IB Mathematics AA – Topic 1: Number & Algebra

Comprehensive Guide to Sequences & Series

Introduction to Sequences & Series

A sequence is an ordered list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term. Sequences can be finite (having a specific number of terms) or infinite (continuing indefinitely).

A series is the sum of the terms in a sequence. When we add up all or some of the terms in a sequence, we create a series. This distinction is crucial: sequences list terms, series add them.

Notation: We typically denote the nth term of a sequence as $u_n$ or $a_n$, where $n$ represents the position of the term in the sequence. The sum of the first $n$ terms is denoted $S_n$.

1. Arithmetic Sequences

Definition & Theory

An arithmetic sequence (also called an arithmetic progression) is a sequence where the difference between consecutive terms is constant. This constant is called the common difference, denoted by $d$.

For example, in the sequence 3, 7, 11, 15, 19, ..., the common difference is $d = 4$ because each term increases by 4.

General form: $u_1, u_1 + d, u_1 + 2d, u_1 + 3d, \ldots$

The common difference can be positive (increasing sequence), negative (decreasing sequence), or zero (constant sequence).

Key Formulas:

nth term: $u_n = u_1 + (n-1)d$
Where $u_1$ is the first term, $n$ is the position, and $d$ is the common difference
Common difference: $d = u_{n+1} - u_n = u_n - u_{n-1}$
Can be found by subtracting any term from the next term
Sum of first n terms (Method 1): $S_n = \frac{n}{2}(2u_1 + (n-1)d)$
Use when you know $u_1$, $n$, and $d$
Sum of first n terms (Method 2): $S_n = \frac{n}{2}(u_1 + u_n)$
Use when you know the first term, last term, and number of terms

⚠ Common Pitfalls:

Don't confuse $n$ (position) with $n-1$ in the formula. The nth term adds $d$ exactly $(n-1)$ times, not $n$ times.
Remember that $d$ can be negative for decreasing sequences.
When solving simultaneous equations, set up two equations carefully using the correct positions.
Always check which sum formula is more efficient based on the given information.

💡 Pro Tips:

To check if a sequence is arithmetic, calculate the difference between consecutive terms—if constant, it's arithmetic.
The sum formula $S_n = \frac{n}{2}(u_1 + u_n)$ can be remembered as: "average of first and last term, multiplied by the number of terms."
For word problems, identify $u_1$ and $d$ from the context before applying formulas.

Example 1: Finding Terms Using Simultaneous Equations

Problem: The fourth term of an arithmetic sequence is 19 and the ninth term is 44. Find:

(a) the first term and common difference

(b) the 20th term

Solution:

(a) Using the nth term formula $u_n = u_1 + (n-1)d$:

For the 4th term: $u_4 = u_1 + 3d = 19$ ... equation (1)

For the 9th term: $u_9 = u_1 + 8d = 44$ ... equation (2)

To solve these simultaneous equations, we subtract equation (1) from equation (2):

$(u_1 + 8d) - (u_1 + 3d) = 44 - 19$

$5d = 25$

$d = 5$

Now substitute $d = 5$ back into equation (1):

$u_1 + 3(5) = 19$

$u_1 + 15 = 19$

$u_1 = 4$

Answer (a): First term $u_1 = 4$, common difference $d = 5$

(b) Now we can find the 20th term using $u_n = u_1 + (n-1)d$:

$u_{20} = 4 + (20-1)(5)$

$u_{20} = 4 + 19 \times 5$

$u_{20} = 4 + 95$

$u_{20} = 99$

(c) Using the sum formula $S_n = \frac{n}{2}(2u_1 + (n-1)d)$:

$S_{15} = \frac{15}{2}(2(4) + (15-1)(5))$

$S_{15} = \frac{15}{2}(8 + 14 \times 5)$

$S_{15} = \frac{15}{2}(8 + 70)$

$S_{15} = \frac{15}{2}(78)$

$S_{15} = 15 \times 39$

$S_{15} = 585$

Note: We could also use $S_n = \frac{n}{2}(u_1 + u_n)$. First find $u_{15} = 4 + 14(5) = 74$, then $S_{15} = \frac{15}{2}(4 + 74) = 585$

Example 2: Finding How Many Terms

Problem: In an arithmetic sequence, $u_1 = 120$ and $d = -7$. Which term is the first negative term?

