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C2.4 · Solve applications involving arithmetic and geometric sequences and series

Learn to solve applications involving arithmetic and geometric sequences and series through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Discrete Functions

Arithmetic and Geometric Applications — MCR3U Expectation C2.4

Many real situations follow a predictable numerical pattern. A salary that increases by the same dollar amount each year, or a population that doubles every decade, are both examples of sequences at work. In this lesson you will connect the formulas you already know for arithmetic and geometric sequences and series to practical problems. You will practise choosing the right formula, substituting carefully, and interpreting your answer in the language of the original question. Every new term is defined when it first appears, and every example is worked out step by step so you can follow the reasoning, not just the arithmetic.

What you will learn

Prerequisite Bridge: Sequences and Series at a Glance

A sequence is an ordered list of numbers. Each number in the list is called a term. The first term is written t1t_1, the second t2t_2, and so on. The term in position nn is written tnt_n and is called the general term.
An arithmetic sequence has a constant difference between consecutive terms. That constant is called the common difference, dd. The general term is tn=t1+(n−1)dt_n = t_1 + (n-1)d. The sum of the first nn terms of an arithmetic sequence is Sn=n2(t1+tn)S_n = \frac{n}{2}(t_1 + t_n), or equivalently Sn=n2(2t1+(n−1)d)S_n = \frac{n}{2}\bigl(2t_1 + (n-1)d\bigr).
A geometric sequence has a constant ratio between consecutive terms. That constant is called the common ratio, rr. The general term is tn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}. The sum of the first nn terms of a geometric sequence is Sn=t1(rn−1)r−1S_n = \frac{t_1(r^n - 1)}{r - 1} when r≠1r \neq 1.
A series is the sum of the terms of a sequence. When a problem asks for a total accumulated amount, you need a series formula. When it asks for a specific value at a particular time or position, you need a sequence formula.

Recognising the Pattern in an Application Problem

The hardest part of an application problem is deciding which type of sequence applies. Look at how the quantity changes from one step to the next. If the change is always the same fixed amount (add or subtract), the situation is arithmetic. If the change is always the same fixed multiplier (multiply or divide), the situation is geometric.
Read the problem once to understand the context, then extract three pieces of information: the starting value, how the value changes at each step, and what the question is actually asking — a specific term, or a total?
Common arithmetic clues in problem language: 'increases by,' 'decreases by,' 'earns an extra,' 'saves an additional,' or any phrase suggesting a fixed dollar or unit change per period.
Common geometric clues: 'doubles,' 'triples,' 'grows by a percentage,' 'depreciates by a percentage,' or 'halves.' A percentage change means you multiply by a fixed ratio each period — for example, growth of 20% per year means r=1.20r = 1.20, and depreciation of 15% per year means r=0.85r = 0.85.
Once you have identified the type, write down t1t_1, dd or rr, and nn explicitly before touching any formula. This habit prevents the most common errors.

Arithmetic Applications: Steady Change Over Time

Suppose someone saves a fixed amount more each month than the previous month. The amounts saved each month form an arithmetic sequence because the increase is constant. If you want to know the total saved after several months, you need the arithmetic series formula.
Consider a scenario: a theatre has 20 seats in the first row, 23 seats in the second row, 26 in the third, and so on. The common difference is d=3d = 3. To find the number of seats in the 15th row, use tn=t1+(n−1)dt_n = t_1 + (n-1)d, giving t15=20+(14)(3)=62t_{15} = 20 + (14)(3) = 62 seats. To find the total seats in the first 15 rows, use S15=152(t1+t15)=152(20+62)=152(82)=615S_{15} = \frac{15}{2}(t_1 + t_{15}) = \frac{15}{2}(20 + 62) = \frac{15}{2}(82) = 615 seats.
Notice that answering 'how many seats in row 15?' required the sequence formula, while 'how many seats in total across 15 rows?' required the series formula. The context drives the formula choice.
In salary and savings problems, check carefully whether the first given value is for period 1 or period 0. This affects the value of nn you use, and getting it wrong by one is the single most common application error.
Sn=n2(t1+tn)S_n = \frac{n}{2}(t_1 + t_n)

