IB Practice / Maths AA HL / Number and algebra

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 dd) and geometric (multiply by a constant rr), 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 dd or rr from two pieces of information, find nn for which a sum or value passes a target, write a sum in sigma notation, or justify convergence with ∣r∣<1|r|<1.

Download the PDFFree, no sign-up.

1. Arithmetic sequences and series

An arithmetic sequence has a constant common difference d=un+1−und = u_{n+1} - u_n. The nnth term is un=u1+(n−1)du_n = u_1 + (n-1)d and the sum of the first nn terms is Sn=n2(2u1+(n−1)d)=n2(u1+un)S_n = \frac{n}{2}\left(2u_1 + (n-1)d\right) = \frac{n}{2}(u_1 + u_n). Both are in the booklet; what you must learn is how to use them. Two given terms give two linear equations in u1u_1 and dd; subtracting them gives dd at once because up−uq=(p−q)du_p - u_q = (p-q)d. A term and a sum also give a pair of simultaneous equations. Note that un=Sn−Sn−1u_n = S_n - S_{n-1} for n≥2n \geq 2, and that for d≠0d \neq 0, SnS_n is a quadratic in nn with no constant term (if d=0d = 0 it is simply Sn=nu1S_n = nu_1), so a condition like Sn=0S_n = 0 or Sn>100S_n > 100 becomes a quadratic equation or inequality in nn. Simple interest is the standard application: a principal PP at r%r\% per year grows by the fixed amount Pr100\frac{Pr}{100} each year, so the values after each year form an arithmetic sequence with u1=P+Pr100u_1 = P + \frac{Pr}{100} and d=Pr100d = \frac{Pr}{100}. Be careful with the term index: the value after nn years is P+nPr100P + \frac{nPr}{100}, which is unu_n only if u1u_1 is the value after the first year.

2. Sigma notation

Sigma notation is not in the booklet. ∑k=1nuk=u1+u2+⋯+un=Sn\sum_{k=1}^{n} u_k = u_1 + u_2 + \dots + u_n = S_n. 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 +1+1), so ∑k=312(2k+1)\sum_{k=3}^{12}(2k+1) has 1010 terms, not 1212. To evaluate it, either write it as a difference S12−S2S_{12} - S_2 of the full series, or treat it as its own arithmetic series with first term 77, last term 2525 and 1010 terms, giving 102(7+25)=160\frac{10}{2}(7+25)=160. To write a given series in sigma notation, identify u1u_1 and dd (or rr), write the general term in terms of kk and fix the limits. For an infinite geometric series the upper limit is ∞\infty: ∑k=1∞u1rk−1\sum_{k=1}^{\infty} u_1 r^{k-1}. 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 r=un+1unr = \frac{u_{n+1}}{u_n}, with r≠0r \neq 0. The nnth term is un=u1rn−1u_n = u_1 r^{n-1} and Sn=u1(rn−1)r−1=u1(1−rn)1−rS_n = \frac{u_1(r^n - 1)}{r-1} = \frac{u_1(1-r^n)}{1-r} for r≠1r \neq 1 (both in the booklet; the second form is convenient when ∣r∣<1|r|<1). Two terms give rr by division: upuq=rp−q\frac{u_p}{u_q} = r^{p-q}. If p−qp-q is even there are two possible values of rr, 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 xx, the arithmetic condition u2−u1=u3−u2u_2 - u_1 = u_3 - u_2 gives a linear equation, while the geometric condition u22=u1u3u_2^2 = u_1 u_3 gives a quadratic in xx; reject any solution that makes a term zero. Applications: compound growth of money, a population growing by a fixed percentage each year (r=1+p100r = 1 + \frac{p}{100}), and depreciation (r=1−p100r = 1 - \frac{p}{100}). Finding nn from a target value needs logarithms: from rn>cr^n > c take ln⁡\ln of both sides, remembering that if 0<r<10 < r < 1 then ln⁡r<0\ln r < 0 and the inequality reverses. On a calculator paper, a table of values of unu_n or SnS_n 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 ∣r∣<1|r| < 1 then rn→0r^n \to 0 as n→∞n \to \infty, so Sn→u11−rS_n \to \frac{u_1}{1-r}. The booklet gives S∞=u11−rS_\infty = \frac{u_1}{1-r} with the condition ∣r∣<1|r|<1, but you must be able to state the condition in words and use it as a reason: a geometric series with u1≠0u_1 \neq 0 converges if and only if −1<r<1-1 < r < 1, endpoints excluded. The R mark in these questions is for writing this condition and applying it, for example when rr is an expression such as x−23\frac{x-2}{3}, leading to the strict interval −1<x<5-1 < x < 5. A recurring decimal is a geometric series: 0.2˙7˙=0.27+0.0027+…0.\dot{2}\dot{7} = 0.27 + 0.0027 + \dots with u1=27100u_1 = \frac{27}{100}, r=1100r = \frac{1}{100}, so it equals 27/10099/100=311\frac{27/100}{99/100} = \frac{3}{11}. When both S∞S_\infty and one term are known, eliminating u1u_1 gives a quadratic in rr; check each root against ∣r∣<1|r|<1 before accepting it. Describe the behaviour of the partial sums if asked: for 0<r<10<r<1 they approach the limit from one side (increasing if u1>0u_1 > 0, decreasing if u1<0u_1 < 0), for −1<r<0-1<r<0 they alternate above and below it.

