1.4 Geometric Sequence (G.S.)
1. The Definition
If the first term of a sequence is $u_1 = 5$, and it is multiplied by a fixed number, say $r=2$, to find the next term, the following sequence is generated: $5, 10, 20, 40, 80, \dots$. Such a sequence is called geometric. In a geometric sequence, the ratio between any two consecutive terms is constant.
Only two elements are needed: The first term $u_1$, and the common ratio $r$.
EXAMPLE 1
- (a) If $u_1 = 1, r=2$, the sequence is $1, 2, 4, 8, 16, 32, 64, \dots$
- (b) If $u_1 = 5, r=10$, the sequence is $5, 50, 500, 5000, \dots$
- (c) If $u_1 = 1, r=-2$, the sequence is $1, -2, 4, -8, 16, \dots$
- (d) If $u_1 = 1, r=\dfrac{1}{2}$, the sequence is $1, \dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}, \dfrac{1}{16}, \dfrac{1}{32}, \dots$
- (e) If $u_1 = 1, r=-\dfrac{1}{2}$, the sequence is $1, -\dfrac{1}{2}, \dfrac{1}{4}, -\dfrac{1}{8}, \dfrac{1}{16}, -\dfrac{1}{32}, \dots$
NOTICE:
- The common ratio $r$ may also be negative. In this case the signs alternate $(+, -, +, -, \dots)$.
- The common ratio $r$ may be between -1 and 1, that is $|r| < 1$. In such a sequence the terms approach 0.
2. QUESTION A: What is the general formula for $u_n$?
If $u_1$ and $r$ are known, then the general formula is:
$$u_n = u_1 r^{n-1}$$
Derivation:
In order to find $u_5$, one starts from $u_1$ and then multiplies 4 times by the ratio $r$.
Hence, $u_5 = u_1 r^4$. Similarly, $u_{10} = u_1 r^9$, and $u_{100} = u_1 r^{99}$.
EXAMPLE 2
In a geometric sequence let $u_1 = 3$ and $r = 2$. Find:
- (a) the first four terms and
- (b) the 100th term.
Solution:
- (a) $3, 6, 12, 24$
- (b) The general formula is needed: $u_{100} = u_1 r^{99} = 3 \cdot 2^{99}$ (This answer is sufficient as it is very large).
EXAMPLE 3
In a geometric sequence let $u_1 = 10$ and $u_{10} = 196830$. Find $u_3$.
Solution:
Since $u_1$ is known, $r$ is needed. The information for $u_{10}$ is exploited first.
$$u_{10} = u_1 r^9 \implies 196830 = 10 \cdot r^9 \implies r^9 = 19683 \implies r = \sqrt[9]{19683} = 3$$
Therefore, $u_3 = u_1 r^2 = 10 \cdot 3^2 = 90$
⚠️ REMEMBER
The first task in a G.S. is to find the basic elements, $u_1$ and $r$, and then everything else!
EXAMPLE 4
A geometric sequence has a fifth term of 3 and a seventh term of 0.75. Find:
- (a) the first term $u_1$ and the common ratio $r$, and
- (b) the 10th term.
Solution:
(a) It is given that $u_5 = 3$ and $u_7 = 0.75$. The formula takes the form:
$$\begin{aligned} u_5 &= u_1 r^4 \\ u_7 &= u_1 r^6 \end{aligned}$$
Divide $u_7$ by $u_5$:
$$\dfrac{u_7}{u_5} = \dfrac{u_1 r^6}{u_1 r^4} \implies \dfrac{0.75}{3} = r^2 \implies r^2 = 0.25 \implies r = \pm 0.5$$
The first equation then gives:
$$3 = u_1 (0.0625) \implies u_1 = \frac{3}{0.0625} = 4.8$$
(b) The 10th term is $u_{10} = u_1 r^9$:
- If $r = 0.5$, then $u_{10} = 4.8(0.5)^9 = 0.09375$
- If $r = -0.5$, then $u_{10} = 4.8(-0.5)^9 = -0.09375$
3. QUESTION B: What is the sum $S_n$ of the first $n$ terms?
Given that $r \neq 1$, the result is given by:
$$S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}$$
Proof for $S_n$ (mainly for Math HL)
By definition:
$$S_n = u_1 + u_1 r + u_1 r^2 + \dots + u_1 r^{n-2} + u_1 r^{n-1} \quad \text{(1)}$$
Multiply both sides by $r$ gives:
$$r \cdot S_n = u_1 r + u_1 r^2 + u_1 r^3 + \dots + u_1 r^{n-1} + u_1 r^n \quad \text{(2)}$$
Subtracting $(2) - (1)$ gives:
$$r \cdot S_n - S_n = u_1 r^n - u_1 \implies (r - 1)S_n = u_1(r^n - 1) \implies S_n = \dfrac{u_1(r^n - 1)}{r - 1}$$
EXAMPLE 5
Consider the sum $2 + 2^2 + 2^3 + \dots + 2^{10}$.