Geometric Applications: Proportional Change Over Time

When a quantity grows or shrinks by the same percentage each period, the amounts form a geometric sequence. The key is converting the percentage change into the common ratio rr. If a quantity increases by 8% each year, then each year's value is 108% of the previous year's, so r=1.08r = 1.08. If it decreases by 8%, then r=0.92r = 0.92.
Geometric series arise when you want to accumulate totals over several periods of percentage change — for example, the total amount of medication absorbed over many doses, or the total distance a bouncing ball travels.
When evaluating rnr^n, always compute the exponent before multiplying by t1t_1. Use the order of operations carefully: t1⋅rnt_1 \cdot r^n means apply the exponent to rr first, then multiply by t1t_1.
One practical trap: if a bank account starts at CAD 1000 and earns 5% annual interest, the amount after 1 year is t2=1000×1.05t_2 = 1000 \times 1.05, not t1t_1. The initial deposit is t1=1000t_1 = 1000, year 1 balance is t2t_2, year 2 balance is t3t_3, and so on. So 'after 6 years' corresponds to t7t_7 — unless the problem explicitly says year 1 is the first interest period, in which case 'after 6 years' is t6t_6. Read carefully and be consistent.
Sn=t1(rn−1)r−1S_n = \frac{t_1(r^n - 1)}{r - 1}

Choosing Your Strategy and Checking Your Answer

Before you write a single formula, answer two questions: (1) Is the change additive or multiplicative? (2) Does the question ask for one specific term or a running total? Your answers to these two questions determine everything.
After calculating, always perform a reasonableness check. For an arithmetic series, the answer should be between n×t1n \times t_1 and n×tnn \times t_n — the total lies between using only the first term and using only the last term for every period. For a geometric series with r>1r > 1, the sum should be larger than n×t1n \times t_1 and should grow quickly; if your sum is smaller than t1t_1, you made an error.
Also check units and rounding. Money problems in a Canadian context are typically rounded to the nearest cent. Counting problems (seats, people, items) must give whole-number answers — if you get a decimal, recheck your setup.
If a problem gives you SnS_n or tnt_n and asks you to find nn, set up the formula, isolate the expression containing nn, and solve. At the MCR3U level, these problems are structured so that nn comes out as a whole number through careful algebra, allowing you to solve without any advanced techniques beyond rearranging the equation and checking whole-number candidates.

Choosing the Right Formula for an Application Problem

Situation typeChange per stepQuestion asks forFormula to use
ArithmeticAdd dd each stepValue at step nntn=t1+(n−1)dt_n = t_1 + (n-1)d
ArithmeticAdd dd each stepTotal of nn termsSn=n2(t1+tn)S_n = \frac{n}{2}(t_1 + t_n)
GeometricMultiply by rr each stepValue at step nntn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}
GeometricMultiply by rr each stepTotal of nn termsSn=t1(rn−1)r−1S_n = \frac{t_1(r^n-1)}{r-1}

Worked example

Arithmetic Series — Total Earnings Over a Contract

A graphic designer signs a 10-year contract. In year 1 she earns CAD 48 000. Each following year her salary increases by CAD 2 400. How much does she earn in total over the entire 10-year contract?
  1. Identify the sequence type and extract known values
    The salary increases by the same fixed amount each year, so this is an arithmetic sequence. The first term is t1=48 000t_1 = 48\,000, the common difference is d=2 400d = 2\,400, and the contract covers n=10n = 10 years. The question asks for the total earnings — a sum — so we need the series formula.
  2. Find the salary in year 10
    Before using the series formula, find t10t_{10} using the general term formula. Substitute n=10n = 10, t1=48 000t_1 = 48\,000, and d=2 400d = 2\,400 into tn=t1+(n−1)dt_n = t_1 + (n-1)d.
    t10=48 000+(10−1)(2 400)t_{10} = 48\,000 + (10-1)(2\,400)
  3. Calculate t10t_{10}
    Evaluate (10−1)=9(10 - 1) = 9, then 9×2 400=21 6009 \times 2\,400 = 21\,600, and finally add to the starting salary.
    t10=48 000+21 600=69 600t_{10} = 48\,000 + 21\,600 = 69\,600
  4. Apply the arithmetic series formula
    The sum of an arithmetic series is Sn=n2(t1+tn)S_n = \frac{n}{2}(t_1 + t_n). Substitute n=10n = 10, t1=48 000t_1 = 48\,000, and t10=69 600t_{10} = 69\,600.
    S10=102(48 000+69 600)S_{10} = \frac{10}{2}(48\,000 + 69\,600)
  5. Calculate the total
    Simplify 102=5\frac{10}{2} = 5 and 48 000+69 600=117 60048\,000 + 69\,600 = 117\,600, then multiply.
    S10=5×117 600=588 000S_{10} = 5 \times 117\,600 = 588\,000
Answer: The designer earns a total of CAD 588 000 over the 10-year contract.
Check: A quick reasonableness check: if she earned her first-year salary every year, the total would be 10×48 000=480 00010 \times 48\,000 = 480\,000. If she earned her last-year salary every year, the total would be 10×69 600=696 00010 \times 69\,600 = 696\,000. The answer CAD 588 000 is exactly halfway between these two extremes, which is correct for an arithmetic series — the average salary is 48 000+69 6002=58 800\frac{48\,000 + 69\,600}{2} = 58\,800 and 10×58 800=588 00010 \times 58\,800 = 588\,000. ✓

