IB Mathematics: Applications and Interpretation SL — Topic 1 Number and algebra
Arithmetic Sequences and Series
Recognise constant additive change, calculate terms and finite sums, use sigma notation, work backwards from partial sums and model constant-increase situations.
Learning goal
Recognise and model arithmetic sequences; calculate individual terms and finite partial sums; use sigma notation; solve term-number and cumulative-sum problems; and interpret arithmetic models.
Syllabus link
AA SL/HL: SL 1.2. Arithmetic sequences and series; nth term; sum of the first n terms; sigma notation; technology; simple interest and approximate arithmetic models.
Big idea
An arithmetic sequence repeatedly adds the same amount. Once the first term \(u_1\) and common difference \(d\) are known, every term and every finite partial sum is determined.
Rounding
Unless stated otherwise, give non-exact numerical answers to three significant figures. Keep full calculator values during working.
This is the full online-textbook version of SL 1.2. The explanations, derivations, worked examples, applications, practice and detailed answers follow the finalized Radford Mathematics handout.
Scope
This lesson treats finite arithmetic sequences and series. Simple interest is included because it is an explicit SL 1.2 application. Compound interest is geometric and is not developed here.
Recognising an arithmetic sequence
A sequence is an ordered list of numbers. An arithmetic sequence is obtained by adding the same constant amount each time. This constant is the common difference.
For an arithmetic sequence, this difference is constant for every pair of consecutive terms.
Worked example 1: is the pattern arithmetic?
Decide whether each sequence is arithmetic. If it is, state the common difference.
(a) \(5,12,19,26,\ldots\) (b) \(42,35,28,21,\ldots\) (c) \(3,8,14,21,\ldots\)
Solution
Compare consecutive differences.
\[(a)\quad12-5=7,\quad19-12=7,\quad26-19=7.\]So (a) is arithmetic with \(d=7\).
\[(b)\quad35-42=-7,\quad28-35=-7,\quad21-28=-7.\]So (b) is arithmetic with \(d=-7\). For (c), \(8-3=5\) but \(14-8=6\), so the differences are not constant and the sequence is not arithmetic.
| Arithmetic | Geometric | |
|---|---|---|
| What stays constant? | Difference \(u_{n+1}-u_n\) | Ratio \(u_{n+1}/u_n\) |
| Typical change | Add/subtract a fixed amount | Multiply by a fixed factor |
| Example | +$500 salary each year | +4% salary each year |
The nth term of an arithmetic sequence
Starting from \(u_1\), each new term adds another copy of \(d\):
\[u_1=u_1,\quad u_2=u_1+d,\quad u_3=u_1+2d,\quad u_4=u_1+3d.\]Worked example 2: find a distant term
The arithmetic sequence begins \(11,15,19,23,\ldots\). Find \(u_{18}\).
Solution
First identify \(u_1=11\) and \(d=4\). Recall \(u_n=u_1+(n-1)d\).
\[u_{18}=11+(18-1)(4)=11+68=\boxed{79}.\]Why is it \(n-1\)?
To reach the nth term from the first term, we add \(d\) exactly \(n-1\) times. Reaching \(u_{10}\), for example, requires nine additions of \(d\). Plotting \((n,u_n)\) gives points on a straight line, but the sequence remains discrete: only integer term numbers are used.
A flexible term formula
If one term \(u_p\) is already known, there is no need to return to \(u_1\). Moving from term \(p\) to term \(n\) requires \(n-p\) additions of \(d\).
Worked example 3: start from a known middle term
An arithmetic sequence has \(u_7=32\) and \(d=-3\). Find \(u_{19}\).
Solution
Use \(u_n=u_p+(n-p)d\) with \(p=7\) and \(n=19\):
\[u_{19}=32+(19-7)(-3)=32-36=\boxed{-4}.\]Finding a term number and crossing a threshold
Worked example 4: find the term number
The sequence is \(14,19,24,29,\ldots\). The number 124 is a term. Find its term number.
Solution
Here \(u_1=14\), \(d=5\). Set \(u_n=124\):
\[124=14+5(n-1)=5n+9\]\[115=5n\Rightarrow\boxed{n=23}.\]So 124 is the 23rd term.
Worked example 5: first term above a target
A savings challenge starts with $18 in week 1 and increases by $11 each week. Find the first week in which the weekly amount is more than $150.
Solution
The model is \(u_n=18+11(n-1)\). We need
\[18+11(n-1)>150\Rightarrow11n+7>150\Rightarrow n>13.\]Since \(n\) is an integer, test the boundary: \(u_{13}=150\), \(u_{14}=161\). Therefore the first week above $150 is week 14.
