Infinite Sum of a Geometric Sequence - Converging Geometric Series

1. This sequence's common ratio is \(r=\frac{1}{3}\).

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2. Since \( -1 < r < 1\), indeed \( -1 < \frac{1}{3} < 1\), we can state without further justification that this sequence has an infinite sum.

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3. To calculate the sum of the first \(10\) terms, we use the formula for the sum of the first \(n\) terms, \(S_n\), to write: \[\begin{aligned} S_{10} & = \frac{u_1\begin{pmatrix}1-r^{10}\end{pmatrix}}{1-r} \\ & = \frac{486 \begin{pmatrix}1-\begin{pmatrix}\frac{1}{3}\end{pmatrix}^{10}\end{pmatrix}}{1-\frac{1}{3}} \\ & = \frac{486 \begin{pmatrix}1 - \begin{pmatrix}\frac{1}{3}\end{pmatrix}^{10}\end{pmatrix}}{\frac{2}{3}} \\ & = 728.98765432098 \\ S_{10} & = 728.99 \quad \text{(rounded to 2 dp)} \end{aligned}\]

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4. The infinite sum is calculated with the formula: \[S_{\infty} = \frac{u_1}{1-r}\] For this sequence this leads to: \[\begin{aligned} S_{\infty} &= \frac{u_1}{1-r} \\ & = \frac{486}{1 - \frac{1}{3}} \\ & = \frac{486}{\frac{2}{3}} \\ & = 486 \times \frac{3}{2} \\ S_{\infty} & = 729 \end{aligned} \]

We know that this sequence has an infinite sum \(S_{\infty} = 20\), in other words we know that the sum of its terms converges towards \(20\).

Furthermore, we're given this sequences's first term \(u_1 = 8\). Consequently we can use the infinite sum formula to write: \[S_{\infty} = \frac{u_1}{1-r}\] Now replacing \(u_1\) by \(8\) and \(S_{\infty}\) by \(20\) we obtain an equation for the unknown common ratio \(r\): \[20 = \frac{8}{1-r}\] To find \(r\) we start by getting away from the denominator, which we do by multiplying both sides of this equation by \(\begin{pmatrix}1-r \end{pmatrix}\): \[20\begin{pmatrix}1-r \end{pmatrix} = 8\] Now that \(r\) is away from the numerator, we can solve the usual way: \[\begin{aligned} & 20 - 20r = 8 \\ & -20r = 8-20 \\ & -20r = -12 \\ & r = \frac{-12}{-20} \\ & r = \frac{-4 \times 3}{-4 \times 5} \\ & r = \frac{3}{5} \quad \begin{pmatrix}r = 0.6 \end{pmatrix} \end{aligned}\]

In this example there are 2 unknowns to find, namely \(u_1\) and \(r\).

This suggests we need 2 distinct equations involving \(u_1\) and \(r\), which we can then solve simultaneously.

To obtain the 2 equations we use the 2 bits of information we're given. This is done as follows:

• We're told that \(u_3=3.2\) and we know that \(u_3 = u_1r^2\). Combining these two results leads to our first equation invvoling the two unknowns we're after: \[3.2 = u_1r^2 \quad \text{: (E1)}\]

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• We're told that \(S_{\infty} = 25\) and we know that \(S_{\infty} = \frac{u_1}{1-r}\). Combining these two results leads to a second equation involving the two unknows we're after: \[25 = \frac{u_1}{1-r} \quad \text{: (E2)} \]

Rearranging (E2), by multiplying both sides by \(\begin{pmatrix}1-r\end{pmatrix}\), we can make \(u_1\) the subject: \[25\begin{pmatrix}1-r\end{pmatrix} = u_1\] that's: \[u_1 = 25\begin{pmatrix}1-r\end{pmatrix} \quad \text{:(E2)}\] Now going back to the other equation we have, \(3.2 = u_1r^2\), and replacing \(u_1\) by the right hand side of that last expression \(u_1 = 25\begin{pmatrix}1-r\end{pmatrix}\), leads to: \[\begin{aligned} 3.2 & = 25\begin{pmatrix}1-r\end{pmatrix} r^2 \\ 3.2 & = 25\begin{pmatrix}r^2-r^3\end{pmatrix} \\ 3.2 & = 25r^2-25r^3 \end{aligned}\] This can be rearranged into the cubic equation: \[25r^3-25r^2+3.2 = 0\]

Now that we know that \(r=0.8\) we can go back to (E2), which was \(u_1 = 25\begin{pmatrix}1-r\end{pmatrix}\), and replace \(r\) by \(0.8\) to find \(u_1\): \[\begin{aligned} u_1 & = 25\begin{pmatrix}1-r \end{pmatrix} \\ & = 25\begin{pmatrix} 1 - 0.8 \end{pmatrix} \\ & = 25 \begin{pmatrix}0.2 \end{pmatrix} \\ u_1 & = 5 \end{aligned}\]

Finally we can state our answers as:

• the first term is \(u_1 = 5\)
• the common ratio is \(r = 0.8\).