What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

Hint: We are asked to find the average of all the multiples of ten from 10 to 190 inclusive. We start to solve the given question by finding out all the multiples of ten from 10 to 190 inclusive. Then, we find the average of the set of numbers to get the desired result.

Complete step-by-step answer:
We are asked to find the average of all the multiples of ten from 10 to 190 inclusive . We will be solving the given question by finding out all the multiples of ten from 10 to 190 inclusive and then finding the average of all the numbers.
The average value of the set of numbers is the middle value or the central value in the set. The average of a set of numbers is the number between the largest and the smallest number in the set.
It is defined as the ratio of the sum of all the values in a set of numbers to the total number of values.
Writing the above lines in the form of the equation, we get,
$\Rightarrow \text{Average = }\dfrac{\text{sum of all the terms}}{\text{total number of terms}}$

As per the question, we need to find the multiples of 10 from 10 to 190.
A given number is divisible by 10 if its last digit is 0.
Following the same,
The multiples of 10 from 10 to 190 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190.
From the above, we know that
$\Rightarrow \text{Average = }\dfrac{\text{sum of all the terms}}{\text{total number of terms}}$
We need to find the sum of all the terms and the total number of terms to calculate the average for the given set of numbers.
The given numbers are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190. So, the total number of terms is 19.
Writing the above line in the form of the equation, we get,
$\therefore \text{total number of terms = 19}$
Now, we need to find the sum of all the numbers.
Finding the sum of all the given numbers, we get,
$\Rightarrow \text{sum of all terms = 10+20+30+40+50+60+70+80+90+100+110+120+130+140+150+160+170+180+190}$
Taking the value of 10 in common from the above expression, we get,
$\Rightarrow \text{sum of all terms = 10}\left( \text{1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19} \right)$
We know that the sum of n terms from 1 to n is given by $\dfrac{n\left( n+1 \right)}{2}$
Here,
n = 19
From the above,
$\Rightarrow \text{1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19 = }\left( \dfrac{19\left( 19+1 \right)}{2} \right)$
Substituting the value in the above equation, we get,
$\Rightarrow \text{sum of all terms = 10}\left( \dfrac{19\left( 19+1 \right)}{2} \right)$
Simplifying the above equation, we get,
$\Rightarrow \text{sum of all terms = 10}\left( \dfrac{19\times 20}{2} \right)$
Canceling the common factors, we get,
$\Rightarrow \text{sum of all terms = 10}\times \text{19}\times \text{10}$
Simplifying it further, we get,
$\therefore \text{sum of all terms = 1900}$
Substituting the values in the formula, we get,
$\Rightarrow \text{Average = }\dfrac{1900}{19}$
Simplifying the above expression, we get,
$\therefore \text{Average = 100}$
$\therefore$ The average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive is 100.

Note: In the given question, the sum of terms 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190 can be found alternatively as follows,
The sum of n terms in an arithmetic progression is given as follows,
 $\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$
Here,
${{S}_{n}}$ is the sum of n terms
n is the number of terms
a is the first term
l is the last term
Following the same,
Here,
a = 10
l = 190
n = 19
Substituting the same, we get,
$\Rightarrow {{S}_{n}}=\dfrac{19}{2}\left( 10+190 \right)$
Simplifying the above equation, we get,
$\Rightarrow {{S}_{n}}=\dfrac{19}{2}\left( 200 \right)$
Canceling the common factors, we get,
$\therefore {{S}_{n}}=1900$