Threshold language matters
“At least 150”, “more than 150”, and “first reaches 150” correspond to different conditions. Translate the wording into \(\ge\), \(>\) or \(=\) before solving.
Finding the common difference and first term from known terms
From \(u_p=u_1+(p-1)d\) and \(u_q=u_1+(q-1)d\), subtract to eliminate \(u_1\):
Worked example 6: reconstruct the sequence
An arithmetic sequence satisfies \(u_4=17\) and \(u_{11}=45\). Find \(d\), then \(u_1\).
Solution
\[45-17=(11-4)d\Rightarrow28=7d\Rightarrow\boxed{d=4}.\]Now use \(u_4=u_1+3d\):
\[17=u_1+3(4)\Rightarrow\boxed{u_1=5}.\]Gradient interpretation
The expression \(d=(u_q-u_p)/(q-p)\) is change in term value divided by change in term number. On a graph of \((n,u_n)\), \(d\) is the gradient of the line through the sequence points.
Three consecutive terms of an arithmetic sequence
For any three consecutive terms \(u_{n-1},u_n,u_{n+1}\), the two consecutive differences are equal.
Worked example 7: algebraic consecutive terms
The first three terms are \(2k+1,\ 5k-2,\ 10k-9\). Find \(k\) and state the first three terms.
Solution
\[(5k-2)-(2k+1)=(10k-9)-(5k-2)\]\[3k-3=5k-7\Rightarrow3k+4=5k\Rightarrow4=2k\Rightarrow\boxed{k=2}.\]The first three terms are \(5,8,11\), with common difference 3.
The sum of the first n terms of an arithmetic sequence
The sum of the first \(n\) terms is \(S_n=u_1+u_2+\cdots+u_n\). There are two equivalent finite-sum formulas.
Worked example 8: use the first and last terms
An arithmetic sequence has \(u_1=7\) and \(u_{30}=94\). Find \(S_{30}\).
Solution
\[S_{30}=\frac{30}{2}(7+94)=15(101)=\boxed{1515}.\]Worked example 9: use the first term and common difference
Find the sum of the first 25 terms of \(12,17,22,27,\ldots\).
Solution
Here \(u_1=12\), \(d=5\), \(n=25\):
\[S_{25}=\frac{25}{2}[2(12)+24(5)]=\frac{25}{2}(144)=\boxed{1800}.\]Why the finite-sum formula works
Write the same sum forwards and backwards:
\[S_n=u_1+(u_1+d)+(u_1+2d)+\cdots+[u_1+(n-1)d]\]\[S_n=[u_1+(n-1)d]+[u_1+(n-2)d]+\cdots+(u_1+d)+u_1.\]Each vertical pair equals \(2u_1+(n-1)d=u_1+u_n\), and there are \(n\) pairs. Hence
\[2S_n=n[2u_1+(n-1)d]=n(u_1+u_n).\]Dividing by 2 gives both formulas above.
From a sequence to its associated series: partial sums
The sequence \((u_n)\) and the partial-sum sequence \((S_n)\) are different objects: \(u_n\) is one term; \(S_n\) is the accumulated total.
Recovering a term from partial sums
\[S_n=u_1+\cdots+u_{n-1}+u_n,\qquad S_{n-1}=u_1+\cdots+u_{n-1}\]Subtract:
\[\boxed{u_n=S_n-S_{n-1}}\quad(n\ge2).\]Also \(u_1=S_1\). If \(S_0=0\), the same relationship works for every \(n\ge1\).
Worked example 10: recover the nth term from \(S_n\)
The sum of the first \(n\) terms is \(S_n=2n^2+3n\). Find \(u_n\). Hence state \(u_1\) and \(d\).
Solution
First find \(S_{n-1}\):
\[S_{n-1}=2(n-1)^2+3(n-1)=2n^2-n-1.\]Therefore
\[u_n=(2n^2+3n)-(2n^2-n-1)=\boxed{4n+1}.\]So \(u_1=5\) and \(d=4\).
Working with finite arithmetic sums
Worked example 11: find the number of terms from \(S_n\)
The sequence begins \(5,8,11,14,\ldots\). The sum of the first \(n\) terms is 670. Find \(n\).
Solution
\[670=\frac n2[2(5)+(n-1)3]\]\[1340=n(3n+7)\Rightarrow3n^2+7n-1340=0\]\[(n-20)(3n+67)=0.\]Since \(n\) is a positive integer, \(\boxed{n=20}\).
