The average of the first 18 positive multiples of 9 is:

Correct Answer: The correct option is 85.5.

Step-by-Step Explanation:

Step 1: Understand the series of numbers

The problem asks for the average of the first 18 positive multiples of 9.

The first 18 positive multiples of 9 are:

$9\times 1=9$

$9\times 2=18$

$9\times 3=27$

...

$9\times 18=162$

Step 2: Formula for the average of an Arithmetic Progression (AP)

Since the difference between consecutive multiples of 9 is constant (9), this sequence forms an Arithmetic Progression (AP).

For any sequence in an arithmetic progression, the average can be found using the formula:

$\text{Average}=\frac{\text{First term}+\text{Last term}}{2}$

Step 3: Calculate the average

Here, the first term is 9 and the 18^th (last) term is 162.

Substituting these values into the formula:

$\text{Average}=\frac{9+162}{2}$

$\text{Average}=\frac{171}{2}$

$\text{Average}=85.5$

Alternative Method:

The sum of the first n natural numbers is given by:

$\text{Sum}=\frac{n(n+1)}{2}$

Therefore, the sum of the first 18 multiples of 9 is:

$\text{Sum}=9\times \frac{18\times 19}{2}=9\times 9\times 19=1539$

To find the average, divide the sum by the total number of terms (18):

$\text{Average}=\frac{1539}{18}=85.5$

Thus, the average of the first 18 positive multiples of 9 is 85.5.