Geometric Series

We have an infinite geometric series:

 

$$a + ar + ar^2 + ar^3 + \dotsb$$

 

The sum of this series can be calculated using the formula for the sum of an infinite geometric series:

 

$$S = \frac{a}{1 - r}$$

 

Given that the sum of the series is $3$, we have:

 

$$3 = \frac{a}{1 - r}$$

 

So, we have one equation:

 

$$a = 3(1 - r)$$

 

Now, let's consider the sum of the squares of all the terms in the series. The sum of the squares of an infinite geometric series is given by:

 

$$S_2 = \frac{a^2}{1 - r^2}$$

 

Given that the sum of the squares of all the terms is $4$, we have:

 

$$4 = \frac{a^2}{1 - r^2}$$

 

Now, let's substitute $a = 3(1 - r)$ into this equation:

 

$$4 = \frac{(3(1 - r))^2}{1 - r^2}$$

 

$$4 = \frac{9(1 - 2r + r^2)}{1 - r^2}$$

 

$$4 = \frac{9 - 18r + 9r^2}{1 - r^2}$$

 

$$4 - 4r^2 = 9 - 18r + 9r^2$$

 

Rearranging terms:

 

$$9r^2 - 4r^2 - 18r + 9 = 0$$

 

$$5r^2 - 18r + 9 = 0$$

 

Now, we can use the quadratic formula to solve for $r$:

 

$$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

 

$$r = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(5)(9)}}{2(5)}$$

 

$$r = \frac{18 \pm \sqrt{324 - 180}}{10}$$

 

$$r = \frac{18 \pm \sqrt{144}}{10}$$

 

$$r = \frac{18 \pm 12}{10}$$

 

Now, we have two possible values for $r$:

 

$$r_1 = \frac{18 + 12}{10} = \frac{30}{10} = 3$$

$$r_2 = \frac{18 - 12}{10} = \frac{6}{10} = \frac{3}{5}$$

 

However, since the common ratio of a geometric series cannot be equal to 1 (as it would lead to a divergent series), the only valid solution is $r = \frac{3}{5}$.