Geometric progression - SS2 Mathematics Lesson Note
If a sequence of numbers is such that the ratio (known as the common ratio, \(r\)) of any term to the term just preceding it, is constant then the sequence is a geometric progression or geometric sequence. In a sequence, \(2,\ 6,\ 18,\ 54\), the common ratio is \(\frac{6}{2} = \frac{18}{6} = \frac{54}{18} = 3\), making this a geometric sequence.
The \(n\)th term of a G.P. is \(T_{n} = ar^{n - 1}\)
The sum or series of \(n\) terms of a G.P. is:
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\(S_{n} = \frac{a(1 - r^{n})}{1 - r}\) when \(r < 1\)
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\(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\ when\ r > 1\ \).
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\(S_{n} = an\) when \(r = 1\)
Example 3
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What is the 12th term of the GP \(2,\ 14,\ 98,\ \ldots\)
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Find the sum of 10 terms of the GP \(4,\ 8,\ 16,\ \ldots\)
Solution
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\(T_{n} = ar^{n - 1}\), where \(n = 12\), \(a = 2\) and \(r = \frac{14}{2} = 7\)
\[T_{12} = (2){(7}^{12 - 1})\]
\(T_{12} = 2{(7}^{11})\) , unless otherwise required it is fine to leave your answer in index form
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Given a GP = \(4,\ 8,\ 16,\ \ldots\), \(r = \frac{8}{4} = 2\)
\[S_{n} = \frac{a(r^{n} - 1)}{r - 1}\ \]
\[S_{10} = \frac{4(2^{10} - 1)}{2 - 1} = \ \frac{4 \times 2^{10} - 4}{1} = 4 \times 2^{10} - 4 = \ 2^{2} \times 2^{10} - 2^{2} = 2^{12} - 2^{2}\]
The geometric mean of a set of numbers is the positive square root of the product of the first and last terms of the progression. That is, GM = \(\sqrt{al}\)
Example 4 Find three geometric means between \(7\frac{1}{2}\) and \(120\)
Solution
Let the geometric means be x, y and z between \(7\frac{1}{2}\) and \(120\), then \(7\frac{1}{2},\ x,\ y,\ z,\ 120\) will be GP. Suppose the common ratio be \(r\), then since \(a\ = 7\frac{1}{2}\) and \(T_{n}\ = \ 120\), we have,
\[T_{n} = ar^{n - 1}\]
\[T_{5} = 120 = (7\frac{1}{2})r^{4}\]
\[\frac{120}{1} = \frac{15r^{4}}{2}\]
\[240 = 15r^{4}\]
\[r^{4} = \frac{240}{15} = 16\]
\[r^{4} = 2^{4}\]
\[r = 2\]
\[x = T_{2} = \left( 7\frac{1}{2} \right)2^{2 - 1} = 15\]
\[y = \ T_{3} = \left( 7\frac{1}{2} \right)2^{3 - 1} = 30\]
\[z = \ T_{4} = \left( 7\frac{1}{2} \right)2^{4 - 1} = 60\]
Hence, the three geometric means between \(7\frac{1}{2}\) and \(120\) are \(15,\ 30\ and\ 60\).