Find the arithmetic mean of :

(i) first eight natural numbers;

(ii) first five prime numbers;

(iii) first six positive even integers;

(iv) first five positive integral multiples of 3;

(v) all factors of 20.

(i) Given,

first eight natural numbers = 1, 2, 3, 4, 5, 6, 7, 8

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} \\[1em] = \dfrac{36}{8} \\[1em] = 4.5$

Hence, mean of given numbers = 4.5.

(ii) Given,

first five prime numbers = 2, 3, 5, 7, 11

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{2 + 3 + 5 + 7 + 11}{5} \\[1em] = \dfrac{28}{5} \\[1em] = 5.6$

Hence, mean of given numbers = 5.6.

(iii) Given,

first six positive even integers; = 2, 4, 6, 8, 10, 12

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{2 + 4 + 6 + 8 + 10 + 12}{6} \\[1em] = \dfrac{42}{6} \\[1em] = 7$

Hence, mean of given numbers = 7.

(iv) Given,

first five positive integral multiples of 3 = 3, 6, 9, 12, 15

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{3 + 6 + 9 + 12 + 15}{5} \\[1em] = \dfrac{45}{5} \\[1em] = 9.$

Hence, mean of given numbers = 9.

(v) Given,

all factors of 20 = 1, 2, 4, 5, 10, 20

Mean = $\dfrac{\sum x_i}{n}$

Substitute values, we get:

$\Rightarrow \text{Mean} = \dfrac{1 + 2 + 4 + 5 + 10 + 20}{6} \\[1em] = \dfrac{42}{6} \\[1em] = 7.$

Hence, mean of given numbers = 7.