Suppose a sequence \(a_1,\,a_2,\,a_3,\,\cdots,a_n,\,\cdots\) has the property that the same number is being multiplied (or divided) to get the next term of the sequence. In other words, \(a_k=a_{k-1}\cdot r\), from a recursive format. When this occurs, the sequence is called an geometric sequence, and the number \(r\) is called the common ratio.
Consider the sequence given by \(12, 36, 108, 324, \cdots\).
To check for a common ratio, since \(a_k=a_{k-1}\cdot r\), solving for \(r\): \(r=\displaystyle{\frac{a_k}{a_{k-1}}}\). This means, start with a term and divide by the term IN FRONT of it. $$\begin{array}{l} r=\frac{36}{12} & \implies r=3 \\ r=\frac{108}{36} & \implies r=3 \\ r=\frac{324}{108} & \implies r=3 \end{array}$$ The common ratio is \(r=3\) and this sequence is geometric.
Since each term is multiplying by \(r=3\), a pattern can be found to help generate an explicit formula for the general term \(a_n\): $$\begin{array}{l} a_1=12 & & \implies a_1=12\cdot3^0 \\ a_2=36 & \implies a_2=12\cdot3 & \implies a_2=12\cdot3^1 \\ a_3=108 & \implies a_3=36\cdot3 & \implies a_3=12\cdot3^2 \\ a_4=324 & \implies a_4=324\cdot3 & \implies a_4=12\cdot3^3 \end{array}$$ The pattern emerges as starting at \(a_1=12\) and multiplying by a exponent of \(r=3\). That exponent is one less than the term number. For example, \(a_4\) starts at \(a_1=12\) and multiplies by \(3\) three times, \(a_3\) starts at \(a_1=12\) and multiplies by \(3\) two times, \(a_2\) starts at \(a_1=12\) and multiplies by \(3\) one time. To finish the pattern, \(a_1\) starts at \(12\) and multiplies by \(3\) zero times. This leads to the explicit formula: \(a_n=12\cdot3^{n-1}\).
The above example shows how to create a more general formula for all geometric sequences. Once the common ratio \(r\) has been found, each term starts at \(a_1\) and multiplies by \(r\;\;n-1\) times. This is expressed in the explicit formula below:
For a geometric sequence with first term \(a_1\) and common ratio \(r\), the general term is given by: $$a_n=a_1\cdot r^{n-1}$$
Consider the series formed from a geometric sequence: \(\displaystyle{S_n=\sum_{k=1}^n\left(a_1\cdot r^{n-1}\right)}\), the \(S_n\) is a quick short-hand notation for this sum. Since the sequence is geometric, a formula to calculate this sum can be obtained that depends on only the first and last terms of the sequence. To illustrate how the formula is derived, consider a geometric sequence of just 5 terms:
| Want the sum: | \(S_5=a_1+a_2+a_3+a_4+a_5\) |
| Applying geometric sequence formula: | \(\begin{array}{l} S_5=a_1+(a_1\cdot r)+(a_1\cdot r^2)+(a_1\cdot r^3)+(a_1\cdot r^4) \\ \implies S_5=a_1\left(1+r+r^2+r^3+r^4\right) \end{array}\) |
| Clever trick of multiplying by \(\displaystyle{\frac{1-r}{1-r}}\): | \(\displaystyle{S_5=a_1\left(1+r+r^2+r^3+r^4\right)\cdot\frac{1-r}{1-r}}\) |
| Use of the Distributive Property & like terms: | \(\begin{array}{l} S_5=a_1\left(\large{\frac{1+r+r^2+r^3+r^4-r-r^2-r^3-r^4-r^5}{1-r}}\right) \\ \implies S_5=a_1\left(\large{\frac{1-r^5}{1-r}}\right)\;\color{green}{\checkmark} \end{array}\) |
This example shows how the following general formula for a geometric series is derived:
For a finite geometric series with first term \(a_1\) and common ratio \(r\) , the sum \(S_n\) is given by: $$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$
The infinite geometric series can only occur provided the property \(|r|< 1\) holds. In other words, the common ratio must have an absolute value less than \(1\). In this case, it can be shown that \(r^n\longrightarrow0\) as \(n\longrightarrow\infty\). This modifies the above finite geometric series formula to the following:
For an infinite geometric series with first term \(a_1\) and common ratio \(|r|< 1\) , the sum \(S\) is given by: $$S=\sum_{k=1}^\infty a_1r^{k-1}=\frac{a_1}{1-r}$$
Find the sum \(\;\displaystyle{\sum_{k=0}^\infty(-10)(0.2)^k}\).
In this example, the infinite sum starts at \(k=0\), but that has no bearing on the calculation of the answer. The common ratio is \(r=0.2\) which meets the criteria that \(|r|< 1\) in the formula for an infinite sum. Here \(a_1=-10\) and directly substituting into the formula gives:
\(\displaystyle{\sum_{k=0}^\infty(-10)(0.2)^k=\frac{-10}{1-(0.2)} \implies\frac{-10}{0.8}\implies -12.5\;\color{green}{\checkmark}}\)