Worked example

Geometric Series — Total Mass of Recycled Materials

A recycling programme collects 400 kg of material in its first month. Each month after that, the amount collected is 1.15 times the amount collected the previous month (a 15% increase). What is the total mass of material collected over the first 8 months? Round your answer to the nearest kilogram.
  1. Identify the sequence type and extract known values
    Each month's collection is multiplied by the same factor, 1.15, so this is a geometric sequence. The first term is t1=400t_1 = 400, the common ratio is r=1.15r = 1.15, and we want the total over n=8n = 8 months. A total means we need the geometric series formula.
  2. Write the geometric series formula and substitute
    The sum of the first nn terms of a geometric series is Sn=t1(rn−1)r−1S_n = \frac{t_1(r^n - 1)}{r - 1}. Substitute t1=400t_1 = 400, r=1.15r = 1.15, and n=8n = 8.
    S8=400(1.158−1)1.15−1S_8 = \frac{400(1.15^8 - 1)}{1.15 - 1}
  3. Evaluate the power 1.1581.15^8
    Calculate 1.1581.15^8 step by step to keep accuracy. First, 1.152=1.32251.15^2 = 1.3225. Then 1.154=1.32252≈1.749006251.15^4 = 1.3225^2 \approx 1.74900625. Finally, 1.158=1.749006252≈3.059023211.15^8 = 1.74900625^2 \approx 3.05902321. Keep several decimal places to avoid rounding error in the middle of the calculation.
    1.158≈3.059023211.15^8 \approx 3.05902321
  4. Simplify the numerator
    Replace 1.1581.15^8 with the approximate value, evaluate 1.158−11.15^8 - 1, and multiply by 400.
    S8=400(3.05902321−1)0.15=400×2.059023210.15S_8 = \frac{400(3.05902321 - 1)}{0.15} = \frac{400 \times 2.05902321}{0.15}
  5. Complete the division
    Multiply 400×2.05902321=823.60928400 \times 2.05902321 = 823.60928, then divide by 0.150.15.
    S8=823.609280.15≈5490.73S_8 = \frac{823.60928}{0.15} \approx 5490.73
  6. Round and state the answer
    The question asks for the nearest kilogram, so round 5490.735490.73 to 54915491.
    S8≈5491 kgS_8 \approx 5491 \text{ kg}
Answer: The recycling programme collects approximately 5 491 kg of material over the first 8 months.
Check: A lower bound: if the programme collected 400 kg every month, the total would be 8×400=3 2008 \times 400 = 3\,200 kg. Since the amounts are growing, the true total must be larger than 3 200 kg, and 5 491 kg is indeed larger. As a spot check, t8=400×1.157≈400×2.66002≈1064t_8 = 400 \times 1.15^7 \approx 400 \times 2.66002 \approx 1064 kg. The average of t1=400t_1 = 400 and t8≈1064t_8 \approx 1064 is about 732, and 8×732≈58568 \times 732 \approx 5856. For a geometric series the weighted average sits slightly below the simple midpoint of first and last, so 5 491 kg is a plausible and consistent result. ✓

Common mistakes and how to avoid them

Using the sequence formula when the question asks for a total, or using the series formula when the question asks for a specific term.
Correction: Ask yourself: does the question want one value at a particular step, or the accumulated total of many steps? One value → tnt_n. Total → SnS_n.
Converting a percentage change to a common ratio incorrectly — for example, writing r=15r = 15 for a 15% increase instead of r=1.15r = 1.15.
Correction: A 15% increase means each term is 115% of the previous one, so r=1+0.15=1.15r = 1 + 0.15 = 1.15. A 15% decrease gives r=1−0.15=0.85r = 1 - 0.15 = 0.85.
Off-by-one error in nn — for example, treating 'after 6 years' as n=6n = 6 when the problem's setup means year 6 is actually t7t_7.
Correction: Clearly define what t1t_1 represents in the context of the problem, then count forward consistently. Writing out the first few terms in a short table often reveals the correct value of nn.
Multiplying t1t_1 by rnr^n instead of rn−1r^{n-1} when finding a specific term, giving an answer that is one ratio too large.
Correction: The general term formula is tn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}, not t1⋅rnt_1 \cdot r^n. The first term requires zero multiplications by rr, so the exponent is n−1n - 1.
Rounding an intermediate result such as rnr^n to too few decimal places, causing a large error in the final answer for geometric series.
Correction: Keep at least four or five decimal places in all intermediate steps. Only round the final answer to the precision the problem requests.

