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[h] IB Mathematics AA HL Flashcards- Arithmetic Sequences & Series

[q] **Arithmetic Sequences & Series**

[a]

The first type of sequences you will see is **arithmetic**. They are sequences with the same difference between each term, this is called the common difference \((d)\).

E.G.1 // \( 1, 3, 5, 7, 9, \dots \) \(\quad d = 2\)

E.G.2 // \( 28, 22, 16, 10, 4, -2, \dots \) \(\quad d = -6\)

[q]

FINDING THE N\(^\text{th}\) TERM // You may be asked to find a specific term later in the sequence, 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 \quad \text{in F.B.}

\]

[a]

E.G.3 // \( 4, 9, 14, 19, \dots \) \(\quad \text{find 9th term}: U_9 = 4 + (9 – 1) \times 5, \quad U_9 = 44\)

SERIES/SUM OF 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)

\]

[q]

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.}

\]

[a]

E.G.4 // Find the sum of the first 9 terms from E.G.3:

\[

S_9 = \frac{9}{2}(4 + 44) = 216

\]

[q]

E.G.5 // \( U_2 = 11, U_5 = 23, \) find \( \sum_{n=1}^8 U_n \): There are 3 common differences between \( U_5 \) & \( U_2 \), so \( 3d = U_5 – U_2 = 23 – 11 = 12 \). Thus, \( d = 4 \). Then,

\[

S_8 = \frac{8}{2}(2(7) + (8-1)4) = 168

\]

APPLICATIONS // One of the most common real-world applications of this is simple interest, where the same % of the initial investment is added each year (as opposed to compound interest in 1.4).

[a]

E.G.6 // Find the value of a £1,200 investment after 6 yrs w/ 4% simple interest:

\[

U_1 = 1200, d = 0.04 \times 1200 = 48.

\]

Need to find \(U_7\) (not \(U_6\)):

\[

U_7 = 1200 + 6 \times 48 = £1,488

\]

The syllabus also states that you may need to be able to approximate values in some arithmetic sequences problems.

[a]

Properties of an arithmetic sequence \(\u_{n}\) :

[q] Solved Example 1

Five pipes labelled, “6 metres in length”, were delivered to a building site. The contractor measured each pipe to check its length (in metres) and recorded the following;

5.96, 5.95, 6.02, 5.95, 5.99.

a. (i) Find the mean of the contractor’s measurements.

a. (ii) Calculate the percentage error between the mean and the stated, **approximate** length of 6 metres.[3]

Calculate \(\sqrt {{{3.87}^5} – {{8.73}^{ – 0.5}}} \), giving your answer

b. (i) correct to the nearest integer,

b.(ii) in the form \(a \times 10^k\), where 1 ≤ *a* < 10, \(k \in {\mathbb{Z}}\) .[3]

[a]

a. (i) Mean = (5.96 + 5.95 + 6.02 + 5.95 + 5.99) / 5 = 5.974 (5.97) **(A1)**

a. (ii) \({\text{% error}} = \frac{{error}}{{actualvalue}} \times 100\% \)

\( = \frac{{6 – 5.974}}{{5.974}} \times 100\% = 0.435\% \)** (M1)(A1)****(ft)**

**(M1)** for correctly substituted formula.

*Allow 0.503% as follow through from 5.97*

*Note: An answer of 0.433% is incorrect. (C3)*

**[3 marks]**number is 29.45728613

b. (i) Nearest integer = 29 **(A1)**

b. (ii) Standard form = 2.95 × 10^{1} (*accept* 2.9 × 10^{1}) *(A1)*(ft)*(A1)*

*Award (A1) for each correct term*

*Award (A1)(A0) for 2.95 × 10 (C3)*

**[3 marks]**\(\frac{2}{4}\)

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