Geometric Progression

Consider the following sequences:

1, 3, 9, 27, 81, 243, \ldots; \quad 2, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots; \quad 5;\; {-}0.5;\; 0.05;\; {-}0.005;\; 0.0005;\; \ldots

They share a characteristic feature: each subsequent term is obtained by multiplying the previous term by the same nonzero number. For the first sequence this number is 3, for the second it is $\dfrac{1}{2}$ , and for the third it is $-0.1$ . Such sequences are called geometric progressions.

Definition

📐Definition — Geometric Progression

A geometric progression is a sequence whose first term is nonzero and each term, starting from the second, equals the previous term multiplied by the same nonzero number.

This number equals the ratio of any consecutive terms of the sequence. It is called the common ratio of the geometric progression and is denoted by $q$ (from the French word quotient — ratio).

If $(b_n)$ is a geometric progression with common ratio $q$ , then

q = \frac{b_2}{b_1} = \frac{b_3}{b_2} = \frac{b_4}{b_3} = \ldots,

that is, for any natural number $n$ we have $\dfrac{b_{n+1}}{b_n} = q$ . This gives the recurrent formula $b_{n+1} = b_n q$ .

ℹNote — Defining a Geometric Progression Recurrently

A geometric progression can be defined recurrently:

b_1 = b, \quad b_{n+1} = b_n q.

Thus, to define a geometric progression, one needs to specify its first term and common ratio.

Some examples:

If $b_1 = 1$ and $q = 3$ : $1, 3, 9, 27, 81, 243, \ldots$
If $b_1 = 2$ and $q = 2$ : $2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, \ldots$
A stationary sequence with all nonzero terms is both an arithmetic and a geometric progression (with common ratio $q = 1$ ).

Formula for the $n$ -th Term

From the definition of a geometric progression:

b_2 = b_1 \cdot q, \quad b_3 = b_1 q^2, \quad b_4 = b_1 q^3, \quad b_5 = b_1 q^4.

These examples lead to the following conclusion:

⚡Theorem — Formula for the n-th Term of a Geometric Progression

b_n = b_1 \cdot q^{n-1}

This can be proved by mathematical induction.

Comparing Growth Rates

Consider two sequences:

Arithmetic progression $(a_n)$ with $a_1 = 1$ , $d = 2$ : $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, \ldots$
Geometric progression $(b_n)$ with $b_1 = 1$ , $q = 2$ : $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, \ldots$

Comparing corresponding terms, we see that the geometric progression grows much faster than the arithmetic one. For example, $a_{20} = 1 + 2 \cdot 19 = 39$ , while $b_{20} = 1 \cdot 2^{19} = 524\,288$ .

Property of Terms of a Geometric Progression

⚡Theorem — Characteristic Property

The square of any term of a geometric progression, except the first (and last, if the sequence is finite), equals the product of its two neighboring terms. That is, for $n > 1$ :

b_n^2 = b_{n-1} \cdot b_{n+1}

If all terms of the geometric progression $(b_n)$ are positive, then:

b_n = \sqrt{b_{n-1} \cdot b_{n+1}}.

⚡Theorem — Theorem 33.1

A sequence $(b_n)$ with more than two terms, all nonzero, is a geometric progression if and only if for every $n \geqslant 2$ the equality $b_n^2 = b_{n-1} b_{n+1}$ holds.

Examples

✎Example — Finding the Fourth Term and Common Ratio

Problem. Find the fourth term and common ratio of the geometric progression $(b_n)$ if $b_3 = 36$ and $b_5 = 49$ .

Solution. By the characteristic property, $b_4^2 = b_3 b_5$ , so $b_4 = \sqrt{36 \cdot 49} = 6 \cdot 7 = 42$ or $b_4 = -\sqrt{b_3 b_5} = -42$ .

If $b_4 = 42$ , then $q = \dfrac{b_4}{b_3} = \dfrac{42}{36} = \dfrac{7}{6}$ ; if $b_4 = -42$ , then $q = -\dfrac{7}{6}$ .

Answer: $b_4 = 42$ , $q = \dfrac{7}{6}$ or $b_4 = -42$ , $q = -\dfrac{7}{6}$ .

✎Example — Product of Terms of a Geometric Progression

Problem. In a geometric progression $(b_n)$ , it is known that $b_{10} = 2$ . Find the product of the first nineteen terms.

Solution. We have:

b_1 b_2 b_3 \cdot \ldots \cdot b_{19} = b_1^{19} \cdot q^{1+2+\ldots+18} = b_1^{19} \cdot q^{171} = (b_1 q^9)^{19} = (b_{10})^{19} = 2^{19}.

Answer: $2^{19}$ .

✎Example — Geometric and Arithmetic Progressions Simultaneously

Problem. Find all triples of numbers that form a geometric progression with the following properties: their sum is 63, and when 7, 18, and 2 are added to these numbers respectively, the result is an arithmetic progression.

Solution. Write the desired numbers as $a$ , $aq$ , $aq^2$ . Then $a + 7$ , $aq + 18$ , $aq^2 + 2$ form an arithmetic progression. This gives $2(aq + 18) = a + 7 + aq^2 + 2$ . Using the conditions:

\begin{cases} a + aq + aq^2 = 63, \\ a - 2aq + aq^2 = 27. \end{cases}

Dividing the equations: $\dfrac{1 + q + q^2}{1 - 2q + q^2} = \dfrac{7}{3}$ . Solving gives $q = 4$ or $q = \dfrac{1}{4}$ .

If $q = 4$ , then $a = 3$ ; if $q = \dfrac{1}{4}$ , then $a = 48$ .

Answer: 3, 12, 48 or 48, 12, 3.

Exercises

✏Exercise — Geometric Progression Problems

33.2. The sixth term of a geometric progression $(b_n)$ is 8 and the common ratio is $-4$ . Find the seventh term.

33.5. In a geometric progression $(y_n)$ , the first term is $y_1 = 64$ and the common ratio is $q = -\dfrac{1}{2}$ . Find: 1) $y_6$ ; 2) $y_{10}$ .

33.7. Find the common ratio and the fifth term of the geometric progression $\dfrac{1}{216}$ , $\dfrac{1}{36}$ , $\dfrac{1}{6}$ , $\ldots$

33.13. Find the common ratio of the geometric progression $(b_n)$ if:

$b_1 = \dfrac{1}{2}$ , $b_8 = 64$
$b_6 = 75$ , $b_8 = 27$