Lesson summary

Check your understanding

Question 1

A landscaping company plants 30 trees in week 1 and plants 5 more trees each subsequent week. How many trees does it plant in week 9?
  1. 70 trees
  2. 75 trees
  3. 65 trees
  4. 360 trees
Show answer and explanation
70 trees
This is arithmetic with t1=30t_1 = 30, d=5d = 5, and n=9n = 9. Using t9=30+(9−1)(5)=30+40=70t_9 = 30 + (9-1)(5) = 30 + 40 = 70. Option B uses n=9n = 9 instead of n−1=8n - 1 = 8 as the multiplier, giving 30+45=7530 + 45 = 75. Option C uses d=4d = 4 by mistake. Option D is the series sum S9S_9, not a single term.

Question 2

A ball is dropped and bounces to 60% of its previous height each time. The first bounce reaches 250 cm. Which expression gives the total height covered by the first 5 bounces (upward only)?
  1. 250×0.604250 \times 0.60^4
  2. 5×250×0.605 \times 250 \times 0.60
  3. 250(0.605−1)0.60−1\frac{250(0.60^5 - 1)}{0.60 - 1}
  4. 250(1−0.605)1−0.60\frac{250(1 - 0.60^5)}{1 - 0.60}
Show answer and explanation
250(1−0.605)1−0.60\frac{250(1 - 0.60^5)}{1 - 0.60}
The bounce heights form a geometric sequence with t1=250t_1 = 250 and r=0.60r = 0.60. The total of 5 terms is S5=250(0.605−1)0.60−1S_5 = \frac{250(0.60^5 - 1)}{0.60 - 1}. Since both numerator and denominator are negative, they cancel and the expression equals 250(1−0.605)1−0.60\frac{250(1 - 0.60^5)}{1 - 0.60}, which is option D. Option C is algebraically equivalent but written with two negatives that must be cancelled — option D is the cleaner standard form. Option A gives only the 5th term. Option B is not a valid sequence or series formula.

Question 3

An arithmetic series has t1=12t_1 = 12, d=4d = 4, and n=20n = 20. What is S20S_{20}?
  1. 480
  2. 1 000
  3. 760
  4. 88
Show answer and explanation
1 000
First find t20=12+(19)(4)=12+76=88t_{20} = 12 + (19)(4) = 12 + 76 = 88. Then S20=202(12+88)=10×100=1 000S_{20} = \frac{20}{2}(12 + 88) = 10 \times 100 = 1\,000. Option A is 20×24=48020 \times 24 = 480, using twice t1t_1 instead of t1+t20t_1 + t_{20}. Option C comes from only summing to n=10n = 10 by mistake. Option D is just t20t_{20} alone, not the series sum.

Question 4

A social media post is shared by 3 people on day 1. Each of those people shares it with 3 new people on day 2, and so on (each day's recipients each share with 3 new people). How many people in total have seen the post by the end of day 5, including the original 3?
  1. 243
  2. 363
  3. 360
  4. 3 906
Show answer and explanation
363
The daily new viewers form a geometric sequence: t1=3t_1 = 3 and r=3r = 3. The total after 5 days is S5=3(35−1)3−1=3(243−1)2=3×2422=7262=363S_5 = \frac{3(3^5 - 1)}{3 - 1} = \frac{3(243 - 1)}{2} = \frac{3 \times 242}{2} = \frac{726}{2} = 363. Option A (243=35243 = 3^5) is only the number of new viewers on day 5. Option C (360) results from an arithmetic error when adding the five terms. Option D is far too large and comes from misapplying the formula with an incorrect ratio.

Key terms

Sequence
An ordered list of numbers where each number is called a term. The position of a term is given by its index, usually written as nn.
Series
The sum of the terms of a sequence. A partial sum SnS_n is the sum of the first nn terms.
Arithmetic sequence
A sequence in which each term is obtained from the previous term by adding a fixed amount called the common difference dd.
Geometric sequence
A sequence in which each term is obtained from the previous term by multiplying by a fixed amount called the common ratio rr.
Common difference (dd)
The constant amount added at each step in an arithmetic sequence. It can be positive (increasing), negative (decreasing), or zero (constant).
Common ratio (rr)
The constant factor multiplied at each step in a geometric sequence. Values of rr between 0 and 1 produce decreasing sequences; values greater than 1 produce increasing sequences.
General term (tnt_n)
A formula that gives the value of any term in a sequence based on its position number nn.
Partial sum (SnS_n)
The result of adding together the first nn terms of a sequence. The formula for SnS_n differs for arithmetic and geometric sequences.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation C2.4. It is a study resource, not an official curriculum publication.

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