Solution:
It is a geometric series of 10 terms with $u_1 = 2$ and $r = 2$. The sum is:
$$S_{10} = \dfrac{u_1(r^{10} - 1)}{r - 1} = \dfrac{2(2^{10} - 1)}{2 - 1} = 2046$$
EXAMPLE 6
Compute the sum:
$$1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dots + \dfrac{1}{2^{10}}$$
Solution:
It is a geometric series of 11 terms with $u_1 = 1$ and $r = 1/2$. The sum is given by:
$$S_{11} = \dfrac{u_1(1 - r^{11})}{1 - r} = \dfrac{1 \cdot (1 - (\dfrac{1}{2})^{11})}{1 - \dfrac{1}{2}} = \dfrac{1 - \dfrac{1}{2048}}{\dfrac{1}{2}} = \dfrac{2047}{1024}$$
EXAMPLE 7
In a G.S. $u_2 = -30$ and $S_2 = -15$. Find $u_1$ and $r$.
Solution:
$$S_2 = u_1 + u_2 \implies -15 = u_1 - 30 \implies u_1 = 15$$
Since $u_1 = 15$ and $u_2 = -30$,
$$r = \frac{u_2}{u_1} = \frac{-30}{15} = -2$$
4. Notice for Consecutive Terms
Let $a, x, b$ be consecutive terms of a geometric sequence. The common ratio is equal to:
$$\frac{x}{a} = \frac{b}{x}$$
For example, if $10, x, 90$ are consecutive terms in a G.S., then:
$$\frac{x}{10} = \frac{90}{x} \implies x^2 = 900 \implies x = \pm 30$$
5. The Sum of Terms in an Infinite G.S.
Consider the sum of the infinite geometric sequence $S_\infty = u_1 + u_2 + u_3 + \dots$ (it never stops!).
The result exists only if $-1 < r < 1$. It is given by the formula:
$$S_\infty = \dfrac{u_1}{1 - r}$$
In this case, the series is said to converge. Otherwise (if $|r| \ge 1$), the series diverges.
Three Proofs of the Formula (mainly for Math HL)
(a) Consider the formula for:
$$S_n = \frac{u_1(r^n - 1)}{r - 1}$$
If $n \to \infty$, then $r^n \to 0$ (since $-1 < r < 1$). Thus:
$$S_n \to \dfrac{u_1(0 - 1)}{r - 1} = \dfrac{u_1}{1 - r}$$
(b) An alternative proof is similar to that for $S_n$:
$$\begin{aligned} S_\infty &= u_1 + u_1 r + u_1 r^2 + u_1 r^3 + \dots \quad \text{(1)} \\ r \cdot S_\infty &= \quad \ + u_1 r + u_1 r^2 + u_1 r^3 + \dots \quad \text{(2)} \end{aligned}$$
Assuming that $S_\infty$ exists, subtracting (1) - (2) gives:
$$S_\infty - r \cdot S_\infty = u_1 \implies (1 - r)S_\infty = u_1 \implies S_\infty = \frac{u_1}{1 - r}$$
(c) A slight modification of proof (b):
$$\begin{aligned} S_\infty &= u_1 + u_1 r + u_1 r^2 + u_1 r^3 + \dots \\ &= u_1 + r(u_1 + u_1 r + u_1 r^2 + u_1 r^3 + \dots) \\ &= u_1 + r S_\infty \end{aligned}$$
Hence, (assuming again that $S_\infty$ exists):
$$S_\infty - r S_\infty = u_1 \implies S_\infty = \frac{u_1}{1 - r}$$
EXAMPLE 8
Show that $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots = 1$.
Solution:
This is an infinite G.S. with $u_1 = \frac{1}{2}$ and $r = \dfrac{1}{2}$.
Since $|r| < 1$, the sum is obtained by:
$$S_\infty = \dfrac{u_1}{1 - r} = \dfrac{\frac{1}{2}}{1 - \dfrac{1}{2}} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$$
EXAMPLE 9
Show that $0.3333\dots = \dfrac{1}{3}$.
Solution:
It can be written as:
$$0.3333\dots = 0.3 + 0.03 + 0.003 + \dots$$
This is in fact an infinite G.S. with $u_1 = 0.3$ and $r = 0.1$. Hence:
$$0.3333\dots = \dfrac{u_1}{1 - r} = \dfrac{0.3}{1 - 0.1} = \dfrac{0.3}{0.9} = \dfrac{1}{3}$$
EXAMPLE 10
Show that $0.9999\dots = 1$.
Solution:
Indeed, using the geometric series formula:
$$0.9999\dots = 0.9 + 0.09 + 0.009 + \dots = \dfrac{u_1}{1 - r} = \dfrac{0.9}{1 - 0.1} = \dfrac{0.9}{0.9} = 1$$
Alternative Proof 1:
It is known that $0.3333\dots = \dfrac{1}{3}$. Multiplying both sides by 3 yields $0.9999\dots = \dfrac{3}{3} = 1$.
Alternative Proof 2:
Note that:
$$\begin{aligned} x &= 0.9999\dots \quad \text{(1)} \\ 10x &= 9.9999\dots \quad \text{(2)} \end{aligned}$$
Subtracting (2) - (1) yields:
$$10x - x = 9 \implies 9x = 9 \implies x = 1 \quad \text{(3)}$$
Thus, (1) and (3) give $0.9999\dots = 1$.