Develop and use finite arithmetic and geometric series formulas

Ontario Grade 11 Mathematics

Discrete Functions

Building and Applying Formulas for Sums of Sequences

You already know how to identify arithmetic and geometric sequences and how to find any single term using a formula. Now comes the natural next question: what if you need to add up many terms at once? Adding 200 terms one by one would take forever, but there is a shortcut — a formula — for each type. In this lesson you will see where those formulas come from, why they work, and how to use them on problems that mix both types.

What you will learn

Explain the difference between a sequence and a series in your own words.
Derive and apply the formula for the sum of a finite arithmetic series.
Derive and apply the formula for the sum of a finite geometric series.
Choose the correct formula for a given problem and interpret the result in context.

Prerequisite Bridge: Sequences vs. Series

A sequence is an ordered list of numbers. An arithmetic sequence increases or decreases by a fixed amount called the common difference,

d

. A geometric sequence multiplies by a fixed amount called the common ratio,

r

. You met both of these in the sequences unit.

A series is what you get when you add the terms of a sequence together. For example, the arithmetic sequence CAD 3, 7, 11, 15 becomes the arithmetic series

3 + 7 + 11 + 15

. The word finite means the series has a specific last term — you are not adding forever.

The sum of a finite series is written

S_n

, where

n

is the number of terms being added. Your goal is to find

S_n

without writing out and adding every single term.

A sequence lists terms; a series adds them.
$S_n$ means the sum of the first $n$ terms.
Arithmetic series: constant difference between terms.
Geometric series: constant ratio between terms.

Developing the Arithmetic Series Formula

Imagine you want to add up the first

n

terms of an arithmetic sequence whose first term is

a

and whose common difference is

d

. Written out, the sum looks like this:

S_n = a + (a+d) + (a+2d) + \cdots + t_n

, where

t_n

is the last term.

Here is the clever trick that turns this into a formula. Write

S_n

a second time, but in reverse — from the last term back to the first. Then add the two versions together term by term. Every pair adds up to exactly

a + t_n

, and there are

n

such pairs. So the combined total is

n(a + t_n)

. Because you added

S_n

twice, you divide by 2.

This gives the first version of the arithmetic series formula. If you also recall that

t_n = a + (n-1)d

, you can replace

t_n

and get a second version that only needs

a

,

n

, and

d

. Both versions are equally valid — pick whichever fits the information you have.

A concrete sanity check: for the four-term series

3 + 7 + 11 + 15

, the first term is

a = 3

, the last term is

t_4 = 15

, and there are

n = 4

terms. The formula gives

S_4 = \frac{4}{2}(3 + 15) = 2 \times 18 = 36

. Adding directly:

3 + 7 + 11 + 15 = 36

. It checks out.

S_n = \frac{n}{2}(a + t_n) = \frac{n}{2}[2a + (n-1)d]

Write the sum forwards and backwards, then add — pairs always equal $a + t_n$ .
$S_n = \frac{n}{2}(a + t_n)$ when the last term is known.
$S_n = \frac{n}{2}[2a + (n-1)d]$ when only $a$ , $d$ , and $n$ are known.
Both forms come from the same pairing argument.

Developing the Geometric Series Formula

Now consider the first

n

terms of a geometric sequence with first term

a

and common ratio

r

(where

r \neq 1

). The sum is

S_n = a + ar + ar^2 + \cdots + ar^{n-1}

.

A different trick is needed here. Multiply both sides by

r

to get

rS_n = ar + ar^2 + ar^3 + \cdots + ar^n

. Now subtract the original

S_n

from

rS_n

. Almost every term cancels — the middle terms appear in both and disappear. Only

ar^n

from the second expression and

a

from the first survive. So

rS_n - S_n = ar^n - a

.

Factor both sides: the left side gives

S_n(r - 1)

and the right side gives

a(r^n - 1)

. Dividing both sides by

(r - 1)

produces the formula. A quick sanity check: for the geometric series

2 + 6 + 18 + 54

we have

a = 2

,

r = 3

,

n = 4

. The formula gives

S_4 = \frac{2(3^4 - 1)}{3 - 1} = \frac{2 \times 80}{2} = 80

. Direct addition:

2 + 6 + 18 + 54 = 80

. Correct.

Note: when

r = 1

, every term equals

a

, so

S_n = na

. This special case is simple enough that the formula is rarely needed.

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

Multiply by $r$ and subtract to cancel interior terms.
$S_n = \frac{a(r^n - 1)}{r - 1}$ for $r \neq 1$ .
Equivalently written as $S_n = \frac{a(1 - r^n)}{1 - r}$ — both forms are the same.
Identify $a$ , $r$ , and $n$ before substituting.

Choosing the Right Formula

Before calculating anything, you must decide which type of series you have. Test consecutive terms: if the difference

t_2 - t_1 = t_3 - t_2

(constant difference), it is arithmetic. If the ratio

\frac{t_2}{t_1} = \frac{t_3}{t_2}

(constant ratio), it is geometric. A series is never both (unless every term is the same, which is a special case).