Solution:

We need to find the smallest value of $n$ for which $u_n < 0$.

Using $u_n = u_1 + (n-1)d$:

$u_n = 120 + (n-1)(-7)$

$u_n = 120 - 7n + 7$

$u_n = 127 - 7n$

We want $u_n < 0$:

$127 - 7n < 0$

$127 < 7n$

$n > \frac{127}{7}$

$n > 18.14...$

Since $n$ must be a positive integer, the smallest value is $n = 19$.

Verification: $u_{18} = 127 - 7(18) = 1$ (still positive)

$u_{19} = 127 - 7(19) = -6$ (negative) ✓

Answer: The 19th term is the first negative term.

Example 3: Sum to a Specific Value

Problem: The sum of the first $n$ terms of an arithmetic sequence is given by $S_n = 3n^2 + 2n$. Find:

(a) the first term and common difference

(b) the 10th term

Solution:

(a) To find $u_1$, we use $S_1$:

$S_1 = 3(1)^2 + 2(1) = 3 + 2 = 5$

Since $S_1 = u_1$, we have $u_1 = 5$

To find $d$, we use the relationship $u_n = S_n - S_{n-1}$ for $n \geq 2$:

$u_2 = S_2 - S_1$

$S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16$

$u_2 = 16 - 5 = 11$

Therefore, $d = u_2 - u_1 = 11 - 5 = $ $6$

Alternative method: Compare $S_n = 3n^2 + 2n$ with $S_n = \frac{n}{2}(2u_1 + (n-1)d)$

$3n^2 + 2n = \frac{n}{2}(2u_1 + (n-1)d) = nu_1 + \frac{n(n-1)d}{2}$

$3n^2 + 2n = nu_1 + \frac{d}{2}n^2 - \frac{d}{2}n$

Comparing coefficients: $\frac{d}{2} = 3$ so $d = 6$, and $u_1 - \frac{d}{2} = 2$ so $u_1 = 5$

(b) Using $u_n = u_1 + (n-1)d$:

$u_{10} = 5 + (10-1)(6) = 5 + 9 \times 6 = 5 + 54$

$u_{10} = 59$

2. Geometric Sequences

Definition & Theory

A geometric sequence (also called a geometric progression) is a sequence where each term is obtained by multiplying the previous term by a constant. This constant is called the common ratio, denoted by $r$.

For example, in the sequence 2, 6, 18, 54, 162, ..., the common ratio is $r = 3$ because each term is multiplied by 3.

General form: $u_1, u_1r, u_1r^2, u_1r^3, \ldots$

The common ratio can be positive, negative, or a fraction (but never zero). When $|r| > 1$, the sequence grows; when $0 < |r| < 1$, it decays; when $r < 0$, terms alternate in sign.

Key Formulas:

nth term: $u_n = u_1 r^{n-1}$
Where $u_1$ is the first term, $n$ is the position, and $r$ is the common ratio
Common ratio: $r = \frac{u_{n+1}}{u_n} = \frac{u_n}{u_{n-1}}$
Found by dividing any term by the previous term (assuming $u_n \neq 0$)
Sum of first n terms: $S_n = \frac{u_1(r^n - 1)}{r - 1}$ or $S_n = \frac{u_1(1 - r^n)}{1 - r}$ for $r \neq 1$
Both forms are equivalent; choose based on convenience. If $r = 1$, then $S_n = nu_1$
Sum to infinity: $S_\infty = \frac{u_1}{1 - r}$ for $|r| < 1$
Only converges when the absolute value of $r$ is less than 1

⚠ Common Pitfalls:

The infinite sum only converges when $|r| < 1$. If $|r| \geq 1$, the series diverges and has no finite sum.
Be careful with negative ratios—the sequence will alternate between positive and negative values.
When $0 < r < 1$, the sequence decreases toward zero; when $-1 < r < 0$, it oscillates toward zero.
Don't forget that $r^{n-1}$ in the nth term formula, not $r^n$.
For the sum formula, choose the version that makes your calculation easier (avoid negative denominators).

