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# Geometric Series

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Let$$a + ar + ar^2 + ar^3 + \dotsb$$be an infinite geometric series. The sum of the series is 3 The sum of the squares of all the terms is 4 Find the common ratio.

Apr 15, 2024

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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 $$r = \frac{3}{5}$$.

Apr 28, 2024