5. Financial applications: compound interest and depreciation

The booklet formula is FV=PV(1+r100k)knFV = PV\left(1 + \frac{r}{100k}\right)^{kn}, where PVPV is the present value, rr the annual nominal rate as a percentage, kk the number of compounding periods per year (k=1k=1 yearly, 22 half-yearly, 44 quarterly, 1212 monthly) and nn the number of years. Each period multiplies the balance by 1+r100k1 + \frac{r}{100k}, so the exponent counts periods, not years. Depreciation is not in the booklet: a value falling by p%p\% per year is Vn=V0(1−p100)nV_n = V_0\left(1 - \frac{p}{100}\right)^n, the same shape with a minus sign, so learn it. Typical tasks: find FVFV 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 1+r/1001+i/100\frac{1 + r/100}{1 + i/100}, which is close to, but not equal to, 1+r−i1001 + \frac{r-i}{100}.

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 NN, the annual rate I%I\%, the present value PVPV, the payment PMTPMT (zero for a lump sum), the future value FVFV, and the compounding settings. Sign convention matters: money paid in and money received have opposite signs, so a deposit of 50005000 with PV=−5000PV = -5000 gives a positive FVFV. 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 1.0031.003) is equally acceptable for 'how many years until' questions; show the two consecutive rows that bracket the target. Graphing software showing y=Sxy = S_x against xx 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

un=u1+(n−1)du_n = u_1 + (n-1)din the booklet
Sn=n2(2u1+(n−1)d)=n2(u1+un)S_n = \frac{n}{2}\left(2u_1 + (n-1)d\right) = \frac{n}{2}(u_1 + u_n)in the booklet
un=u1rn−1u_n = u_1 r^{n-1}in the booklet
Sn=u1(rn−1)r−1=u1(1−rn)1−r,  r≠1S_n = \frac{u_1(r^n - 1)}{r-1} = \frac{u_1(1-r^n)}{1-r}, \; r \neq 1in the booklet
S∞=u11−r,  ∣r∣<1S_\infty = \frac{u_1}{1-r}, \; |r|<1in the booklet
FV=PV(1+r100k)knFV = PV\left(1 + \frac{r}{100k}\right)^{kn}in the booklet
Vn=V0(1−p100)nV_n = V_0\left(1 - \frac{p}{100}\right)^n (depreciation)learn it
∑k=1nuk=Sn\sum_{k=1}^{n} u_k = S_n, number of terms = upper −- lower +1+1learn it
Convergence: a geometric series with u1≠0u_1 \neq 0 converges if and only if −1<r<1-1 < r < 1learn it
rn>c⇒nln⁡r>ln⁡cr^n > c \Rightarrow n \ln r > \ln c (for r,c>0r, c > 0); divide by ln⁡r\ln r and reverse the inequality if 0<r<10<r<1learn it
un=Sn−Sn−1u_n = S_n - S_{n-1} for n≥2n \geq 2learn it