💡 Pro Tips:

To check if a sequence is geometric, divide consecutive terms—if the ratio is constant, it's geometric.
For convergence questions, always check $|r| < 1$ before calculating $S_\infty$.
When solving for $r$, you may get multiple solutions; check which one(s) make sense in context.
Use logarithms when solving for $n$ in exponential equations like $r^{n-1} = k$.

Example 4: Finding Terms and Sum

Problem: The second term of a geometric sequence is 12 and the fifth term is 96. Find:

(a) the common ratio and first term

(b) the 10th term

Solution:

(a) Using the nth term formula $u_n = u_1 r^{n-1}$:

For the 2nd term: $u_2 = u_1 r = 12$ ... equation (1)

For the 5th term: $u_5 = u_1 r^4 = 96$ ... equation (2)

Divide equation (2) by equation (1) to eliminate $u_1$:

$\frac{u_1 r^4}{u_1 r} = \frac{96}{12}$

$r^3 = 8$

$r = \sqrt[3]{8}$

$r = 2$

Now substitute $r = 2$ back into equation (1):

$u_1(2) = 12$

$u_1 = 6$

Answer (a): Common ratio $r = 2$, first term $u_1 = 6$

Verification: Sequence is 6, 12, 24, 48, 96, ... ✓

(b) Using $u_n = u_1 r^{n-1}$:

$u_{10} = 6 \times 2^{10-1} = 6 \times 2^9$

$u_{10} = 6 \times 512$

$u_{10} = 3072$

(c) Using $S_n = \frac{u_1(r^n - 1)}{r - 1}$:

$S_8 = \frac{6(2^8 - 1)}{2 - 1}$

$S_8 = \frac{6(256 - 1)}{1}$

$S_8 = 6 \times 255$

$S_8 = 1530$

Example 5: Sum to Infinity

Problem: A geometric series has first term 18 and common ratio $\frac{2}{3}$. Find:

(a) the sum to infinity

(b) the smallest value of $n$ for which $S_\infty - S_n < 0.5$

Solution:

(a) First, check convergence:

$|r| = \left|\frac{2}{3}\right| = \frac{2}{3} < 1$ ✓ (converges)

Since $|r| < 1$, we can use the sum to infinity formula:

Using $S_\infty = \frac{u_1}{1 - r}$:

$S_\infty = \frac{18}{1 - \frac{2}{3}}$

$S_\infty = \frac{18}{\frac{1}{3}}$

$S_\infty = 18 \times 3$

$S_\infty = 54$

(b) We need $S_\infty - S_n < 0.5$:

First, find $S_n$ using $S_n = \frac{u_1(1 - r^n)}{1 - r}$:

$S_n = \frac{18\left(1 - \left(\frac{2}{3}\right)^n\right)}{1 - \frac{2}{3}} = \frac{18\left(1 - \left(\frac{2}{3}\right)^n\right)}{\frac{1}{3}} = 54\left(1 - \left(\frac{2}{3}\right)^n\right)$

Now, $S_\infty - S_n = 54 - 54\left(1 - \left(\frac{2}{3}\right)^n\right)$:

$S_\infty - S_n = 54 - 54 + 54\left(\frac{2}{3}\right)^n = 54\left(\frac{2}{3}\right)^n$

We require $54\left(\frac{2}{3}\right)^n < 0.5$:

$\left(\frac{2}{3}\right)^n < \frac{0.5}{54} = \frac{1}{108}$

Taking logarithms of both sides:

$n \ln\left(\frac{2}{3}\right) < \ln\left(\frac{1}{108}\right)$

Note: $\ln\left(\frac{2}{3}\right) < 0$, so when dividing, the inequality reverses:

$n > \frac{\ln\left(\frac{1}{108}\right)}{\ln\left(\frac{2}{3}\right)} = \frac{-\ln(108)}{\ln(2) - \ln(3)}$

$n > \frac{-4.682}{-0.405} \approx 11.56$

Answer: The smallest value is $n = 12$

Example 6: Geometric Sequence with Fractional Ratio

Problem: The third term of a geometric sequence is 24 and the sixth term is 3. Find the first term and the common ratio, given that all terms are positive.