Worked example 12: add a middle block
An arithmetic sequence has \(u_1=4\) and \(d=3\). Find \(u_{21}+u_{22}+\cdots+u_{50}\).
Solution
Use \(S_{50}-S_{20}\):
\[S_{50}=\frac{50}{2}[8+49(3)]=3875,\qquad S_{20}=\frac{20}{2}[8+19(3)]=650.\]\[S_{50}-S_{20}=\boxed{3225}.\]Worked example 13: remove every multiple of 3
Find the sum of all positive integers from 1 to 300 that are not divisible by 3.
Solution
Start with all integers:
\[1+2+\cdots+300=\frac{300}{2}(1+300)=45150.\]The multiples of 3 are \(3,6,\ldots,300\), or three times \(1,2,\ldots,100\):
\[3\left[\frac{100}{2}(1+100)\right]=15150.\]Subtract:
\[45150-15150=\boxed{30000}.\]Sigma notation for arithmetic sums
Need a sigma-notation refresher?
Use the separate Reading and Writing Sigma Notation reference lesson for the anatomy of sigma notation, shifted indices and expansion.
Number of terms
\[\sum_{k=a}^{b}f(k)\quad\text{contains}\quad\boxed{b-a+1}\text{ terms}.\]Worked example 14: read and evaluate a sigma sum
Evaluate \(\sum_{k=4}^{18}(5k-2)\).
Solution
First term: \(5(4)-2=18\). Last term: \(5(18)-2=88\). Number of terms: \(18-4+1=15\).
\[\sum_{k=4}^{18}(5k-2)=\frac{15}{2}(18+88)=\boxed{795}.\]Worked example 15: write a series in sigma notation
Write \(7+11+15+\cdots+[7+4(n-1)]\) in sigma notation.
Solution
The kth term is \(7+4(k-1)=4k+3\). Therefore
\[\boxed{\sum_{k=1}^{n}[7+4(k-1)]}=\boxed{\sum_{k=1}^{n}(4k+3)}.\]Simple interest and other constant-increase models
A fixed amount added each period creates an arithmetic sequence. With simple interest, the interest is always calculated from the original principal, so the same amount is added each year.
If principal \(P\) earns simple interest at annual rate \(r\) (decimal), yearly interest is \(I=Pr\). After \(n\) years:
\[\boxed{A_n=P+nPr=P(1+nr)}.\]Worked example 16: simple interest
An account contains $5000 initially and earns simple interest at 3.2% per year. Find the balance after 8 years.
Solution
Each year the same amount is added:
\[I=5000(0.032)=160.\]After 8 years:
\[A_8=5000+8(160)=\boxed{\$6280}.\]Indexing: time 0 versus term 1
Use \(u_1\) when the first value naturally belongs to period 1:
\[u_n=u_1+(n-1)d,\qquad S_n=\frac n2[2u_1+(n-1)d].\]Use \(u_0\) when there is a genuine initial value at time 0:
\[\boxed{u_n=u_0+nd}.\]If the sum includes the initial value, it has \(n+1\) terms:
\[\boxed{u_0+u_1+\cdots+u_n=\frac{n+1}{2}(2u_0+nd)}.\]If the sum begins at \(u_1\), then
\[\boxed{u_1+\cdots+u_n=\frac n2[2u_0+(n+1)d]}.\]Worked example 17: when \(u_1\) is natural
A salary is $42 000 in year 1 and increases by $1250 each year. (a) Find the salary in year 10. (b) Find total salary over the first 10 years.
Solution
\(u_1=42000\), \(d=1250\).
\[u_{10}=42000+9(1250)=\boxed{\$53250}.\]\[S_{10}=\frac{10}{2}[2(42000)+9(1250)]=\boxed{\$476250}.\]Worked example 18: when \(u_0\) is natural
A fundraiser receives $200 on launch day, day 0. The amount received each day then increases by $25. Find (a) day 6 amount, (b) total from days 0–6, (c) total from days 1–6.
Solution
\(u_0=200\), \(d=25\).
\[u_6=200+6(25)=\boxed{350}.\]Days 0–6 contain 7 terms:
\[\frac72[2(200)+6(25)]=\boxed{1925}.\]Days 1–6 contain 6 terms and exclude \(u_0\):
\[\frac62[2(200)+7(25)]=\boxed{1725}.\]Arithmetic models that are only approximate
Real data may not have exactly equal differences. SL 1.2 expects analysis, interpretation and prediction when the arithmetic model is only approximate.
Worked example 19: approximate the common difference
A measured quantity is \(52,57,61,66,70\). Use an arithmetic model to estimate term 10.