Once you know the type, list out what you know —

a

,

d

or

r

,

n

, and possibly

t_n

or

S_n

— and match those to the formula. Sometimes

n

is not given directly; you may need to use the general term formula

t_n = a + (n-1)d

or

t_n = ar^{n-1}

first to find it.

Context clues matter. A problem about equal monthly deposits into a savings account most likely involves an arithmetic series. A problem about a bouncing ball whose height decreases by a fixed percentage each bounce involves a geometric series.

Check for constant difference → arithmetic; constant ratio → geometric.
List known values ( $a$ , $d$ or $r$ , $n$ , $t_n$ , $S_n$ ) before choosing a formula.
Find a missing $n$ using the general term formula if needed.
Context often signals which type of series applies.

Putting It Together: Mixed Applications

Many problems require more than one step. You might be given

S_n

and asked to find

n

, or given enough information to find

a

before computing the sum. Work systematically: identify the series type, write down all known values, decide which formula version to use, substitute carefully, and check that your answer is reasonable.

Pay attention to units and context. If a problem says 'the total amount saved over 12 months', then

n = 12

and

S_{12}

is the answer. If it asks 'how many terms must be added to reach a total of 500', then

S_n = 500

and you solve for

n

.

Always verify your answer by checking at least one of its properties — for example, confirm that the sum of the first and last term matches what the formula predicts, or compute a few terms by hand and compare a partial sum.

Multi-step problems: find missing values first, then apply the series formula.
Match 'total' language in the problem to $S_n$ .
Verify answers against direct calculations when possible.
Keep track of units and interpret $S_n$ in the context of the problem.

Arithmetic vs. Geometric Series: Formula Comparison

Feature	Arithmetic Series	Geometric Series
What stays constant?	Common difference $d$	Common ratio $r$
General term	$t_n = a + (n-1)d$	$t_n = ar^{n-1}$
Sum formula (version 1)	$S_n = \frac{n}{2}(a + t_n)$	$S_n = \frac{a(r^n-1)}{r-1}$
Sum formula (version 2)	$S_n = \frac{n}{2}[2a+(n-1)d]$	Same formula, also written $\frac{a(1-r^n)}{1-r}$
Quick check method	Add first and last term, multiply by $\frac{n}{2}$	Multiply by $r$ , subtract, and factor

Worked example

Arithmetic Series: Seating in a Theatre

A theatre has 20 rows of seats. The first row has 15 seats and each row after it has 3 more seats than the one in front. How many seats are in the theatre in total?

Identify the series type and list known values
Each row has 3 more seats than the previous one, so the number of seats per row forms an arithmetic sequence. The first term is $a = 15$ , the common difference is $d = 3$ , and the number of terms (rows) is $n = 20$ .
$a = 15, d = 3, n = 20$
Choose the formula version
The last term $t_{20}$ is not directly given, but $a$ , $d$ , and $n$ are all known. Use the version $S_n = \frac{n}{2}[2a + (n-1)d]$ so you do not need to find $t_{20}$ separately.
$S_n = \frac{n}{2}[2a + (n-1)d]$
Substitute the known values
Replace $n$ with $20$ , $a$ with $15$ , and $d$ with $3$ in the formula.
$S_{20} = \frac{20}{2}[2(15) + (20-1)(3)]$
Simplify inside the brackets
Calculate $2 \times 15 = 30$ and $(20 - 1) \times 3 = 19 \times 3 = 57$ . Add those together to get $30 + 57 = 87$ .
$S_{20} = 10 × [30 + 57] = 10 × 87$
Compute the final answer
Multiply $10 \times 87$ to get the total number of seats.
$S_{20} = 870$

Answer: The theatre has 870 seats in total.

Check: The last row has

t_{20} = 15 + (19)(3) = 15 + 57 = 72

seats. Using the other formula version:

S_{20} = \frac{20}{2}(15 + 72) = 10 \times 87 = 870

. Both versions agree.

Worked example

Geometric Series: Doubling Prize Money

A game show offers prize money that doubles each round. The first round prize is CAD 500. A contestant wins every round. After 8 rounds, what is the total prize money the contestant has won?

Identify the series type and list known values
Each round's prize is double the previous one, so the prizes form a geometric sequence. The first term is $a = 500$ , the common ratio is $r = 2$ , and the number of terms is $n = 8$ .
$a = 500, r = 2, n = 8$
Write the geometric series formula
Since $r \neq 1$ , use the geometric series formula.
$S_n = \frac{a(r^n - 1)}{r - 1}$
Substitute the known values
Replace $a$ with $500$ , $r$ with $2$ , and $n$ with $8$ .
$S_8 = \frac{500(2^8 - 1)}{2 - 1}$
Evaluate the power and simplify
Calculate $2^8 = 256$ , so $2^8 - 1 = 255$ . The denominator is $2 - 1 = 1$ , so dividing by 1 changes nothing.
$S_8 = \frac{500 × 255}{1} = 500 × 255$
Compute the final answer
Multiply $500 \times 255$ to get the total winnings.
$S_8 = 127{,}500$

Answer: The contestant won a total of CAD 127,500 over 8 rounds.

