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

$u_n = u_1 r^{n-1}$, where $r$ is the common ratio.

Formula for the sum of the first $n$ terms of a geometric sequence

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

How do you find $r$ from two non-consecutive terms, e.g. $u_2$ and $u_5$?

$u_5 = u_2 r^{3}$, so $r = \sqrt[3]{u_5 / u_2}$. The power is the gap between the term numbers.

A quantity grows by 6% per year. What is $r$?

$r = 1.06$. A decrease of 6% gives $r = 0.94$.

SL 1.3 Geometric sequences and series — IB Mathematics: Applications and Interpretation notes and practice questions

Sequences that change by a constant factor each step, their terms and sums, and growth or decay applications. Study guide, 3 practice questions with markschemes and 4 flashcards. Free.

Common ratio and nth term · Sum of a finite geometric series · Sigma notation · Growth, decay and spread-type applications · Finding unknown parameters

Study guide

The idea

A geometric sequence is multiplied by the same factor every step. A post shared by each viewer to three new people, a population growing 4% per year, a filter removing 20% of the dust each pass: each new value is a fixed multiple of the last. That multiple is the common ratio $r = \dfrac{u_{n+1}}{u_n}$.

Growth of $p\%$ per step: $r = 1 + \dfrac{p}{100}$ (15% growth gives $r = 1.15$).
Decay of $p\%$ per step: $r = 1 - \dfrac{p}{100}$ (20% loss gives $r = 0.8$).

The nth term

Starting from $u_1$ and multiplying $n-1$ times:

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

The sum of n terms

For $r \ne 1$,

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

The two versions are identical; the first is tidier when $r > 1$, the second when $0 < r < 1$.

Worked example

A short video gets 1200 views on day 1, and daily views grow by 15% per day.

Views on day 6: $u_6 = 1200 \times 1.15^{5} \approx 2414$.
Total views in the first 6 days: $S_6 = \dfrac{1200(1.15^{6} - 1)}{0.15} \approx 10\,500$ (10 504.5 before rounding).
On which day do daily views first exceed 10 000? Solve $1200 \times 1.15^{\,n-1} > 10\,000$. With a table or solver, $1.15^{15} \approx 8.14$ and $1.15^{16} \approx 9.36$, while we need more than $8.33$, so $n - 1 = 16$ and $n = 17$: day 17.

Finding unknown parameters

If you know two terms, divide them to remove $u_1$. For example, $u_3 = 45$ and $u_6 = 1215$ give $r^{3} = 27$, so $r = 3$ and $u_1 = 5$. When the unknown is the number of terms, use a GDC table, the solver, or logarithms.

GDC tips

Use the table of $y = 1200 \times 1.15^{x-1}$ to spot the first integer value that passes a target.
Store $r$ in memory when it is not exact, to avoid rounding drift.

Common mistakes

Writing $r = 0.15$ for 15% growth; the ratio is 1.15.
Using the power $n$ instead of $n-1$ in the nth term.
Mixing up "the value on day 6" (a term) and "the total over 6 days" (a sum).
Rounding $r$ to two decimal places early, which can change the final answer noticeably.

Exam tips

Show the substituted formula before the calculator value.
When the answer is a count of days or people, round in the direction that the context demands and say why.
Check reasonableness: a decaying quantity should never become negative or exceed its start.

Cheat sheet

Quantity	Formula
nth term	$u_n = u_1 r^{n-1}$
Sum ($r \ne 1$)	$S_n = \dfrac{u_1(r^n - 1)}{r - 1}$
Common ratio	$r = u_{n+1} / u_n$

$p\%$ growth → $r = 1 + p/100$; $p\%$ decay → $r = 1 - p/100$
Two terms known → divide them to get $r$
Unknown $n$ → table, solver or logs; $n$ must be a positive integer
Term = one value; series = running total

Key definitions

Practice questions

SLPaper 1Foundation

Question 1

A chain message is started by 5 people on day 1. On each following day, the number of new people receiving it is three times the number on the previous day.

(a) [1]

Write down the common ratio.

Markscheme

1 $r = 3$

(b) [2]

Find the number of new people who receive the message on day 6.

Markscheme

1 $5 \times 3^{5}$
1 $1215$

(c) [2]

Find the total number of people who have received the message by the end of day 6.

Markscheme

1 $\frac{5(3^6 - 1)}{3 - 1}$
1 $1820$

SLPaper 2Standard

Question 2

A short video gets 1200 views on its first day online. Each day after that, the number of views that day is 1.35 times the number of views on the previous day.

(a) [2]

Find the number of views on day 8. Give your answer to the nearest whole number.

Markscheme

1 $1200 \times 1.35^{7}$
1 $9807$ views

(b) [2]

Find the total number of views in the first 8 days.

Markscheme

1 $\frac{1200(1.35^{8} - 1)}{1.35 - 1}$
1 $34\,400$ (accept $34\,397$)

(c) [1]

Suggest one reason why this model may not be suitable for predicting views on day 60.

Markscheme

1 Valid reason, e.g. growth cannot continue for ever / the number of possible viewers is limited / interest falls over time

SLPaper 2Challenging

Question 3

After each cleaning cycle, the amount of a chemical left in a pond is a fixed fraction of the amount before the cycle. The amounts, in grams, form a geometric sequence $u_1, u_2, u_3, \ldots$ where $u_2 = 48$ and $u_5 = 24.576$.

(a) [3]

Find the common ratio and $u_1$.

Markscheme

1 $r^{3} = \frac{24.576}{48}$
1 $r = 0.8$
1 $u_1 = 60$

(b) [2]

Find $\sum_{n=1}^{10} u_n$.

Markscheme

1 $\frac{60(1 - 0.8^{10})}{1 - 0.8}$
1 $268$ (267.787...)

(c) [3]

Find the smallest value of $n$ for which $u_n < 5$.

Markscheme

1 $60 \times 0.8^{n-1} < 5$
1 Uses GDC/logs or table: $u_{12} = 5.15$, $u_{13} = 4.12$
1 $n = 13$

Quantity	Formula
nth term	\(u_n = u_1 r^{n-1}\)
Sum (\(r \ne 1\))	\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\)
Common ratio	\(r = u_{n+1} / u_n\)

SL 1.3 Geometric sequences and series

Study guide

The idea

The nth term

The sum of n terms

Worked example

Finding unknown parameters

GDC tips

Common mistakes

Exam tips

Cheat sheet

Key definitions

Practice questions

Question 1

Question 2

Question 3

Key facts