Solution:

Using $u_n = u_1 r^{n-1}$:

$u_3 = u_1 r^2 = 24$ ... equation (1)

$u_6 = u_1 r^5 = 3$ ... equation (2)

Divide equation (2) by equation (1):

$\frac{u_1 r^5}{u_1 r^2} = \frac{3}{24}$

$r^3 = \frac{1}{8}$

$r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

Substitute back into equation (1):

$u_1 \left(\frac{1}{2}\right)^2 = 24$

$u_1 \times \frac{1}{4} = 24$

$u_1 = 96$

Answer: First term $u_1 = 96$, common ratio $r = \frac{1}{2}$

Verification: 96, 48, 24, 12, 6, 3, ... ✓

3. Sigma Notation

Definition & Theory

Sigma notation (also called summation notation) uses the Greek letter sigma (Σ) to represent the sum of a sequence of terms. It provides a concise way to write long sums.

The general form is $\displaystyle\sum_{i=m}^{n} f(i)$, which reads as "the sum of $f(i)$ from $i = m$ to $i = n$".

Components:

$i$ is the index of summation (can be any variable)
$m$ is the lower bound (starting value)
$n$ is the upper bound (ending value)
$f(i)$ is the general term (expression being summed)

Example: $\displaystyle\sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$

Key Properties & Standard Results:

Constant: $\displaystyle\sum_{i=1}^{n} c = nc$ (where $c$ is a constant)
Adding the same number $n$ times gives $n \times c$
Sum of first n natural numbers: $\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$
Example: $1 + 2 + 3 + ... + 100 = \frac{100(101)}{2} = 5050$
Sum of first n squares: $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$
This formula is provided in the IB formula booklet
Sum of first n cubes: $\displaystyle\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$
Notice this equals $\left(\sum i\right)^2$
Linearity (sum rule): $\displaystyle\sum_{i=1}^{n} (a_i + b_i) = \sum_{i=1}^{n} a_i + \sum_{i=1}^{n} b_i$
You can split sums term by term
Constant multiple: $\displaystyle\sum_{i=1}^{n} ka_i = k\sum_{i=1}^{n} a_i$
Constants can be factored out

💡 Sigma Notation Tips:

For arithmetic series: $\displaystyle\sum_{i=1}^{n} (a + (i-1)d) = S_n = \frac{n}{2}(2a + (n-1)d)$
For geometric series: $\displaystyle\sum_{i=0}^{n-1} ar^i = \frac{a(r^n-1)}{r-1}$ or $\displaystyle\sum_{i=1}^{n} ar^{i-1} = \frac{a(r^n-1)}{r-1}$
Break complex expressions into simpler sums using linearity properties.
Check the starting index carefully—it may not always start at 1.

Example 7: Evaluating Sigma Notation (Arithmetic)

Problem: Evaluate $\displaystyle\sum_{r=1}^{10} (3r - 2)$.

Solution:

Method 1: Arithmetic Series Approach

This is an arithmetic series. Find the first and last terms:

When $r = 1$: $u_1 = 3(1) - 2 = 1$

When $r = 10$: $u_{10} = 3(10) - 2 = 28$

Common difference: $d = 3$, Number of terms: $n = 10$

Using $S_n = \frac{n(u_1 + u_n)}{2}$:

$S_{10} = \frac{10(1 + 28)}{2} = \frac{10 \times 29}{2} = 5 \times 29 = 145$

Method 2: Using Sigma Properties

$\displaystyle\sum_{r=1}^{10} (3r - 2) = \sum_{r=1}^{10} 3r - \sum_{r=1}^{10} 2$

$= 3\sum_{r=1}^{10} r - \sum_{r=1}^{10} 2$

$= 3 \times \frac{10(11)}{2} - 2(10)$

$= 3 \times 55 - 20$

$= 165 - 20$

$= 145$

Example 8: Geometric Series in Sigma Notation

Problem: Express and evaluate $\displaystyle\sum_{k=0}^{5} 2(3)^k$.

