What is index error in the term formula?

The nth term is ar^n-1 (the power is n - 1, not n).

Find the 8th term of the GP 3, 6, 12, [2 marks]

A GP has a = 5 and r = -12. Find the sum to infinity. [2 marks]

Explain why the series 1 + 2 + 4 + 8 + has no sum to infinity. [2 marks]

SingaporeMathsSyllabus dot point

How do geometric progressions grow, and when does their sum converge?

Use the formulae for the nth term and the sum of a geometric progression, determine convergence, and find the sum to infinity of a convergent geometric series

A focused answer to the H2 Mathematics outcome on geometric progressions. The nth term and sum formulae, the condition for convergence, the sum to infinity, and applications to growth and decay problems.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use the geometric progression (GP) formulae for the $n$ th term and the sum of $n$ terms, decide whether an infinite geometric series converges, and find its sum to infinity when it does. This connects to exponential growth and decay throughout the syllabus.

The answer

Definition and the nth term

A geometric progression has a constant common ratio $r$ between consecutive terms. With first term $a$ :

u_n = ar^{n-1}

Each term is the previous one multiplied by $r$ , giving exponential rather than linear growth.

The sum of n terms

S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a(r^n - 1)}{r - 1}, \qquad r \neq 1

The two forms are equal; use whichever keeps the arithmetic tidy (the first when $|r| < 1$ , the second when $r > 1$ ).

Convergence and the sum to infinity

As $n \to \infty$ , $r^n \to 0$ only when $|r| < 1$ . In that case the partial sums settle to a finite limit:

S_\infty = \frac{a}{1 - r}, \qquad |r| < 1

If $|r| \geq 1$ the terms do not shrink to zero and the series diverges (no sum to infinity). This convergence condition is the central new idea.

Finding a and r

As with APs, two conditions give two equations. Dividing one by the other usually eliminates $a$ and isolates a power of $r$ , which you then solve.

Worked example

A geometric series has first term $12$ and common ratio $\dfrac{2}{3}$ . Find the sum to infinity and the least number of terms whose sum exceeds $35$ .

Step 1: Confirm convergence and find the sum to infinity

$|r| = \dfrac{2}{3} < 1$ , so $S_\infty = \dfrac{12}{1 - \frac{2}{3}} = \dfrac{12}{\frac{1}{3}} = 36$ .

Step 2: Write the partial sum

S_n = \frac{12\left(1 - \left(\frac{2}{3}\right)^n\right)}{1 - \frac{2}{3}} = 36\left(1 - \left(\tfrac{2}{3}\right)^n\right)

Step 3: Set up the inequality

36\left(1 - \left(\tfrac{2}{3}\right)^n\right) > 35 \implies 1 - \left(\tfrac{2}{3}\right)^n > \frac{35}{36} \implies \left(\tfrac{2}{3}\right)^n < \frac{1}{36}

Step 4: Solve using logarithms

$n\ln\dfrac{2}{3} < \ln\dfrac{1}{36}$ . Dividing by the negative $\ln\frac{2}{3}$ reverses the inequality: $n > \dfrac{\ln(1/36)}{\ln(2/3)} = \dfrac{-3.584}{-0.405} \approx 8.84$ .

Step 5: State the result

The least integer is $n = 9$ terms.

Examples in context

Example 1. Compound interest. $1000 invested at$ 5% $per year grows geometrically: the balance after$ n $years is$ 1000(1.05)^n $. Because$ r = 1.05 > 1$ the series of yearly balances diverges, matching the fact that money keeps growing without bound.

Example 2. A bouncing ball. A ball dropped from $2$ m rebounds to $0.6$ of its previous height each bounce. The total distance travelled is $2 + 2(2 \times 0.6 + 2 \times 0.6^2 + \cdots) = 2 + 2 \times \dfrac{2 \times 0.6}{1 - 0.6} = 2 + 6 = 8$ m, using a convergent GP for the rebounds.

Try this

Q1. Find the $8$ th term of the GP $3, 6, 12, \ldots$ [2 marks]

Cue. $a = 3$ , $r = 2$ , $u_8 = 3 \times 2^7 = 384$ .

Q2. A GP has $a = 5$ and $r = -\dfrac{1}{2}$ . Find the sum to infinity. [2 marks]

Cue. $|r| < 1$ , so $S_\infty = \dfrac{5}{1 - (-\frac{1}{2})} = \dfrac{5}{\frac{3}{2}} = \dfrac{10}{3}$ .

Q3. Explain why the series $1 + 2 + 4 + 8 + \cdots$ has no sum to infinity. [2 marks]

Cue. The common ratio is $2$ with $|r| \geq 1$ , so the terms grow and the partial sums increase without limit.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA geometric progression has second term

6

and fifth term

48

. Find the first term and the common ratio.

Show worked answer →

Using $u_n = ar^{n-1}$ : $ar = 6$ and $ar^4 = 48$ .

Divide: $\dfrac{ar^4}{ar} = r^3 = \dfrac{48}{6} = 8$ , so $r = 2$ .

Then $a(2) = 6$ , so $a = 3$ .

Markers reward writing both conditions, dividing to eliminate $a$ and find $r$ , and substituting back for $a$ .

Original4 marksA geometric series has first term

a

and common ratio

r = 0.8

. The sum to infinity is

40

. Find

a

and the sum of the first

5

terms.

Show worked answer →

Since $|r| < 1$ , the sum to infinity is $S_\infty = \dfrac{a}{1 - r} = \dfrac{a}{0.2} = 40$ , so $a = 8$ .

Sum of first $5$ terms: $S_5 = \dfrac{a(1 - r^5)}{1 - r} = \dfrac{8(1 - 0.8^5)}{0.2}$ .

$0.8^5 = 0.32768$ , so $S_5 = \dfrac{8(0.67232)}{0.2} = \dfrac{5.37856}{0.2} = 26.8928 \approx 26.9$ .

Markers reward using the sum to infinity to find $a$ , the finite sum formula, and a correct numerical answer.