Worked examples

Example 1

(Paper 1 style, 8 marks) An arithmetic sequence has u4=−3u_4 = -3 and u10=9u_{10} = 9. (a) Find the common difference and the first term. (b) Find the value of nn, with n>1n>1, for which Sn=0S_n = 0. (c) Find the least value of nn for which Sn>100S_n > 100.

Solution
  1. (a) Subtract the term formulae: u10−u4=6du_{10} - u_4 = 6d, so 6d=126d = 12 and d=2d = 2 (M1 A1).
  2. Then u4=u1+3du_4 = u_1 + 3d gives −3=u1+6-3 = u_1 + 6, so u1=−9u_1 = -9 (A1).
  3. (b) Sn=n2(2(−9)+(n−1)(2))=n2(2n−20)=n(n−10)S_n = \frac{n}{2}\left(2(-9) + (n-1)(2)\right) = \frac{n}{2}(2n - 20) = n(n-10) (M1 A1).
  4. n(n−10)=0n(n-10) = 0 with n>1n>1 gives n=10n = 10 (A1). Check: the terms −9,−7,…,9-9, -7, \dots, 9 cancel in pairs.
  5. (c) n(n−10)>100n(n-10) > 100, so n2−10n−100>0n^2 - 10n - 100 > 0 (M1). The positive root of n2−10n−100=0n^2 - 10n - 100 = 0 is n=5+125=5+55≈16.2n = 5 + \sqrt{125} = 5 + 5\sqrt{5} \approx 16.2.
  6. Since nn is a positive integer and the quadratic is positive beyond its larger root, the least nn is 1717. Check: S16=16×6=96S_{16} = 16 \times 6 = 96, S17=17×7=119S_{17} = 17 \times 7 = 119 (A1).

Answer: (a) d=2d = 2, u1=−9u_1 = -9. (b) n=10n = 10. (c) n=17n = 17.

Example 2

(Paper 1 style, 8 marks) The sum to infinity of a geometric series is 4848 and its second term is 99. (a) Show that the common ratio rr satisfies 16r2−16r+3=016r^2 - 16r + 3 = 0. (b) Find the two possible values of rr 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
  1. (a) From the booklet, S∞=u11−r=48S_\infty = \frac{u_1}{1-r} = 48, so u1=48(1−r)u_1 = 48(1-r) (M1). The second term is u1r=9u_1 r = 9 (A1).
  2. Substitute: 48r(1−r)=948r(1-r) = 9, so 48r−48r2−9=048r - 48r^2 - 9 = 0. Divide by −3-3: 16r2−16r+3=016r^2 - 16r + 3 = 0 (A1 AG). Write every line; the final mark of a 'show that' is usually withheld unless the expansion is shown.
  3. (b) Factorise: (4r−1)(4r−3)=0(4r - 1)(4r - 3) = 0, so r=14r = \frac{1}{4} or r=34r = \frac{3}{4} (M1 A1).
  4. Both satisfy ∣r∣<1|r| < 1, so both series converge and both are valid (R1).
  5. First terms: u1=48(1−14)=36u_1 = 48\left(1 - \frac{1}{4}\right) = 36 when r=14r = \frac{1}{4}, and u1=48(1−34)=12u_1 = 48\left(1 - \frac{3}{4}\right) = 12 when r=34r = \frac{3}{4} (A1).
  6. (c) The larger first term is 3636 with r=14r = \frac{1}{4}: the series is ∑k=1∞36(14)k−1\sum_{k=1}^{\infty} 36\left(\frac{1}{4}\right)^{k-1} (A1).