Solution:

This is a geometric series. Let's identify the components:

When $k = 0$: first term $u_1 = 2(3)^0 = 2$

Common ratio: $r = 3$

Number of terms: from $k = 0$ to $k = 5$ gives 6 terms

Using $S_n = \frac{u_1(r^n - 1)}{r - 1}$:

$S_6 = \frac{2(3^6 - 1)}{3 - 1}$

$S_6 = \frac{2(729 - 1)}{2}$

$S_6 = \frac{2 \times 728}{2}$

$S_6 = 728$

Answer: 728

Verification: $2 + 6 + 18 + 54 + 162 + 486 = 728$ ✓

Example 9: Using Standard Summation Formulas

Problem: Evaluate $\displaystyle\sum_{k=1}^{20} (2k^2 + 3k - 1)$.

Solution:

Break into separate sums using linearity:

$\displaystyle\sum_{k=1}^{20} (2k^2 + 3k - 1) = \sum_{k=1}^{20} 2k^2 + \sum_{k=1}^{20} 3k - \sum_{k=1}^{20} 1$

Factor out constants:

$= 2\sum_{k=1}^{20} k^2 + 3\sum_{k=1}^{20} k - \sum_{k=1}^{20} 1$

Apply standard formulas:

$= 2 \times \frac{20(21)(41)}{6} + 3 \times \frac{20(21)}{2} - 20$

$= 2 \times \frac{17220}{6} + 3 \times 210 - 20$

$= 2 \times 2870 + 630 - 20$

$= 5740 + 630 - 20$

$= 6350$

4. Applications of Sequences & Series

Real-World Context

Sequences and series appear extensively in real-world scenarios, particularly in finance, science, and business. Understanding which type of sequence models a situation is crucial for solving application problems.

Key Applications by Type:

Arithmetic Sequences:

Simple Interest: Fixed amount added each period
Linear Depreciation: Value decreases by constant amount
Salary Increments: Fixed raise each year
Production Plans: Constant increase/decrease in output
Savings Plans: Depositing same amount regularly (without interest)

Geometric Sequences:

Compound Interest: Interest calculated on principal and accumulated interest
Population Growth: Increases by constant percentage
Radioactive Decay: Decreases by constant percentage
Viral Spread: Each person infects constant number of others
Depreciation (reducing balance): Value decreases by constant percentage
Chain Reactions: Exponential growth patterns

Key Application Formulas:

Simple Interest: $A = P(1 + rt)$ or $A = P + Prt$
Forms an arithmetic sequence with $d = Pr$
Compound Interest (annual): $A = P(1 + r)^t$
Forms a geometric sequence with ratio $(1 + r)$
Population Growth: $P_n = P_0(1 + r)^n$
Where $r$ is the growth rate (as a decimal)
Exponential Decay: $N = N_0(1 - r)^t$ or $N = N_0 e^{-\lambda t}$
Common in radioactive decay, drug elimination, depreciation

💡 Application Problem Strategy:

Identify if the problem involves arithmetic or geometric growth/decay
Define variables clearly (what is $u_1$, what is $n$, etc.)
Write the appropriate formula
Substitute known values
Solve for the unknown
Check your answer makes sense in context

Example 10: Population Growth Application

Problem: A town's population is 20,000 and increases by 5% each year.

(a) Find the population after 10 years.

(b) After how many complete years will the population first exceed 30,000?

Solution:

This is a geometric sequence because population increases by a constant percentage (5%).

