IB Mathematics AA SL Geometric sequences and series Study Notes

IB Mathematics AA SL Geometric sequences and series Study Notes Study Notes Offer a clear explanation of Geometric sequences and 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 Geometric sequences and series

GEOMETRIC SERIES

A geometric sequence is a sequence such that the next term is generated by multiplying or being divided by the same number from the previous term.

Suppose you open a savings account by investing $1000 with an interest rate of 0.5% per annum.

After 1 year:
Interest earned: $5
Total amount: $1005
After 2 years:
Interest earned: $5.25
Total amount: $1010.25

A geometric series involves a constant ratio (r). In this example, the ratio is:

$ r = 1.005 $

Geometric Series nth Term

The nth term of a geometric series can be expressed as:

$ A_n = A \cdot r^{(n-1)} $

Using the example above, the value after 10 years is calculated as follows:

$ A_{10} = 1000 \cdot (1.005)^9 = \$1045.91 $

Now, if the interest rate were 3.5%, the calculation would be:

$ A_{10} = 1000 \cdot (1.035)^9 = \$1362.90 $

Example Sequence

Consider the sequence: 32, 16, 8, 4, 2, …

$ U_n = 8 $
$ a = 32 $

Sum Formula

The sum of a geometric series can be calculated using the formula:

$ S_n = \frac{a(1 – r^n)}{1 – r} $

For the case when $ |r| > 1 $:

$ S_n = \frac{a(r – 1)}{r – 1} $

Properties of a Geometric Sequence

The terms of a geometric sequence can be represented as $ (u_1, u_2, u_3, \ldots) $:

1. First term: $ u_1 $

2. Common ratio: $ r = \frac{u_n}{u_{n-1}} = \frac{u_2}{u_1} $

3. General term (nth term): $ u_n = u_1 \cdot r^{n-1} $

4. Sum of the first $ n $ terms: $ S_n = \frac{u_1(1 – r^n)}{1 – r} \quad (r \neq 1) $

5. Sum of an infinite number of terms (Sum to infinity): $ S_{\infty} = \frac{u_1}{1 – r} \quad \text{valid only when} \quad |r| < 1 $

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

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

Filicia baked a very large apple pie that she cuts into slices to share with her friends. The smallest

slice is cut first. The volume of each successive slice of pie forms a geometric sequence.

The second smallest slice has a volume of 30 cm³. The fifth smallest slice has a volume

of 240 cm³.

1. Find the common ratio of the sequence. [2]
2. Find the volume of the smallest slice of pie. [2] The apple pie has a volume of 61 425 cm³.
3. Find the total number of slices Filicia can cut from this pie. [2]

▶️Answer/Explanation

Ans:

(a)

$u_{1}r= 30 \:and\: u_{1}r^{4}$ = 240,

OR $30r^{3}= 240 (r^{3}$=8 )

r=2

(b)

$u_{1}\times 2= 30 \: OR \: u_{1} \times 2^{4}$= 240

$u_{1}$= 15

(c)

$\frac{15 (2^{n}-1)}{2-1}$= 61425

n= 12 (slices)

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IB Mathematics AA SL Geometric sequences and series Study Notes