Question

Find the average of the first five multiples of 10.

Answer 1

Hint: We solve this question by using the concept of average or mean. Average or mean is nothing but the sum of all the elements divided by the number of elements. Mean is calculated using the formula $\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}$ , where $\overline{x}$ is the average or mean, ${{x}_{i}}$ is the ith element in the group of elements and n is the total number of elements. Using this formula for the first five multiples of 10, we find its average.

Complete step by step solution:
In order to solve this question, let us first calculate the first five multiples of 10. We know that multiples of 10 means the next term differing from the previous term by 10. Hence, the first five multiples of 10 are 10, 20, 30, 40 and 50. Now we are required to calculate the mean of these numbers. We do this by using the formula $\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}$ , where $\overline{x}$ is the average or mean, ${{x}_{i}}$ is the ith element in the group of elements and n is the total number of elements. This basically means the sum of all the terms divided by the number of terms. Here, 10 is ${{x}_{1}},$ 20 is ${{x}_{2}},$ 30 is ${{x}_{3}},$ 40 is ${{x}_{4}},$ and 50 is ${{x}_{5}}.$
$\Rightarrow \overline{x}=\dfrac{\sum\limits_{i=1}^{5}{{{x}_{i}}}}{5}$
Expanding this,
$\Rightarrow \overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}}{5}$
Substituting in this formula,
$\Rightarrow \overline{x}=\dfrac{10+20+30+40+50}{5}$
Adding all the terms in the numerator and dividing by 5,
$\Rightarrow \overline{x}=\dfrac{150}{5}=30$
Hence, the average of the first five multiples of 10 is 30.

Note: We need not use the formula for mean as $\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}.$ We can also solve this question by just using the basic definition of average which means the sum of all the terms divided by the number of terms. By using the formula, we are just representing this is a mathematical way.