Find the value of r such that the mean of the first r odd natural numbers is $$\frac{r^2}{16}$$​.

Correct option is A

Given:

the mean of the first r odd natural numbers is equal to $$\frac{r^2}{16}$$​​

Concept Used:

First r Odd Natural Numbers:

The sequence of the first r odd natural numbers is:

​$$1, 3, 5, \ldots, (2r - 1)$$​​

Sum of First r Odd Natural Numbers:

It is a known result that the sum of the first r odd natural numbers is:

S = $$r^2$$​​

Mean of First r Odd Natural Numbers:

​$$\mu = \frac{S}{r} = \frac{r^2}{r} = r$$​​

Solution:

From the concepts above, we have:

​$$r = \frac{r^2}{16}$$​​

​$$16r = r^2$$​​

​$$r^2 - 16r = 0$$​​

r(r - 16) = 0

​$$r = 0 \quad \text{or} \quad r = 16$$​​

r = 0 is not valid because we cannot have zero terms in this context.

r = 16 is the valid solution.