IB Practice / Maths AA HL / Number and algebra
Free study notes · Maths AA HL · SL 1.2, 1.3, 1.4, 1.8
Sequences and series: arithmetic, geometric and financial applications: study notes
A sequence is an ordered list of numbers generated by a rule; a series is the sum of its terms. This topic covers the two standard rules, arithmetic (add a constant ) and geometric (multiply by a constant ), their term and sum formulae, sigma notation, the infinite geometric sum, and the money contexts built on them: simple interest, compound interest with several compounding periods per year, and annual depreciation. Exam questions are short and formulaic: find or from two pieces of information, find for which a sum or value passes a target, write a sum in sigma notation, or justify convergence with .
1. Arithmetic sequences and series
An arithmetic sequence has a constant common difference . The th term is and the sum of the first terms is . Both are in the booklet; what you must learn is how to use them. Two given terms give two linear equations in and ; subtracting them gives at once because . A term and a sum also give a pair of simultaneous equations. Note that for , and that for , is a quadratic in with no constant term (if it is simply ), so a condition like or becomes a quadratic equation or inequality in . Simple interest is the standard application: a principal at per year grows by the fixed amount each year, so the values after each year form an arithmetic sequence with and . Be careful with the term index: the value after years is , which is only if is the value after the first year.
2. Sigma notation
Sigma notation is not in the booklet. . The variable under the sign is a dummy index; the lower and upper limits give the first and last values of the index. Count the number of terms as (upper limit lower limit ), so has terms, not . To evaluate it, either write it as a difference of the full series, or treat it as its own arithmetic series with first term , last term and terms, giving . To write a given series in sigma notation, identify and (or ), write the general term in terms of and fix the limits. For an infinite geometric series the upper limit is : . Examiners accept any correct dummy variable, but the expression inside must produce exactly the listed terms.
3. Geometric sequences and series
A geometric sequence has a constant common ratio , with . The th term is and for (both in the booklet; the second form is convenient when ). Two terms give by division: . If is even there are two possible values of , positive and negative, and the question will usually tell you which to take or ask for both. When three consecutive terms are given as linear expressions in a parameter , the arithmetic condition gives a linear equation, while the geometric condition gives a quadratic in ; reject any solution that makes a term zero. Applications: compound growth of money, a population growing by a fixed percentage each year (), and depreciation (). Finding from a target value needs logarithms: from take of both sides, remembering that if then and the inequality reverses. On a calculator paper, a table of values of or on the GDC is an accepted method provided you write down the two values either side of the target.
4. Infinite geometric series and convergence
If then as , so . The booklet gives with the condition , but you must be able to state the condition in words and use it as a reason: a geometric series with converges if and only if , endpoints excluded. The R mark in these questions is for writing this condition and applying it, for example when is an expression such as , leading to the strict interval . A recurring decimal is a geometric series: with , , so it equals . When both and one term are known, eliminating gives a quadratic in ; check each root against before accepting it. Describe the behaviour of the partial sums if asked: for they approach the limit from one side (increasing if , decreasing if ), for they alternate above and below it.
5. Financial applications: compound interest and depreciation
The booklet formula is , where is the present value, the annual nominal rate as a percentage, the number of compounding periods per year ( yearly, half-yearly, quarterly, monthly) and the number of years. Each period multiplies the balance by , so the exponent counts periods, not years. Depreciation is not in the booklet: a value falling by per year is , the same shape with a minus sign, so learn it. Typical tasks: find to two decimal places; find the rate or the initial amount by rearranging; find the number of years (or months) for the value to exceed, or fall below, a target. For the last type set up the inequality, solve with logarithms or by the finance solver, then round up to the next whole period and state the answer in the units the question demands (complete years, months). An inflation adjustment compares two growth factors: the real growth factor is , which is close to, but not equal to, .
6. Technology: GDC, finance solver and spreadsheets
On Paper 2 the finance solver on the GDC is an accepted method. Its inputs are the number of periods , the annual rate , the present value , the payment (zero for a lump sum), the future value , and the compounding settings. Sign convention matters: money paid in and money received have opposite signs, so a deposit of with gives a positive . Write the inputs out in full on the page: a visible, correct setup is what the method mark rewards, and if the final answer is wrong the listed inputs are the only evidence that the method was right. A sequence or table on the GDC (or a spreadsheet with a cell formula such as previous cell times ) is equally acceptable for 'how many years until' questions; show the two consecutive rows that bracket the target. Graphing software showing against helps to see where an arithmetic sum is maximised: the vertex of the parabola need not fall at an integer, and when it does not you must test the integers on each side rather than rounding. Paper 1 is a non-calculator paper, so technology is for revision only there: use it at home to check your exact Paper 1-style work, but in the exam the exact fraction or surd form, reached by hand, is what earns the marks.