Solution
The differences are 5, 4, 5, 4, so the data are not perfectly arithmetic. A sensible average difference over the full interval is
\[d\approx\frac{70-52}{5-1}=\frac{18}{4}=4.5.\]Using \(u_1\approx52\):
\[u_n\approx52+4.5(n-1),\qquad u_{10}\approx52+4.5(9)=\boxed{92.5}.\]How to comment on an approximate model
State that the common difference is approximate. Consider whether the observed differences fluctuate around a typical value, whether a constant additive change is plausible, and whether the prediction is close to or far beyond the observed data. Long-range extrapolation is usually less reliable.
Linear-function connection
\[u_n=u_1+(n-1)d=dn+(u_1-d).\]Plotting \(u_n\) against \(n\) gives a straight line with gradient \(d\). The sequence is still discrete because only integer \(n\) correspond to terms.
Past-paper-informed exam patterns: worked sister questions
These are original sister questions, not copied IB questions. They reproduce common SL 1.2 skills and multi-part structure.
Exam-style 1: two known terms and a finite sum
[6 marks]An arithmetic sequence satisfies \(u_2=9\), \(u_7=34\). Find (a) \(d\), (b) \(u_1\), (c) \(S_{25}\).
Solution
\[34-9=(7-2)d\Rightarrow d=5.\]\[9=u_1+5\Rightarrow u_1=4.\]\[S_{25}=\frac{25}{2}[2(4)+24(5)]=\boxed{1600}.\]Exam-style 2: partial sums
[8 marks]For a sequence, \(S_n=n^2+5n\). Find \(S_6\), \(u_6\), \(u_1\), an expression for \(u_n\), and \(d\).
Solution
\(S_6=66\), \(S_5=50\), so \(u_6=16\). Also \(u_1=S_1=6\).
\[u_n=(n^2+5n)-[(n-1)^2+5(n-1)]=\boxed{2n+4},\]hence \(d=2\).
Exam-style 3: three consecutive algebraic terms
[7 marks]The first three terms are \(2k+1,5k-2,10k-9\). Find \(k\), the first three terms, and \(S_{40}\).
Solution
Equal differences give \(k=2\), so the terms are \(5,8,11\), with \(d=3\).
\[S_{40}=\frac{40}{2}[10+39(3)]=20(127)=\boxed{2540}.\]Exam-style 4: simple-interest sequence
[8 marks]$6000 is invested at simple interest of 3% per year. Find the yearly interest, the balance after 7 years, and the sum of the end-of-year balances for years 1–10.
Solution
Yearly interest \(=6000(0.03)=180\). Balance after 7 years \(=6000+7(180)=7260\). The end-of-year balance in year 1 is 6180 and year 10 is 7800, so
\[S_{10}=\frac{10}{2}(6180+7800)=\boxed{69900}.\]Practice
Try these without the answer key
- Decide whether each is arithmetic and state \(d\) if it is: (a) 8,13,18,23,… (b) 40,32,24,16,… (c) 3,7,12,18,…
- \(u_1=6,d=4\). Find \(u_{17}\).
- \(u_5=31,d=-2\). Find \(u_{18}\).
- The sequence 14,19,24,… contains 124. Find its term number.
- \(u_1=18,d=11\). Find the first term greater than 150.
- \(u_4=22,u_{10}=58\). Find \(d\) and \(u_1\).
- The first three terms are \(2k+1,5k-2,10k-9\). Find \(k\).
- \(u_1=-7,d=5\). Find \(u_{30}\).
- \(u_1=4,d=5\). Find \(S_{25}\).
- \(u_1=30,d=-3\). Find \(S_{12}\).
- \(u_1=2,d=3\). Find \(u_{15}+\cdots+u_{40}\).
- Evaluate \(\sum_{k=3}^{15}(4k+1)\).
- Write \(12+19+26+\cdots+[12+7(n-1)]\) in sigma notation.
- \(S_n=3n^2+2n\). Find \(u_n\), \(u_1\) and \(d\).
- $8000 earns simple interest at 2.5% per year. Find the balance after 9 years.
- A salary is $38 500 in year 1 and rises by $1250 per year. Find year-12 salary and total salary for years 1–12.
- Data: 41,46,50,55,59. Use an approximate arithmetic model to predict term 8.
- Find the sum of positive integers from 1 to 300 that are not divisible by 3.
- The sequence has \(u_1=6,d=4\), and \(S_n=880\). Find \(n\).
- Find the sum of the first 50 positive odd integers.
- Explain one strength and one limitation of fitting an arithmetic model to data whose differences are approximately, but not exactly, constant.
