarrow {{S}_{n}}=3\left( \dfrac{n\left( n+1 \right)}{2} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{3n\left( n+1 \right)}{2} \\
\end{align}\]
Here, we got the sum of first \['n'\] multiples of 3.
But, we are asked to find the sum of the first five multiples of 3.
By substituting \[n=5\] in above result we get
\[\begin{align}
  & \Rightarrow {{S}_{5}}=\dfrac{3\left( 5 \right)\left( 5+1 \right)}{2} \\
 & \Rightarrow {{S}_{5}}=45 \\
\end{align}\]
Let us assume that the mean of the first five multiples of 3 as \[X\].
We know that the formula of mean is given as
\[\text{mean}=\dfrac{\text{sum of all the numbers}}{\text{number of numbers}}\]
By substituting the required numbers in above formula we get
\[\begin{align}
  & \Rightarrow X=\dfrac{{{S}_{5}}}{n} \\
 & \Rightarrow X=\dfrac{45}{5} \\
 & \Rightarrow X=9 \\
\end{align}\]
Therefore the mean of the first five multiples of 3 is 9.

Note: This problem can be solved in another method.
We know that the first five multiples of 3 are 3, 6, 9, 12, 15.
Let us find the sum of numbers as
\[\begin{align}
  & \Rightarrow S=3+6+9+12+15 \\
 & \Rightarrow S=45 \\
\end{align}\]
Here, there are 5 numbers.
Let us assume that the mean of the first five multiples of 3 as \[X\].
We know that the formula of mean as
\[\text{mean}=\dfrac{\text{sum of all the numbers}}{\text{number of numbers}}\]
By substituting the required numbers in above formula we get
\[\begin{align}
  & \Rightarrow X=\dfrac{S}{n} \\
 & \Rightarrow X=\dfrac{45}{5} \\
 & \Rightarrow X=9 \\
\end{align}\]
Therefore the mean of the first five multiples of 3 is 9.