1.2

Arithmetic Sequences and Series

  • An arithmetic sequence has a constant difference $d$ between consecutive terms: $$d = u_{n} - u_{n-1} = u_2 - u_1$$
  • The $n$-th term is given by: $$u_n = u_1 + (n - 1)d$$
  • The sum of the first $n$ terms (series) is given by: $$S_n = \frac{n}{2}(2u_1 + (n - 1)d) = \frac{n}{2}(u_1 + u_n)$$
  • Sigma Notation: $\sum_{k=1}^n u_k = u_1 + u_2 + \dots + u_n = S_n$.

Solved Example 1: Arithmetic Progression

An arithmetic sequence has $u_4 = 19$ and $u_9 = 44$. Find the first term $u_1$, the common difference $d$, and the sum of the first 20 terms $S_{20}$.

Step 1: Set up simultaneous equations: $$u_4 = u_1 + 3d = 19$$ $$u_9 = u_1 + 8d = 44$$
Step 2: Subtract equations to find $d$ and $u_1$: $$(u_1 + 8d) - (u_1 + 3d) = 44 - 19 \implies 5d = 25 \implies d = 5$$ $$u_1 = 19 - 3(5) = 19 - 15 = 4$$
Step 3: Calculate $S_{20}$: $$S_{20} = \frac{20}{2}[2(4) + (20 - 1)(5)] = 10[8 + 95] = 10(103) = 1030$$

Final Answer: $u_1 = 4$, $d = 5$, $S_{20} = 1030$

1.3

Geometric Sequences and Series

  • A geometric sequence has a constant ratio $r$ between consecutive terms: $$r = \frac{u_n}{u_{n-1}} = \frac{u_2}{u_1} \quad (r \ne 0)$$
  • The $n$-th term is given by: $$u_n = u_1 r^{n-1}$$
  • The sum of the first $n$ terms is given by: $$S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r} \quad (r \ne 1)$$
  • Growth vs. Decay: When $r > 1$, terms grow exponentially. When $0 < r < 1$, terms decrease exponentially towards zero.

Solved Example 2: Geometric Modeling

A bacteria colony begins with 500 cells ($u_1 = 500$) and doubles every hour ($r = 2$).

  1. Find the population after 6 hours (i.e. at the start of the 7th hour, $u_7$).
  2. Determine the first hour in which the total population exceeds $1,000,000$.

(a) Calculate $u_7$: $$u_7 = 500(2)^{7-1} = 500(2)^6 = 500(64) = 32,000\text{ cells}$$
(b) Solve $u_n > 1,000,000$: $$500(2)^{n-1} > 1,000,000 \implies 2^{n-1} > 2000$$ $$\ln(2^{n-1}) > \ln(2000) \implies (n-1)\ln(2) > \ln(2000)$$ $$n - 1 > \frac{\ln(2000)}{\ln(2)} \approx 10.966 \implies n > 11.966$$ Since $n$ must be an integer, $n = 12$.

Final Answer: (a) $32,000$ cells, (b) During the 12th hour ($n = 12$)

Select a model and check the index

  • Equal additions suggest arithmetic; equal percentage changes suggest geometric. For a value at time0 use $A(t)=A_0r^t$, whereas a first listed term is $u_n=u_1r^{n-1}$.
  • For r1, a geometric sequence is constant and $S_n=nu_1$; do not divide by r−1. For negative r, signs alternate. Ratio comparisons need a nonzero preceding term.
  • For $\sum_{k=p}^q u_k$, the number of terms is q−p+1, not q. Recover $u_n=S_n-S_{n-1}$.
  • A term-count solution must be a positive integer. For threshold inequalities compare adjacent integer times using unrounded values. Growth models may cease to be realistic when resources are limited.

10 Practice Exercises

Complete with $[M][A][R]$ markschemes covering arithmetic, geometric, and sigma calculations.

Exercise 1 (Paper 1 — 3 Marks): Consider the arithmetic sequence $7, 11, 15, 19, \dots$.
  1. Find the 25th term $u_{25}$.
  2. Find the sum of the first 25 terms $S_{25}$.
$u_1 = 7, d = 11 - 7 = 4$.
(a) $u_{25} = 7 + (25 - 1)(4) = 7 + 24(4) = 7 + 96 = 103$ [M1][A1]
(b) $S_{25} = \frac{25}{2}(7 + 103) = 12.5(110) = 1375$ [M1][A1]

Answer: (a) $103$, (b) $1375$

Exercise 2 (Paper 1 — 4 Marks): In a geometric sequence, the second term is $12$ and the fifth term is $324$. Find the common ratio $r$, the first term $u_1$, and the sum $S_6$.
$u_2 = u_1 r = 12$ and $u_5 = u_1 r^4 = 324$.
Divide: $\frac{u_1 r^4}{u_1 r} = \frac{324}{12} \implies r^3 = 27 \implies r = 3$ [M1][A1]
$u_1 = \frac{12}{3} = 4$ [A1]
$S_6 = \frac{4(3^6 - 1)}{3 - 1} = \frac{4(729 - 1)}{2} = 2(728) = 1456$ [A1]

