Geometric Sequences Revision Notes for AQA A-Level Mathematics

4.4.1 Geometric Sequences

Geometric Sequences

A geometric sequence is one that has a common ratio or multiplier between terms.

e.g. $2, 4, 8, 16, 32, \ldots$

(a = 2, \quad r = 2) \quad \text{where } r \text{ is the common ratio between terms}

e.g.\ 100, 50, 25, 12.5, \ldots

(a = 100, \quad r = \frac{1}{2})

Note

We can obtain the common ratio between terms by dividing a term by its previous term.

$n^{th}$ Term of a Geometric Sequence

If ' $a$ ' denotes the first term and ' $r$ ' the common ratio between terms, then:

Note

\begin{align*} u_1 &= a \\ u_2 &= ar \\ u_3 &= ar^2 \\ &\vdots \\ u_n &= ar^{n-1} \end{align*}

For a geometric sequence:

Note

u_n = ar^{n-1} \quad \text{(Must remember this)}

Sum of n Terms of a Geometric Sequence

Note

S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{or} \quad \frac{a(r^n - 1)}{r - 1} \quad \text{(Given in formulae)}

Proof

Note

\begin{align*} S_n &= a + ar + ar^2 + \ldots + ar^{n-1} \\ rS_n &= ar + ar^2 + \ldots + ar^{n-1}+ar^n \\ S_n - rS_n &= a + 0 + 0 + \ldots + 0 - ar^n \\ \Rightarrow S_n - rS_n &= a - ar^n \\ \Rightarrow S_n (1 - r) &= a (1 - r^n) \\ \Rightarrow S_n &= \frac{a (1 - r^n)}{1 - r} \quad \text{as required} \end{align*}

Note

For each of the following geometric progressions, find an expression for the $n^{th}$ term. a) $1, 5, 25, 125, \ldots$

a = 1, \quad r = 5

\Rightarrow u_n = 1 \times 5^{n-1} = 5^{n-1}

b) $3, -12, 48, -192, \ldots$

a = 3, \quad r = -4

u_n = 3 \times (-4)^{n-1}

\sum_{r=1}^{6} 8^{r+1} = 64 + 512 + \ldots

a = 64, \quad r = 8

S_6 = \frac{64(1-8^6)}{1-8}

= 2396736 \quad \text{(Check)}

Note

The second and fifth terms of a geometric series are 0.5 and 32 respectively. a. Find the first term and common ratio of the series.

b. Find the number of terms of the series that are smaller than 10,000.

Note

Solution:

Given:

ar = 0.5 \quad \text{(A)}

ar^4 = 32 \quad \text{(B)}

To find the ratio, divide (B) by (A) because (B) has the higher power of $r$ :

\frac{\cancel ar^4}{\cancel ar} = \frac{32}{0.5}\quad B \div A

\Rightarrow r^3 = 64 \Rightarrow r = \sqrt[3]{64} = 4

Using $ar = 0.5$ :

a(4) = 0.5 \Rightarrow a = \frac{1}{8}

b) Finding the number of terms:

ar^{n-1} < 10,000

\Rightarrow \frac{1}{8}(4)^{n-1} < 10,000

4^{n-1} < 80,000 \quad \text{(Requires logarithms)}

\Rightarrow \log_4(4^{n-1}) < \log_4(80,000)

\Rightarrow n-1 < \log_4(80,000)

\Rightarrow n < \log_4(80,000) + 1

\Rightarrow n < 9.14

\Rightarrow n = 9 \quad \text{(n must be an integer)}

Geometric Sequences (AQA A-Level Mathematics): Revision Notes

4.4.1 Geometric Sequences