- A planned study time is 15 minutes on day 1 and rises by 4 minutes per day. Find the first day on which it exceeds 50 minutes.
Detailed answer key
Answers 1–5: recognition, terms and thresholds
1. (a) Arithmetic, \(d=5\). (b) Arithmetic, \(d=-8\). (c) Not arithmetic: differences 4,5,6.
2. \(u_{17}=6+16(4)=\boxed{70}\).
3. \(u_{18}=u_5+(18-5)d=31+13(-2)=\boxed{5}\).
4. \(124=14+5(n-1)=5n+9\Rightarrow n=\boxed{23}\).
5. \(18+11(n-1)>150\Rightarrow n>13\). Since \(u_{13}=150\), the first term greater than 150 is \(u_{14}=161\).
Answers 6–14: reconstruction, sums and sigma notation
6. \(58-22=(10-4)d\Rightarrow d=6\). Then \(22=u_1+3(6)\Rightarrow u_1=4\).
7. Equal differences give \(3k-3=5k-7\Rightarrow\boxed{k=2}\).
8. \(u_{30}=-7+29(5)=\boxed{138}\).
9. \(S_{25}=\frac{25}{2}[8+24(5)]=\boxed{1600}\).
10. \(S_{12}=\frac{12}{2}[60+11(-3)]=\boxed{162}\).
11. \(S_{40}=2420\), \(S_{14}=301\), so the block sum is \(\boxed{2119}\).
12. First term 13, last term 61, 13 terms: \(\frac{13}{2}(13+61)=\boxed{481}\).
13. \(\boxed{\sum_{k=1}^{n}[12+7(k-1)]}\).
14. \(S_{n-1}=3n^2-4n+1\), so \(u_n=S_n-S_{n-1}=\boxed{6n-1}\). Thus \(u_1=5,d=6\).
Answers 15–22: applications and interpretation
15. Yearly interest \(=8000(0.025)=200\). Balance after 9 years \(=8000+9(200)=\boxed{\$9800}\).
16. \(u_{12}=38500+11(1250)=\boxed{\$52250}\). \(S_{12}=\frac{12}{2}[2(38500)+11(1250)]=\boxed{\$544500}\).
17. \(d\approx(59-41)/(5-1)=4.5\), so \(u_8\approx41+7(4.5)=\boxed{72.5}\).
18. \(45150-15150=\boxed{30000}\).
19. \(880=\frac n2[12+4(n-1)]=2n(n+2)\). Thus \(n^2+2n-440=0\Rightarrow(n-20)(n+22)=0\), so \(\boxed{n=20}\).
20. The first 50 odd integers are 1 to 99: \(S_{50}=\frac{50}{2}(1+99)=\boxed{2500}\).
21. Strength: the differences may fluctuate around a stable typical value, making a constant additive model useful over the observed interval. Limitation: the differences are not exactly constant and long-range extrapolation assumes the trend continues.
22. \(15+4(n-1)>50\Rightarrow4n+11>50\Rightarrow n>9.75\), so the first integer solution is \(\boxed{10}\).
Formula summary and checklist
| Common difference | \(d=u_{n+1}-u_n\) |
|---|---|
| nth term from \(u_1\) | \(u_n=u_1+(n-1)d\) |
| nth term from \(u_0\) | \(u_n=u_0+nd\) |
| Flexible term | \(u_n=u_p+(n-p)d\) |
| Two known terms | \(u_q-u_p=(q-p)d\) |
| Three consecutive terms | \(u_{n+1}-u_n=u_n-u_{n-1}\) |
| Term from partial sums | \(u_n=S_n-S_{n-1}\) |
| Finite sum | \(S_n=\frac n2(u_1+u_n)=\frac n2[2u_1+(n-1)d]\) |
| Block sum | \(u_p+\cdots+u_q=S_q-S_{p-1}\) |
| Sigma form | \(S_n=\sum_{k=1}^{n}[u_1+(k-1)d]\) |
| Simple interest | \(A_n=P(1+nr)\) |
- Recognise constant differences.
- Distinguish arithmetic from geometric change.
- Use both nth-term forms.
- Choose \(u_0\) or \(u_1\) consistently.
- Find term numbers and solve thresholds.
- Recover \(d\) and \(u_1\) from known terms.
- Use equal consecutive differences.
- Distinguish \(u_n\) from \(S_n\).
- Recover terms from partial sums.
- Derive and use both sum formulas.
- Use block sums and sigma notation.
- Model simple interest.
- Approximate a common difference and critique the model.