## IB Mathematics AA SL Arithmetic Sequences & Series Study Notes

IB Mathematics AA SL Arithmetic Sequences & Series Study Notes Offer a clear explanation of Arithmetic Sequences & Series , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Arithmetic Sequences & Series

**What is Arithmetic Sequences & Series**

An arithmetic sequence is a sequence where each term differs by a constant value called the common

difference (d).

The formula for the n^{th} term of an arithmetic sequence is given some info such as the first term \((U_1)\), or \(d\).

\[

U_1 \quad \xrightarrow{+d} \quad U_2 \quad \xrightarrow{+d} \quad U_3 \quad \dots \quad U_n

\]

Here, we see that you must add \(d\), \( (n-1) \) times, in other words:

\[

U_n = U_1 + (n – 1)d

\]

For Example in series of number \( 4, 9, 14, 19, \dots \) \(\quad \text{find 9th term}\)

Solution: \(U_9 = 4 + (9 – 1) \times 5, \quad U_9 = 44\)

* Properties of an arithmetic sequence *\((u_1, u_2, u_3, \dots)\):

1. First term: \( u_1 \)

2. Common difference : \( d = u_2 – u_1 = u_n – u_{n-1} \)

3. General term (nth term) : \( u_n = u_1 + (n – 1)d \)

4. Sum of the first \( n \) terms : \( S_n = \frac{n}{2} [2u_1 + (n – 1)d] = \frac{n}{2} [u_1 + u_n] \)

**Sum Of Arithmetic Terms**

If we have a formula for each term, we should also be able to find the sum of the first \(n\) terms:

\[

S_n = (U_1 + d) + (U_1 + 2d) + \dots + (U_1 + (n-1)d)

\]

By simplifying:

\[

S_n = \frac{n}{2}(2U_1 + (n-1)d) \quad \text{or} \quad S_n = \frac{n}{2}(U_1 + U_n) \quad \text{both in F.B.}

\]

*Practical Question*

Q. I start saving money, putting away $10 in week one, $15 in week two, $20 in week three, and so on.

1. How much do I save on week 8?

2. How much do I save on week 30?

3. How much have I saved in total in one year?

4. How much will I have in total after 5 years?

**▶️Answer/Explanation**

Answers:

1. For week 8:

\( a_n = 5n + 5 \)

\( a_8 = 5(8) + 5 = 40 + 5 = 45 \)

2. For week 30:

\( a_{30} = 5(30) + 5 = 150 + 5 = 155 \)

3. Total saved in one year (52 weeks):

\( a_{52} = 5(52) + 5 = 260 + 5 = 265 \)

\( a_1 = 10 \)

\( S_{52} = \frac{52(10 + 265)}{2} = 26 \times 275 = 7150 \)

4. Total saved after 5 years (260 weeks):

\( 52 \times 5 = 260 \)

\( S_{260} = \frac{260(20 + 259 \times 5)}{2} = 130(20 + 1295) \)

\( = 130 \times 1315 = 170950 \)

### Arithmetic Sequences & Series Exam Style Worked Out Questions

*Question*

Consider the arithmetic sequence \(u_1\) , \(u_2\) , \(u_3\) , ….

The sum of the first n terms of this sequence is given by \(S_n = n^2 + 4n\).

(a) (i) Find the sum of the first five terms.

(ii) Given that \(S_6\) = 60, find \(u_6\).

(b) Find \(u_1\).

(c) Hence or otherwise, write an expression for \(u_n\) in terms of n.

Consider a geometric sequence, \(v_n\), where \(v_2 = u_1\) and \(v_4 = u_6\).

(d) Find the possible values of the common ratio, r.

(e) Given that \(v_{99} < 0\), find \(v_5\).

**Answer/Explanation**

**Answer:**

(a) (i) recognition that n = 5

\(S_5\) = 45

(ii) METHOD 1

recognition that \(S_5 + u_6 = S_6\).

\(u_6\) = 15

METHOD 2

recognition that \(60 = \frac{6}{2} (S_1 + u_6)\)

\(u_6\) = 15

METHOD 3

substituting their \(u_1\) and d values into \(u_1\) + (n – 1)d

\(u_6\) = 15

(b) recognition that \(u_1 = S_1\) (may be seen in (a)) OR substituting their \(u_6\) into \(S_6\)

OR equations for \(S_5\) and \(S_6\) in temrs of \(u_1\) and d

1 + 4 OR 60 = \(\frac{6}{2}\) (\(u_1\) + 15)

\(u_1\) = 5

(c) EITHER

valid attempt to find d (may be seen in (a) or (b))

d = 2

OR

valid attempt to find \(S_n – S_{n-1}\)

\(n^2 + 4n – (n^2 – 2n + 1 + 4n – 4)\)

OR

equating \(n^2 + 4n = \frac{n}{2}(5+u_n)\)

2n + 8 = 5 + \(u_n\) (or equivalent)

THEN

\(u_n\) = 5 + 2(n – 1) OR \(u_n\) = 2n + 3

(d) recognition that \(v_2r^2 = v_4\) OR \((v_3)^2 = v_2 \times v_4\)

\(r^2 = 3\) OR \(v_3 = (±)5\sqrt{3}\)

r = ±\(\sqrt{3}\)

(e) recognition that r is negative

\(v_5\) = -15\(\sqrt{3}\) (=-\(\frac{45}{\sqrt{3}}\))

*Question*

The n^{th} term of an arithmetic sequence is given by u_{n} = 15 – 3n .

(a) State the value of the first term, u_{1} .

(b) Given that the n^{th} term of this sequence is -33 , find the value of n .

(c) Find the common difference, d .

**Answer/Explanation**

Ans

(a) \(u_1=12\)

(b) \(15-3n =-33 \therefore n =16\)

(c) valid approach to find d

\(u_2-u_1 = 9-12 =-3\)

\(-33=12+15d\)

or \(d =-3\)