Initial population $P_0 = 20000$, growth rate $r = 0.05$, common ratio $= 1.05$

(a) After 10 years, we need the 11th term (year 0 is initial):

Using $P_n = P_0(1 + r)^n$:

$P_{10} = 20000(1.05)^{10}$

$P_{10} = 20000 \times 1.62889...$

$P_{10} \approx 32,578$ people

(b) We need to solve $20000(1.05)^n > 30000$:

$(1.05)^n > \frac{30000}{20000}$

$(1.05)^n > 1.5$

Take logarithms of both sides:

$n \ln(1.05) > \ln(1.5)$

$n > \frac{\ln(1.5)}{\ln(1.05)}$

$n > \frac{0.4055}{0.04879}$

$n > 8.31$

Answer: After 9 complete years (at start of year 10)

Check: $P_8 = 20000(1.05)^8 \approx 29,549$ (not yet), $P_9 = 20000(1.05)^9 \approx 31,027$ (exceeds) ✓

(c) Total growth = Population at year 5 - Initial population:

$P_5 = 20000(1.05)^5 = 20000 \times 1.27628 = 25,526$

Total growth $= 25,526 - 20,000 = $ 5,526 people

Alternatively: Growth = Sum of increases = $(P_1-P_0) + (P_2-P_1) + ... + (P_5-P_4)$

Example 11: Comparing Simple vs Compound Interest

Problem: $10,000 is invested for 5 years. Calculate the final amount if interest is:

(a) 6% simple interest per annum

(b) 6% compound interest per annum

Solution:

(a) Simple Interest (Arithmetic):

Using $A = P(1 + rt)$:

$A = 10000(1 + 0.06 \times 5)$

$A = 10000(1 + 0.3)$

$A = 10000(1.3)$

$A = \$13,000$

Sequence: 10000, 10600, 11200, 11800, 12400, 13000 (adds $600 each year)

(b) Compound Interest (Geometric):

Using $A = P(1 + r)^t$:

$A = 10000(1.06)^5$

$A = 10000 \times 1.33823$

$A \approx \$13,382.26$

Sequence: 10000, 10600, 11236, 11910.16, 12624.77, 13382.26 (multiplies by 1.06 each year)

(c) Difference:

$\text{Difference} = 13382.26 - 13000 = $ $382.26

Conclusion: Compound interest earns $382.26 more because interest is earned on previously accumulated interest.

5. Compound Interest (Detailed)

Theory & Concepts

Compound interest is interest calculated on both the initial principal and the accumulated interest from previous periods. Unlike simple interest, which forms an arithmetic sequence, compound interest forms a geometric sequence because the amount grows by a constant percentage.

The fundamental principle: each period, interest is added to the principal, and in the next period, interest is calculated on this new, larger amount. This creates exponential growth.

Key Formulas & Variables:

General Compound Interest Formula:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

Where:

$A$ = final amount (future value)
$P$ = principal (initial investment or loan amount)
$r$ = annual interest rate (as a decimal, e.g., 5% = 0.05)
$n$ = number of times interest is compounded per year
$t$ = time in years

Special Cases:

Annual compounding ($n = 1$): $A = P(1 + r)^t$
Semi-annual ($n = 2$): $A = P\left(1 + \frac{r}{2}\right)^{2t}$
Quarterly ($n = 4$): $A = P\left(1 + \frac{r}{4}\right)^{4t}$
Monthly ($n = 12$): $A = P\left(1 + \frac{r}{12}\right)^{12t}$
Daily ($n = 365$): $A = P\left(1 + \frac{r}{365}\right)^{365t}$
Continuous compounding: $A = Pe^{rt}$ (where $e \approx 2.71828$)

Depreciation (Decreasing Value):

$V = P(1 - r)^t$

Used when value decreases by a constant percentage (e.g., car depreciation)

⚠ Common Pitfalls:

Always convert percentage rates to decimals (e.g., 5% = 0.05) before substituting into formulas.
Ensure time units match the compounding period. If interest is monthly and time is in years, convert appropriately.
Remember $n$ is how many times per year, while $nt$ is the total number of compounding periods.
When finding time using logarithms, pay attention to whether you need complete periods or can use fractional values.
For depreciation, use $(1 - r)$ not $(1 + r)$.

