IB Mathematics AA SL Flashcards- Geometric sequences and series

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[h] IB Mathematics AA SL Flashcards- Geometric Sequences and Series

[q] Geometric sequences and series

[a]

The next type of sequence is geometric. In these, the ratio between consecutive terms is always equal. In other words, we always multiply by the same number to move from one term to the next → this is called the common ratio \( (r) \).

E.G.1 // \( 1, 3, 9, 27, \dots \quad r = 3 \)

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E.G.2 // \( 96, -48, 24, -12, 6, \dots \quad r = -\frac{1}{2} \)

Note that you can always find \( r \) by doing (any term) ÷ (the previous term)

[a]

FINDING THE N\(^\text{th}\) TERM // Just like arithmetic sequences, we may be given some info, then asked to find a specific term.
\[
U_1 \quad \xrightarrow{\times r} \quad U_2 \quad \xrightarrow{\times r} \quad U_3 \quad \dots \quad U_n
\]

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Here, we see that you must multiply by \( r \), \( (n-1) \) times. Therefore:
\[
U_n = U_1 \cdot r^{n-1} \quad \text{in F.B.}
\]

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E.G.3 // \( 0.3, 1.5, 7.5, \dots \quad \text{find } U_6: \quad U_6 = 0.3 \times 5^5 = 937.5 \)

SUM OF TERMS // Again, we can add together the terms:
\[
S_n = U_1 + U_1r + U_1r^2 + U_1r^3 + \dots + U_1r^{n-1}, \quad \text{which equals } U_1(1 + r + r^2 + \dots + r^{n-1})
\]

[q]

Due to the fact that \( 1 + r + r^2 + \dots + r^{n-1} = \frac{r^n – 1}{r – 1} \), we get the formulae:
\[
S_n = \frac{U_1(r^n – 1)}{r – 1} \quad \text{or} \quad S_n = \frac{U_1(1 – r^n)}{1 – r} \quad \text{in F.B.}
\]

[a]

E.G.4 // Find \( \sum_{n=1}^6 U_n \) from E.G.3:
\[
S_6 = \frac{0.3(5^6 – 1)}{5 – 1} = 1,171.8 \approx 1,170
\]

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E.G.5 // If the sum of the first 4 terms is \( \frac{26}{27} \) and \( r = \frac{1}{3} \), find \( U_6 \):
\[
S_4 = \frac{26}{27}, \quad U_1 \left( \frac{1 – \left( \frac{1}{3} \right)^4}{2/3} \right) = U_1 \times \frac{80}{27} = U_1 \times \frac{40}{27} = \frac{26}{27}, \quad \text{so } U_1 = \frac{13}{20}
\]

[a]

Thus:
\[
U_n = \frac{13}{20} \left( \frac{1}{3} \right)^{n-1}, \quad \text{so } U_5 = \frac{13}{20} \left( \frac{1}{3} \right)^4 = \frac{13}{20 \times 81} = \frac{13}{1620}
\]

When \( |r| < 1 \), we are able to compute the sum of infinite terms.

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APPLICATIONS // The main application is compound interest \( (1.4) \), but you may also see it applied to exponential population growth and decay, spread of disease, depreciation in the value of assets, and many more.

[a] Standard form:

A number in the format , \( \pm a \times10^k \) , where \(1\leqslant a\leqslant 10 \) and \(k\in Z\) (k is an integer)

(enter text or “Add Media”; select text to format)

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