Formula for the $n$th term of a geometric sequence

$u_n = u_1 r^{n-1}$, where $r$ is the common ratio ($r = \frac{u_{n+1}}{u_n}$).

Sum of the first $n$ terms of a geometric series

$S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}$, valid for $r \ne 1$.

How do you find the first term greater than some value $V$ in a geometric sequence?

Solve $u_1 r^{n-1} > V$ using logs or a GDC table, then take the next whole number $n$. Check the terms either side.

Common mistake with a negative ratio such as $r = -2$

Forgetting brackets: $(-2)^4 = 16$ but $-2^4 = -16$. Always bracket a negative ratio when raising it to a power.

SL 1.3 Geometric sequences and series — IB Mathematics: Analysis and Approaches notes and practice questions

Geometric sequences multiply by a constant ratio each step, giving exponential growth or decay with neat formulas for terms and sums. Study guide, 3 practice questions with markschemes and 4 flashcards. Free.

Common ratio and nth term formula · Sum of a finite geometric series · Sigma notation for geometric sums · Finding r, n or the first term from conditions · Growth and decay applications

Study guide

The big idea

In a geometric sequence you multiply by the same number every step. That number is the common ratio $r = \frac{u_{n+1}}{u_n}$. Doubling bacteria, a ball losing 20% of its height each bounce, and a population growing 3% a year are all geometric. After $(n-1)$ multiplications,

\[u_n = u_1 r^{\,n-1}.\]

If $r > 1$ the terms grow, if $0 < r < 1$ they shrink towards zero, and if $r < 0$ they alternate in sign.

Summing a geometric series

Write $S_n = u_1 + u_1 r + \dots + u_1 r^{n-1}$, multiply by $r$, and subtract: almost everything cancels, leaving $S_n(r - 1) = u_1(r^n - 1)$. So

\[S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}, \quad r \neq 1.\]

The two forms are equal; choose the one that keeps the denominator positive.

Worked example

The sequence $3, 6, 12, \dots$ has $u_1 = 3$, $r = 2$.

$u_{10} = 3 \times 2^9 = 1536$.
$S_{10} = \dfrac{3(2^{10} - 1)}{2 - 1} = 3 \times 1023 = 3069$.

Smallest $n$ with $S_n > 10\,000$. We need $3(2^n - 1) > 10\,000$, so $2^n > 3334.3$. Since $2^{11} = 2048$ and $2^{12} = 4096$, the answer is $n = 12$. (With logs: $n > \frac{\ln 3334.3}{\ln 2} \approx 11.7$.)

Two terms given. If $u_3 = 12$ and $u_6 = 96$, divide: $r^3 = 8$, so $r = 2$ and $u_1 = 12 / 4 = 3$. Dividing removes $u_1$, just as subtracting removed it for arithmetic sequences.

Growth and decay

A quantity that changes by a fixed percentage is geometric. A 6% increase means $r = 1.06$; a 6% decrease means $r = 0.94$. A town of 40 000 growing 2% a year has $40\,000 \times 1.02^{n}$ people after $n$ years. Be careful with indexing: "after $n$ years" uses power $n$, while "in year $n$" with year 1 as the start uses $n - 1$.

Sigma notation

$\sum_{k=1}^{6} 5(0.5)^{k-1}$ is geometric with $u_1 = 5$, $r = 0.5$, six terms: $S_6 = \frac{5(1 - 0.5^6)}{0.5} = 9.84375$.

Common mistakes

Writing $r^n$ instead of $r^{n-1}$ in the $n$th term.
Turning a 6% decrease into $r = 0.06$ instead of $0.94$.
Taking only the positive root when $r^2$ is known: $r$ could be negative.
Dividing by $\ln r$ with $r < 1$ and forgetting the inequality flips because $\ln r < 0$.

Exam tips

Check that the ratio, not the difference, is constant before using these formulas.
For inequalities in $n$, a GDC table of $S_n$ is quick and avoids sign errors.
Give exact values when the question works in integers or fractions; otherwise three significant figures.

Cheat sheet

Fact	Formula
Ratio	$r = u_{n+1} / u_n$
$n$th term	$u_n = u_1 r^{n-1}$
Sum of $n$ terms	$S_n = \dfrac{u_1(r^n - 1)}{r - 1}$, $r \ne 1$
Equivalent form	$S_n = \dfrac{u_1(1 - r^n)}{1 - r}$
$p\%$ increase	$r = 1 + p/100$
$p\%$ decrease	$r = 1 - p/100$

Two terms known: divide them to eliminate $u_1$.
Solve $r^n > k$ with logs or a GDC table.

Key definitions

Practice questions

SLPaper 1Foundation

Question 1

A geometric sequence begins $5, 15, 45, \ldots$

(a) [1]

Write down the common ratio.

Markscheme

1 $r = 3$

(b) [2]

Find $u_6$.

Markscheme

1 Substitutes into $u_n = u_1 r^{n-1}$, e.g. $5 \times 3^5$
1 $u_6 = 1215$

(c) [2]

Find the sum of the first 6 terms.

Markscheme

1 Substitutes into $S_n = \frac{u_1(r^n - 1)}{r - 1}$, e.g. $\frac{5(3^6 - 1)}{2}$
1 $S_6 = 1820$

SLPaper 1Standard

Question 2

In a geometric sequence, $u_2 = 12$ and $u_5 = -96$.

(a) [4]

Find the common ratio and the first term.

Markscheme

1 Divides the terms, e.g. $\frac{u_1 r^4}{u_1 r} = \frac{-96}{12}$
1 $r^3 = -8$
1 $r = -2$
1 $u_1 = -6$

(b) [2]

Find $S_8$.

Markscheme

1 Substitutes into the sum formula, e.g. $\frac{-6((-2)^8 - 1)}{-2 - 1}$
1 $S_8 = 510$

SLPaper 2Standard

Question 3

A geometric sequence has first term $2$ and common ratio $1.5$.

(a) [2]

Find the sum of the first 10 terms.

Markscheme

1 Substitutes into the sum formula, e.g. $\frac{2(1.5^{10} - 1)}{0.5}$
1 $S_{10} = 227$ (accept $226.66$)

(b) [4]

Find the first term of the sequence that is greater than 1000. Give its position $n$ and its value.

Markscheme

1 Sets up inequality, e.g. $2(1.5)^{n-1} > 1000$
1 Uses logs or a table, e.g. $n - 1 > \frac{\ln 500}{\ln 1.5} = 15.3$
1 $n = 17$
1 $u_{17} = 1310$ (accept $1313.68$)

Fact	Formula
Ratio	\(r = u_{n+1} / u_n\)
\(n\)th term	\(u_n = u_1 r^{n-1}\)
Sum of \(n\) terms	\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\), \(r \ne 1\)
Equivalent form	\(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\)
\(p\%\) increase	\(r = 1 + p/100\)
\(p\%\) decrease	\(r = 1 - p/100\)

SL 1.3 Geometric sequences and series

Study guide

The big idea

Summing a geometric series

Worked example

Growth and decay

Sigma notation

Common mistakes

Exam tips

Cheat sheet

Key definitions

Practice questions

Question 1

Question 2

Question 3

Key facts