💡 Pro Tips:

More frequent compounding yields higher returns (all else equal), but the difference decreases as $n$ increases.
The "Rule of 72": To estimate doubling time, divide 72 by the interest rate percentage (e.g., at 6%, money doubles in approximately 72÷6 = 12 years).
When solving for $r$ or $t$, use logarithms to "bring down" the exponent.
For comparison problems, calculate both options before determining which is better.

Example 12: Different Compounding Frequencies

Problem: $5,000 is invested at 6% per annum for 4 years. Find the final value if interest is compounded:

(a) annually

(b) quarterly

(d) Compare the results and calculate the additional earnings from monthly vs annual compounding.

Solution:

Given: $P = 5000$, $r = 0.06$, $t = 4$

(a) Annual compounding: $n = 1$

$A = 5000(1 + 0.06)^4$

$A = 5000(1.06)^4$

$A = 5000 \times 1.26248$

$A = \$6,312.38$

(b) Quarterly compounding: $n = 4$

$A = 5000\left(1 + \frac{0.06}{4}\right)^{4 \times 4}$

$A = 5000(1.015)^{16}$

$A = 5000 \times 1.26899$

$A \approx \$6,344.93$

(c) Monthly compounding: $n = 12$

$A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \times 4}$

$A = 5000(1.005)^{48}$

$A = 5000 \times 1.27049$

$A \approx \$6,352.45$

(d) Comparison:

Difference (Monthly vs Annual) $= 6352.45 - 6312.38 = $ $40.07

Note: More frequent compounding leads to higher returns, but the marginal benefit decreases as compounding frequency increases.

Example 13: Finding Time (How Long to Reach a Goal)

Problem: How long will it take for an investment of $8,000 to grow to $12,000 at 7% per annum compounded annually?

Solution:

Given: $P = 8000$, $A = 12000$, $r = 0.07$, $n = 1$, find $t$

Using $A = P(1 + r)^t$:

$12000 = 8000(1.07)^t$

Divide both sides by 8000:

$(1.07)^t = \frac{12000}{8000}$

$(1.07)^t = 1.5$

Take natural logarithm of both sides:

$\ln((1.07)^t) = \ln(1.5)$

Use logarithm property $\ln(a^b) = b\ln(a)$:

$t \ln(1.07) = \ln(1.5)$

$t = \frac{\ln(1.5)}{\ln(1.07)}$

$t = \frac{0.4055}{0.0677}$

$t \approx 5.99 \text{ years}$

Answer: Approximately 6 years

Verification: $8000(1.07)^6 = 8000 \times 1.5007 \approx 12006$ ✓

Example 14: Depreciation Application

Problem: A car is purchased for $25,000 and depreciates at 15% per year.

(a) Find the value after 5 years.

(b) After how many complete years will the car be worth less than $10,000?

Solution:

This is depreciation, so we use $V = P(1 - r)^t$

Given: $P = 25000$, $r = 0.15$

(a) After 5 years:

$V = 25000(1 - 0.15)^5$

$V = 25000(0.85)^5$

$V = 25000 \times 0.4437$

$V \approx \$11,093.24$

(b) We need $25000(0.85)^t < 10000$:

$(0.85)^t < \frac{10000}{25000}$

$(0.85)^t < 0.4$

Take logarithms:

$t \ln(0.85) < \ln(0.4)$

Note: $\ln(0.85) < 0$, so inequality reverses when dividing:

$t > \frac{\ln(0.4)}{\ln(0.85)}$

$t > \frac{-0.9163}{-0.1625}$

$t > 5.64$

Answer: After 6 complete years

Check: After 5 years: $11,093 (still above $10,000), After 6 years: $9,429 (below $10,000) ✓

📋 Complete Formula Reference

Arithmetic Sequences:

$u_n = u_1 + (n-1)d$

$d = u_{n+1} - u_n$

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

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

Geometric Sequences:

$u_n = u_1 r^{n-1}$

$r = \frac{u_{n+1}}{u_n}$

$S_n = \frac{u_1(r^n - 1)}{r - 1}$ for $r \neq 1$

$S_\infty = \frac{u_1}{1-r}$ for $|r| < 1$

Sigma Notation:

$\displaystyle\sum_{i=1}^{n} c = nc$

$\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

$\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

Compound Interest:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

$A = P(1 + r)^t$ (annual)

$V = P(1 - r)^t$ (depreciation)

🎯 IB Exam Strategy & Tips

Before You Start:

Read the question carefully to identify if it's arithmetic or geometric
List all given information and what you need to find
Check your calculator is in the correct mode

Common Question Types:

Finding terms: Use $u_n$ formula directly
Finding $u_1$ and $d$ (or $r$): Set up simultaneous equations
Finding sums: Choose the most efficient $S_n$ formula based on given information
Finding $n$: Solve equations (may need logarithms for geometric)
Applications: Model the situation, then apply appropriate formulas

Show Your Working:

Write down the formula you're using
Substitute values clearly
Show intermediate steps, especially for logarithms
Round only at the final answer (use 3 significant figures unless stated otherwise)
Include units in your final answer for application problems

Time-Saving Techniques:

For geometric sequences, use your calculator's power function efficiently
Store intermediate values in calculator memory to avoid rounding errors
For sigma notation, recognize when it forms an arithmetic or geometric series
When finding $n$, set up the equation first, then use logarithms systematically

Final Checks:

Does your answer make sense? (e.g., population shouldn't be negative)
Have you answered all parts of the question?
Are your units correct?
Is your answer in the requested form (exact vs decimal)?

🔍 Arithmetic vs Geometric: Quick Decision Guide

Question Contains...	It's Likely...	Key Words
Constant amount added/subtracted	Arithmetic	increases by $50, decreases by 3
Constant percentage/ratio	Geometric	increases by 5%, doubles, halves
Simple interest	Arithmetic	simple interest, linear growth
Compound interest	Geometric	compound interest, exponential
Each person infects n others	Geometric	multiplies, spreads, reproduces
Fixed payment/deposit each period	Arithmetic	saves $100 monthly, pays $50 weekly

🧮 Calculator Tips for IB Exams

Essential Calculator Functions:

For sequences: Use sequence mode (if available) or store formulas
For powers: Master the $x^y$ button and negative exponents
For logarithms: Know both ln and log buttons; use change of base if needed
Memory functions: Store $u_1$, $d$, or $r$ in memory (A, B, C, etc.)
ANS button: Use to avoid retyping long numbers
Fractions: Keep answers in fraction form when possible for exact values

⚠️ Common Calculator Mistakes:

Forgetting parentheses: $(1.05)^{10}$ not $1.05^{10}$
Order of operations: Use brackets liberally
Rounding too early: Keep full precision until final answer
Not clearing previous calculations

📝 Practice Problem Checklist

To master this topic, ensure you can:

✓ Identify whether a sequence is arithmetic or geometric from description or terms
✓ Find the nth term given two other terms (using simultaneous equations)
✓ Calculate sums using both formula methods
✓ Convert between sigma notation and expanded form
✓ Evaluate sums using standard formulas ($\sum i$, $\sum i^2$, etc.)
✓ Determine convergence and find sum to infinity
✓ Solve for n using logarithms in geometric sequences
✓ Model real-world situations with appropriate sequences
✓ Compare compound vs simple interest scenarios
✓ Handle depreciation and decay problems

⚡ Most Common IB Exam Question Types

Type 1: Finding First Term & Common Difference/Ratio

Given two terms, set up simultaneous equations and solve.

Difficulty: Medium | Frequency: Very High

Type 2: Sum Problems

Calculate $S_n$ or find n given a target sum.

Difficulty: Medium | Frequency: Very High

Type 3: Real-World Applications

Finance, population, depreciation problems requiring modeling.

Difficulty: Medium-Hard | Frequency: High

Type 4: Sigma Notation

Evaluate summations using properties and standard results.

Difficulty: Medium | Frequency: High

🎓 You've Got This!

Sequences and series are fundamental to IB Mathematics AA. Master the formulas, practice identifying patterns, and always show your working clearly. With consistent practice and understanding of when to apply each type, you'll excel in this topic!

Remember: Understanding > Memorization