Check: List the eight prizes: 500, 1000, 2000, 4000, 8000, 16000, 32000, 64000. Their sum is

500 + 1000 + 2000 + 4000 + 8000 + 16000 + 32000 + 64000 = 127{,}500

. This matches.

Common mistakes and how to avoid them

Confusing the number of terms

n

with the value of the last term

t_n

.

Correction: Always establish

n

(how many terms you are adding) separately from

t_n

(the value of the last term). They are different quantities with different roles in the formulas.

Using the arithmetic series formula on a geometric series, or vice versa.

Correction: Before substituting, check whether the difference or the ratio is constant. Only then choose the matching formula.

Forgetting to subtract 1 in

(n-1)d

or

(n-1)

in the exponent

r^{n-1}

.

Correction: Remember that the first term already accounts for the starting value, so the adjustment applies

n-1

times, not

n

times.

Applying the geometric series formula when

r = 1

, which causes division by zero.

Correction: When

r = 1

, every term equals

a

, so simply use

S_n = na

.

Misreading a problem and solving for a single term

t_n

instead of the sum

S_n

.

Correction: Look for words like 'total', 'altogether', or 'sum' to confirm that

S_n

is needed. Words like 'the value in the

n

th year' or 'the

n

th term' signal a single-term calculation instead.

Lesson summary

A series is the sum of the terms of a sequence; a finite series has a definite last term.
An arithmetic series sums terms with a constant difference $d$ ; use $S_n = \frac{n}{2}[2a+(n-1)d]$ or $S_n = \frac{n}{2}(a + t_n)$ .
A geometric series sums terms with a constant ratio $r$ ; use $S_n = \frac{a(r^n-1)}{r-1}$ for $r \neq 1$ .
Both formulas were developed by a clever algebraic trick: pairing terms (arithmetic) or multiplying by $r$ and subtracting (geometric).
Identify the series type first, list known values, then substitute — and always verify your answer.
Multi-step problems may require finding a missing value such as $n$ or $a$ before applying the series formula.

Check your understanding

Question 1

Which expression gives the sum of the arithmetic series

5 + 9 + 13 + 17 + 21

?

$S_5 = \frac{5}{2}(5 + 21)$
$S_5 = 5 \times 21$
$S_5 = \frac{5}{2}(5 \times 21)$
$S_5 = 5 + 4(5)$

Show answer and explanation

S_5 = \frac{5}{2}(5 + 21)

There are

n = 5

terms, the first is

a = 5

, and the last is

t_5 = 21

. The formula

S_n = \frac{n}{2}(a + t_n)

gives

S_5 = \frac{5}{2}(5 + 21) = \frac{5}{2}(26) = 65

. You can verify:

5 + 9 + 13 + 17 + 21 = 65

.

Question 2

A geometric series has

a = 4

,

r = 3

, and

n = 5

. What is

S_5

?

$484$
$60$
$972$
$324$

Show answer and explanation

484

S_5 = \frac{4(3^5 - 1)}{3 - 1} = \frac{4(243 - 1)}{2} = \frac{4 \times 242}{2} = \frac{968}{2} = 484

. The series is

4 + 12 + 36 + 108 + 324 = 484

.

Question 3

A series begins

100, 95, 90, \ldots

and continues for 10 terms. What is the total?

$775$
$950$
$550$
$850$

Show answer and explanation

775

This is arithmetic with

a = 100

,

d = -5

,

n = 10

. Apply

S_{10} = \frac{10}{2}[2(100) + (10-1)(-5)] = 5[200 - 45] = 5 \times 155 = 775

.

Question 4

A problem asks for the 'total amount saved over 6 months'. Which value does this correspond to?

$S_6$ , the sum of the first 6 terms
$t_6$ , the 6th term only
$6 \times t_1$ , six copies of the first term
$a + 6d$ , the term after the 6th

Show answer and explanation

S_6

, the sum of the first 6 terms

The word 'total' signals that all terms must be added together. That is the definition of a series sum, so you need

S_6

, not a single term.

Key terms

Sequence: An ordered list of numbers, each called a term.
Series: The sum of the terms of a sequence.
Finite series: A series with a fixed number of terms and therefore a definite last term.
$S_n$: The sum of the first $n$ terms of a series.
Common difference ( $d$ ): The fixed amount added from one term to the next in an arithmetic sequence.
Common ratio ( $r$ ): The fixed number each term is multiplied by to get the next term in a geometric sequence.
Arithmetic series: A series formed by adding the terms of an arithmetic sequence.
Geometric series: A series formed by adding the terms of a geometric sequence.

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