+280 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.
+551 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 \(r = \frac{3}{5}\).
