Sequences and series — arithmetic and geometric progressions (KCSE Mathematics Form 4)

TL;DR

Sequences are ordered lists of numbers, and series are the sums of those numbers. Arithmetic progressions have a constant difference between terms, while geometric progressions have a constant ratio. Understanding their formulas helps you find any term or sum in these patterns.

1. The Mental Model

Think of sequences as a line of numbers following a specific rule, like steps on a ladder. A series is simply adding up all those steps. We're focusing on two main types of rules: adding the same number each time, or multiplying by the same number each time.

2. The Core Material

What is a Sequence?

A sequence is just an ordered list of numbers. Each number in the list is called a "term". We often use $T_n$ to represent the $n^{th}$ term. For example, in the sequence 2, 4, 6, 8, ...
$T_1 = 2$
$T_2 = 4$
$T_3 = 6$

What is a Series?

A series is the sum of the terms in a sequence. If you have a sequence $T_1, T_2, T_3, ...$, the corresponding series would be $T_1 + T_2 + T_3 + ...$. We use $S_n$ to denote the sum of the first $n$ terms.

Arithmetic Progression (AP)

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

Finding the $n^{th}$ term of an AP:
The formula for the $n^{th}$ term ($T_n$) of an AP is:
$T_n = a + (n-1)d$
where:
* $a$ is the first term
* $n$ is the term number you're looking for
* $d$ is the common difference

Example: In the AP 3, 7, 11, 15, ...
$a = 3$
$d = 7 - 3 = 4$ (or $11 - 7 = 4$)
To find the 5th term ($T_5$):
$T_5 = 3 + (5-1)4 = 3 + 4 \times 4 = 3 + 16 = 19$

Sum of the first $n$ terms of an AP ($S_n$):
There are two main formulas:
1. $S_n = \frac{n}{2}(a + L)$
where $L$ is the last term (which is $T_n$)
2. $S_n = \frac{n}{2}(2a + (n-1)d)$
This second formula is often more useful as you don't always know the last term directly.

Example: Find the sum of the first 10 terms of the AP 3, 7, 11, 15, ...
$a = 3$, $d = 4$, $n = 10$
$S_{10} = \frac{10}{2}(2 \times 3 + (10-1)4)$
$S_{10} = 5(6 + 9 \times 4)$
$S_{10} = 5(6 + 36)$
$S_{10} = 5(42) = 210$

Geometric Progression (GP)

A geometric progression (GP) is a sequence where the ratio between consecutive terms is constant. This constant ratio is called the common ratio, denoted by $r$.

Finding the $n^{th}$ term of a GP:
The formula for the $n^{th}$ term ($T_n$) of a GP is:
$T_n = ar^{n-1}$
where:
* $a$ is the first term
* $n$ is the term number you're looking for
* $r$ is the common ratio

Example: In the GP 2, 6, 18, 54, ...
$a = 2$
$r = \frac{6}{2} = 3$ (or $\frac{18}{6} = 3$)
To find the 4th term ($T_4$):
$T_4 = 2 \times 3^{4-1} = 2 \times 3^3 = 2 \times 27 = 54$

Sum of the first $n$ terms of a GP ($S_n$):
There are two formulas, depending on the value of $r$:
1. If $r > 1$: $S_n = \frac{a(r^n - 1)}{r - 1}$
2. If $r < 1$: $S_n = \frac{a(1 - r^n)}{1 - r}$
(If $r=1$, the sum is simply $na$ as all terms are the same.)

Example: Find the sum of the first 5 terms of the GP 2, 6, 18, 54, ...
$a = 2$, $r = 3$, $n = 5$
Since $r > 1$, use the first formula:
$S_5 = \frac{2(3^5 - 1)}{3 - 1}$
$S_5 = \frac{2(243 - 1)}{2}$
$S_5 = 242$

Infinite Geometric Series

For a GP, if the common ratio $r$ is between -1 and 1 (i.e., $|r| < 1$), the sum of an infinite number of terms approaches a finite value. This is called the sum to infinity, $S_\infty$.

Sum to infinity ($S_\infty$):
$S_\infty = \frac{a}{1 - r}$ (only if $|r| < 1$)

Example: Find the sum to infinity of the GP 10, 5, 2.5, 1.25, ...
$a = 10$
$r = \frac{5}{10} = 0.5$
Since $|r| = 0.5 < 1$, the sum to infinity exists:
$S_\infty = \frac{10}{1 - 0.5} = \frac{10}{0.5} = 20$

Identifying AP vs. GP

graph TD
    A[Start with a sequence] --> B{Are terms increasing/decreasing by a constant amount?};
    B -- Yes --> C[It's an Arithmetic Progression (AP)];
    B -- No --> D{Are terms increasing/decreasing by a constant factor?};
    D -- Yes --> E[It's a Geometric Progression (GP)];
    D -- No --> F[It's neither AP nor GP (or a different type of sequence)];
    C --> G[Use AP formulas: Tn = a + (n-1)d, Sn = n/2(2a + (n-1)d)];
    E --> H[Use GP formulas: Tn = ar^(n-1), Sn = a(r^n - 1)/(r-1) or a(1-r^n)/(1-r)];
    H -- If |r|<1 --> I[Consider Sum to Infinity: S_inf = a/(1-r)];

3. Worked Example

Problem: The third term of an arithmetic progression is 10, and the sixth term is 22.
a) Find the first term and the common difference.
b) Calculate the sum of the first 15 terms of this progression.

Solution:

a) Find the first term ($a$) and the common difference ($d$).
We know the formula for the $n^{th}$ term of an AP is $T_n = a + (n-1)d$.
Given:
$T_3 = 10 \implies a + (3-1)d = 10 \implies a + 2d = 10$ (Equation 1)
$T_6 = 22 \implies a + (6-1)d = 22 \implies a + 5d = 22$ (Equation 2)

Now we have a system of two linear equations:
1. $a + 2d = 10$
2. $a + 5d = 22$

Subtract Equation 1 from Equation 2:
$(a + 5d) - (a + 2d) = 22 - 10$
$3d = 12$
$d = \frac{12}{3}$
$d = 4$

Substitute $d=4$ into Equation 1:
$a + 2(4) = 10$
$a + 8 = 10$
$a = 10 - 8$
$a = 2$

So, the first term is 2 and the common difference is 4.

b) Calculate the sum of the first 15 terms ($S_{15}$).
We use the formula $S_n = \frac{n}{2}(2a + (n-1)d)$.
We have $a = 2$, $d = 4$, and $n = 15$.

$S_{15} = \frac{15}{2}(2 \times 2 + (15-1)4)$
$S_{15} = \frac{15}{2}(4 + 14 \times 4)$
$S_{15} = \frac{15}{2}(4 + 56)$
$S_{15} = \frac{15}{2}(60)$
$S_{15} = 15 \times 30$
$S_{15} = 450$

The sum of the first 15 terms is 450.

4. Key Takeaways

• An arithmetic progression has a constant common difference ($d$) between terms.
• A geometric progression has a constant common ratio ($r$) between terms.
• Use $T_n = a + (n-1)d$ for the $n^{th}$ term of an AP.
• Use $S_n = \frac{n}{2}(2a + (n-1)d)$ for the sum of the first $n$ terms of an AP.
• Use $T_n = ar^{n-1}$ for the $n^{th}$ term of a GP.
• Use $S_n = \frac{a(r^n - 1)}{r - 1}$ (for $r>1$) or $S_n = \frac{a(1 -