Answer: (a) Shown. (b) r=14r = \frac{1}{4}, u1=36u_1 = 36 or r=34r = \frac{3}{4}, u1=12u_1 = 12; both valid because ∣r∣<1|r|<1. (c) ∑k=1∞36(14)k−1\sum_{k=1}^{\infty} 36\left(\frac{1}{4}\right)^{k-1}.

Example 3

(Paper 2 style, 9 marks) Maria invests 50005000 euros in an account paying a nominal annual interest rate of 3.6%3.6\%, compounded monthly. (a) Find the value of the investment after 55 years, to the nearest cent. (b) Find the number of complete years required for the investment to exceed 80008000 euros. (c) At the same time Maria buys a car for 2400024000 euros which depreciates by 15%15\% per year. Find the value of the car after 44 years, to the nearest euro.

Solution
  1. (a) Use FV=PV(1+r100k)knFV = PV\left(1 + \frac{r}{100k}\right)^{kn} with PV=5000PV = 5000, r=3.6r = 3.6, k=12k = 12, n=5n = 5: FV=5000(1+3.61200)60=5000(1.003)60FV = 5000\left(1 + \frac{3.6}{1200}\right)^{60} = 5000(1.003)^{60} (M1 A1).
  2. FV=5984.47FV = 5984.47 euros (A1). On the finance solver: N=60N = 60, I%=3.6I\% = 3.6, PV=−5000PV = -5000, PMT=0PMT = 0, P/Y=C/Y=12P/Y = C/Y = 12; write these inputs down to earn the method mark.
  3. (b) Require 5000(1.003)12n>80005000(1.003)^{12n} > 8000, so (1.003)12n>1.6(1.003)^{12n} > 1.6 (M1 A1).
  4. Take logarithms: 12nln⁡1.003>ln⁡1.612n \ln 1.003 > \ln 1.6, so n>ln⁡1.612ln⁡1.003=13.07…n > \frac{\ln 1.6}{12 \ln 1.003} = 13.07\ldots (A1). Since ln⁡1.003>0\ln 1.003 > 0 the inequality does not reverse.
  5. The investment first exceeds 80008000 euros during the 1414th year, so 1414 complete years are needed (A1). Check with a table: after 1313 years the value is 7978.407978.40, after 1414 years it is 8270.408270.40.
  6. (c) Depreciation: V=24000(1−0.15)4=24000(0.85)4V = 24000(1 - 0.15)^4 = 24000(0.85)^4 (M1). 0.854=0.522006250.85^4 = 0.52200625, so V=12528.15≈12528V = 12528.15 \approx 12528 euros (A1).

Answer: (a) 5984.475984.47 euros. (b) 1414 years. (c) 1252812528 euros.

Can you...

  • Find dd or rr 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 nn, 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 ∣r∣<1|r|<1 in words, apply it to a ratio containing a parameter to obtain a strict interval, and use S∞S_\infty for recurring decimals?
  • Use the compound interest formula with kk 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 nnth geometric term is u1rn−1u_1 r^{n-1}, not u1rnu_1 r^n, while the value after nn years of compound growth is PV(1+i)nPV(1+i)^n. Decide first whether u1u_1 is the starting value or the value after one period.
  • Using the number of years as the exponent when compounding is monthly: with k=12k = 12 the exponent is 12n12n and the rate per period is r1200\frac{r}{1200}. Write the booklet formula with all four quantities labelled before substituting.
  • Claiming convergence without stating ∣r∣<1|r|<1, or giving the interval with closed endpoints. The R mark needs the strict condition written down and checked against each candidate value of rr.
  • Rounding a continuous answer the wrong way when nn must be an integer: a bound such as n>13.07n > 13.07 means n=14n = 14, and the integer nn 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 ln⁡r\ln r: if 0<r<10<r<1 then ln⁡r\ln r is negative and the direction changes. Alternatively keep everything positive by writing (1r)n\left(\frac{1}{r}\right)^n.
  • Using rounded intermediate values such as (1.003)60≈1.20(1.003)^{60} \approx 1.20 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.

Find out what is costing the marks.

Twenty minutes, one to one, online. No charge, no commitment.

Request an initial assessment