Answer: $r = 3$, $u_1 = 4$, $S_6 = 1456$

Exercise 3 (Paper 1 — 3 Marks): Evaluate $\sum_{k=1}^{15} (3k + 2)$.
First term $u_1 = 3(1) + 2 = 5$ [A1]
Last term $u_{15} = 3(15) + 2 = 47$ [A1]
Number of terms $n = 15$.
$S_{15} = \frac{15}{2}(5 + 47) = 7.5(52) = 390$ [M1][A1]

Answer: $390$

Exercise 4 (Paper 2 — 5 Marks): An athlete runs $2.0\text{ km}$ on Day 1. Each day thereafter, she increases her distance by $8\%$ compared to the previous day.
  1. Find the distance she runs on Day 10.
  2. Find the total cumulative distance run over the first 14 days.
  3. On which day will she first run more than $10.0\text{ km}$ in a single session?
$u_1 = 2.0, r = 1 + 0.08 = 1.08$.
(a) $u_{10} = 2.0(1.08)^9 \approx 3.998 \approx 4.00\text{ km}$ [M1][A1]
(b) $S_{14} = \frac{2.0(1.08^{14} - 1)}{1.08 - 1} = \frac{2.0(2.9372 - 1)}{0.08} \approx 48.4\text{ km}$ [M1][A1]
(c) $2.0(1.08)^{n-1} > 10.0 \implies 1.08^{n-1} > 5.0 \implies (n-1)\ln(1.08) > \ln(5)$
$n - 1 > 20.91 \implies n > 21.91 \implies \text{Day } 22$ [M1][A1]

Answer: (a) $4.00\text{ km}$, (b) $48.4\text{ km}$, (c) Day 22

Exercise 5 (Paper 1 — 3 Marks): The sum of the first $n$ terms of an arithmetic sequence is given by $S_n = 2n^2 + 5n$. Find the first term $u_1$ and the common difference $d$.
$u_1 = S_1 = 2(1)^2 + 5(1) = 7$ [A1]
$S_2 = 2(2)^2 + 5(2) = 8 + 10 = 18$
$u_2 = S_2 - S_1 = 18 - 7 = 11$ [M1]
$d = u_2 - u_1 = 11 - 7 = 4$ [A1]

Answer: $u_1 = 7, d = 4$

Exercise 6

Evaluate $\sum_{k=3}^{8}(2k-1)$.

Terms5,7,9,11,13,15: six terms; sum $6(5+15)/2=60$.

Exercise 7

Find the sum of eight terms of a geometric sequence with first term7and ratio1.

All terms are7, so S8=56. The usual quotient form is0/0 here; use the constant-sequence case.

Exercise 8

For u1=8,r=−1/2, find u4andS4. Explain why a population model would be unsuitable.

u4=−1; terms8,−4,2,−1 sum5. Negative population values are infeasible.

Exercise 9

Two savings plans start with1000. One gains50each year; the other grows5%annually. Compare after3years.

At time0both1000. At year3: arithmetic1150; geometric1000(1.05)^3=1157.625. They agree after1year but growth methods differ.

Exercise 10

Given $S_n=2n^2+3n$, find un and determine whether the sequence is arithmetic.

$S_n-S_{n-1}=4n+1$forn≥2; u1=S1=5 also matches. Difference4, so arithmetic.

Formulae

Official IB AI SL Formula Booklet Equations

Arithmetic Sequences & Series:

$$u_n = u_1 + (n - 1)d$$ $$S_n = \frac{n}{2}(2u_1 + (n - 1)d) = \frac{n}{2}(u_1 + u_n)$$

Geometric Sequences & Series:

$$u_n = u_1 r^{n-1}$$ $$S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}, \quad r \ne 1$$
Examiner Pitfalls

Common Exam Traps in Sequences

  • Off-by-One Error in Geometric Growth: If an amount increases by $r$ at the end of each year, the value after $t$ years is $u_{t+1} = u_1 r^t$. Be very careful whether Day/Year 1 is $u_1$ or $t=0$.
  • Percentage Increase Factor: An increase of $8\%$ corresponds to $r = 1.08$, not $r = 0.08$. A decrease of $15\%$ corresponds to $r = 0.85$, not $r = 0.15$.
Topic 1.2 & 1.3 Check

Exam Mini Test (5 Multiple Choice Questions)

Score: 0 / 5

1. An arithmetic sequence begins $14, 11, 8, 5, \dots$. What is the 10th term $u_{10}$?

2. In a geometric sequence, $u_1 = 5$ and $r = -2$. What is the sum of the first 6 terms $S_6$?

3. Evaluate the sigma sum $\sum_{n=1}^4 3(2)^{n-1}$.

4. A sequence has terms $u_1 = 3, u_2 = 6, u_3 = 12$. What is the formula for the $n$-th term?

5. If the sum of an arithmetic series is $S_n = 50$, $u_1 = 2$, and $u_n = 18$, how many terms are there ($n$)?