Multiply each time, and the sum can stop growing.

Multiply by the same number.

A sequence is a geometric progression when each term is the one before multiplied by a fixed number, the common ratio \(r\). Divide any term by the term before it to find \(r\): in \(5, 15, 45, 135, \dots\) the ratio is \(\dfrac{15}{5} = 3\). The ratio can be a fraction or negative, as in \(64, -32, 16, \dots\) where \(r = -\dfrac{1}{2}\) and the signs alternate. Starting from \(a\), the terms are \(a, ar, ar^{2}, ar^{3}, \dots\), so the \(n\)th term has only \(n - 1\) factors of \(r\).

\[ u_{n} = ar^{n-1} \]

• Test for a GP: \(\dfrac{u_{2}}{u_{1}} = \dfrac{u_{3}}{u_{2}}\). For three consecutive terms \(x, y, z\) this is \(y^{2} = xz\).
• Two terms given: write each as \(ar^{\text{power}}\) and DIVIDE. The \(a\) cancels and leaves a power of \(r\).
• AP or GP? An AP has a common difference (subtract); a GP has a common ratio (divide).

Sum of n terms, and the least n.

To add the first \(n\) terms, write the sum, multiply it by \(r\) and subtract. Every term except the first and the last cancels:

\[ S_{n} = a + ar + ar^{2} + \dots + ar^{n-1} \]

\[ rS_{n} = ar + ar^{2} + \dots + ar^{n-1} + ar^{n} \]

\[ S_{n} - rS_{n} = a - ar^{n} \]

\[ S_{n} = \frac{a(1 - r^{n})}{1 - r} \quad (r \ne 1) \]

• This is the form on the formula list. When \(r > 1\) you may prefer \(\dfrac{a(r^{n} - 1)}{r - 1}\), which is the same thing with both signs flipped.
• Least \(n\): the unknown is a power, so take \(\lg\) of both sides (Topic 6) and use \(\lg r^{n} = n \lg r\).
• If \(0 < r < 1\), then \(\lg r < 0\). Dividing an inequality by a negative number REVERSES it.
• \(n\) counts terms, so it is a whole number: round the boundary UP to the next integer, then check either side.

The sum to infinity.

In Fig. 1 each new term of \(8 + 4 + 2 + \dots\) fills half of the gap that is left, so the sums close in on 16 but never pass it. Algebra shows the same: in \(S_{n} = \dfrac{a(1 - r^{n})}{1 - r}\), if \(|r| < 1\) then \(r^{n} \to 0\) as \(n \to \infty\), so the sums approach a limit.

\[ S_{\infty} = \frac{a}{1 - r} \quad \text{provided } |r| < 1, \text{ that is } -1 < r < 1 \]

• Explaining 'has a sum to infinity': state the value of \(r\) and that \(|r| < 1\), so \(r^{n} \to 0\).
• Explaining 'has no sum to infinity': \(|r| \ge 1\), so the terms do not get smaller and the sums do not settle. For \(r = 2\) they grow without limit; for \(r = -1\) they swap between \(a\) and \(0\).
• Recurring decimals: \(0.\dot{7} = 0.7 + 0.07 + 0.007 + \dots\) is a GP with \(a = 0.7\), \(r = 0.1\), so it equals \(\dfrac{0.7}{0.9} = \dfrac{7}{9}\).