Formulas
| in the booklet | |
| in the booklet | |
| in the booklet | |
| in the booklet | |
| in the booklet | |
| in the booklet | |
| (depreciation) | learn it |
| , number of terms = upper lower | learn it |
| Convergence: a geometric series with converges if and only if | learn it |
| (for ); divide by and reverse the inequality if | learn it |
| for | learn it |
Worked examples
Example 1
(Paper 1 style, 8 marks) An arithmetic sequence has and . (a) Find the common difference and the first term. (b) Find the value of , with , for which . (c) Find the least value of for which .
Solution
- (a) Subtract the term formulae: , so and (M1 A1).
- Then gives , so (A1).
- (b) (M1 A1).
- with gives (A1). Check: the terms cancel in pairs.
- (c) , so (M1). The positive root of is .
- Since is a positive integer and the quadratic is positive beyond its larger root, the least is . Check: , (A1).
Answer: (a) , . (b) . (c) .
Example 2
(Paper 1 style, 8 marks) The sum to infinity of a geometric series is and its second term is . (a) Show that the common ratio satisfies . (b) Find the two possible values of and the corresponding first terms, stating why both are valid. (c) For the case with the larger first term, write the series using sigma notation.
Solution
- (a) From the booklet, , so (M1). The second term is (A1).
- Substitute: , so . Divide by : (A1 AG). Write every line; the final mark of a 'show that' is usually withheld unless the expansion is shown.
- (b) Factorise: , so or (M1 A1).
- Both satisfy , so both series converge and both are valid (R1).
- First terms: when , and when (A1).
- (c) The larger first term is with : the series is (A1).
Answer: (a) Shown. (b) , or , ; both valid because . (c) .
Example 3
(Paper 2 style, 9 marks) Maria invests euros in an account paying a nominal annual interest rate of , compounded monthly. (a) Find the value of the investment after years, to the nearest cent. (b) Find the number of complete years required for the investment to exceed euros. (c) At the same time Maria buys a car for euros which depreciates by per year. Find the value of the car after years, to the nearest euro.
Solution
- (a) Use with , , , : (M1 A1).
- euros (A1). On the finance solver: , , , , ; write these inputs down to earn the method mark.
- (b) Require , so (M1 A1).
- Take logarithms: , so (A1). Since the inequality does not reverse.
- The investment first exceeds euros during the th year, so complete years are needed (A1). Check with a table: after years the value is , after years it is .
- (c) Depreciation: (M1). , so euros (A1).
Answer: (a) euros. (b) years. (c) euros.
Can you...
- Find or from any two given terms by subtracting or dividing the term formulae, and solve the resulting pair of simultaneous equations?
- Convert a verbal condition ('first exceeds', 'is zero', 'falls below half') into an equation or inequality in , solve it exactly or with logarithms, and justify the integer answer?
- Write a given series in sigma notation with correct limits and general term, and count the number of terms in a sum written in sigma notation?
- State the condition in words, apply it to a ratio containing a parameter to obtain a strict interval, and use for recurring decimals?
- Use the compound interest formula with periods per year and the depreciation formula, and record finance solver inputs or a GDC table so the method is visible?
Common errors
- Confusing the term number with the exponent: the th geometric term is , not , while the value after years of compound growth is . Decide first whether is the starting value or the value after one period.
- Using the number of years as the exponent when compounding is monthly: with the exponent is and the rate per period is . Write the booklet formula with all four quantities labelled before substituting.
- Claiming convergence without stating , or giving the interval with closed endpoints. The R mark needs the strict condition written down and checked against each candidate value of .
- Rounding a continuous answer the wrong way when must be an integer: a bound such as means , and the integer that maximises a sum is found by testing both integers either side of the vertex, not by rounding the vertex.
- Reversing, or failing to reverse, the inequality when dividing by : if then is negative and the direction changes. Alternatively keep everything positive by writing .
- Using rounded intermediate values such as and then quoting a final answer to the nearest cent. Keep full calculator accuracy until the last line and round only once.
Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: Number and algebra practice.
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