Geometric Progressions

A geometric progression (GP), also called a geometric sequence, is a sequence of numbers which differ from each other by a common ratio. For example, the sequence $2, 4, 8, 16, \dots$ is a geometric sequence with common ratio $2$.

We can find the common ratio of a GP by finding the ratio between any two adjacent terms.

The following sequence is a geometric progression with initial term $10$ and common ratio $3$:

$$\LARGE \color{#3D99F6}{10} \underbrace{\quad \quad }_{\times 3} \color{#D61F06}{30} \underbrace{\quad \quad }_{\times 3} \color{#20A900}{90} \underbrace{\quad \quad }_{\times 3} \color{cyan}{270} \underbrace{\quad \quad }_{\times 3} \color{orangered}{810} \underbrace{\quad \quad }_{\times 3} \color{grey}{2430}$$

Important terminology

• Initial term: In a geometric progression, the first number is called the "initial term."
• Common ratio: The ratio between a term in the sequence and the term before it is called the "common ratio."

Recursive Formula

We can describe a geometric sequence with a recursive formula, which specifies how each term relates to the one before. Since in a geometric progression, each term is given by the product of the previous term and the common ratio, we can write a recursive description as follows:

$$\text{Term} = \text{Previous term} \times \text{Common ratio}.$$

More concisely, with the common ratio $r$, we have

$$a_n=a_{n-1} \times r.$$

Explicit Formula

While the recursive formula above allows us to describe the relationship between terms of the sequence, it is often helpful to be able to write an explicit description of the terms in the sequence, which would allow us to find any term.

If we know the initial term, the following terms are related to it by repeated multiplication of the common ratio. Thus, the explicit formula is

$$\text{Term} = \text{Initial term} \times \underbrace{\text{Common ratio} \times \dots \times \text{Common ratio}}_{\text{Number of steps from the initial term}}.$$

Using exponents, we can write this with common ratio $r$, as

$$a_n = a_1 \times r^{n-1}.$$

Note: It is sometimes easier to compute values in a geometric progression based on a term in the middle rather than the initial term. When we begin our calculations from the $k^{\text{th}}$ term, the $n^{\text{th}}$ term in the geometric progression is given by

$$a_n = a_k \times r ^ {n-k}.$$

Now let's work out some basic examples that can familiarize you with the above definitions.

What is the explicit formula for the geometric sequence $4, 12, 36, 108, \dots?$

Reveal Answer

The initial term is $4$. Since each subsequent term is the product of the previous term and $3$, the common ratio is $3$. Thus the formula describing this sequence is

$$a_n = 4 \times 3^{n-1}.\ _\square$$

If the fourth term of a geometric progression with common ratio equal to half the initial term is $32,$ what is the $15^{\text{th}}$ term?

Reveal Answer

Let the initial term be $a,$ and the common ratio $r$. As stated in the problem, the $4^{\text{th}}$ term is $ar^{3} = 32$ and the initial term is $a=2r$. Solving these two simultaneous equations gives

$$2r^{4}=32 \implies r=2 \implies a=4.$$

Hence, the $15^{\text{th}}$ term is

$$a_{15}=a \times r^{14}=4 \times 2^{14}=2^{16} . \ _\square$$

We sometimes want to find the sum of some terms of a geometric progression. When the number of terms we want to add is large, it can be difficult to add them all one at a time. The problem below illustrates a method that can be developed into a general technique:

Find the sum of the first $10$ terms of the following geometric progression:

$$3,\ 15,\ 75,\ 375,\ 1875,\, \ldots.$$

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Let the sum of the first $10$ terms of the given series be $A,$ then

$$A=3+3 \cdot 5 +3 \cdot 5^2+ \cdots +3 \cdot 5^9. \qquad (1)$$

Multiplying $A$ by $5,$ we get

$$5A= 3 \cdot 5 +3 \cdot 5^2+3 \cdot 5^3+\cdots+3 \cdot 5^{10}. \qquad (2)$$

Taking $(1)-(2)$ gives

$$\begin{array} { rllll} A&= 3+3 \cdot 5+3 \cdot 5^2&+\cdots+3 \cdot 5^{9} \\ 5A&= 0 +3\cdot 5+3 \cdot 5^2&+\cdots+3 \cdot 5^{9}&+3 \cdot 5^{10} \\ \hline A(1-5)& =3+0~\quad +0&+\cdots+0 &-3 \cdot 5^{10} \\ -4A&=3-3 \cdot 5^{10}\\\\ A&=\dfrac{3 \cdot 5^{10}-3}{4}. \ _\square \end{array}$$

In the example above, we multiplied the sum of the geometric progression by its common ratio and then subtracted the result from the original sum, finding that all the terms cancel out except the first and last ones. Now we can use the same approach to find the general formula for the sum.

For a geometric progression with initial term $a$ and common ratio $r,$ the sum of the first $n$ terms is

$$S_n = \begin{cases}\begin{array}{ll} a \cdot \left( \frac{ r^n -1 } { r - 1 } \right) && \text{for }r \neq 1 \\ a \cdot n && \text{for }r = 1.\end{array} \end{cases}$$

Suppose we wanted to add the first $n$ terms of a geometric progression. If $r = 1$, then we have a constant sequence, and hence the sum is just $n a$. Now, let's suppose that $r \neq 1,$ then we would obtain

$$S_n = a + a \cdot r + a \cdot r^2 + \cdots + a \cdot r^{n-2} + a \cdot r ^ {n-1}. \qquad (1)$$

Multiplying both sides by $r$ gives

$$r S_n = a \cdot r + a \cdot r^2 + \cdots + a \cdot r^{n-1} + a \cdot r ^ {n}. \qquad (2)$$

Taking $(1)-(2),$ we get

$$\begin{array} { rlllllllll} S_n&= a + a \cdot r& + a \cdot r^2& + \cdots + a \cdot r^{n-2}& + a \cdot r ^ {n-1} \\ r S_n& =0+a \cdot r & + a \cdot r^2 & + \cdots + a \cdot r^{n-2}& + a \cdot r ^ {n-1}& + a \cdot r^n \\ \hline S_n(1-r)& =a+0&+0&+\cdots+0& +0& - a \cdot r^{n} \\ (1-r) S_n &= a-a r^n. \end{array}$$

Therefore,

$$S_n = a \times \left( \frac{ r^n - 1 } { r - 1 } \right) ~\text{for } r \neq 1 . \ _\square$$

What is the sum of the first $10$ terms of a geometric progression with initial term $2$ and common ratio $3?$

Reveal Answer

Applying the above formula for the sum of geometric progression terms, we have

$$2 \times \frac{ 3^{10 } - 1 } { 3 - 1 } = 3^{10} - 1 = 59048. \ _\square$$

Now that we know how to find the sum of finitely many terms, let's move on to find the sum of infinitely many terms of a geometric progression. This is done in a similar way, and we do an example first.

Calculate the following geometric series:

$$5+ \dfrac 53 +\dfrac 59 +\dfrac{5}{27}+\cdots.$$

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Let the given sum be $S,$ then

$$S=5+ \dfrac 53 +\dfrac 59 +\dfrac{5}{27}+\cdots. \qquad (1)$$

Multiplying $S$ by $\frac 13$, we get

$$\dfrac 13 S= \dfrac 53 +\dfrac 59 +\dfrac{5}{27}+\dfrac{5}{81}+\cdots. \qquad (2)$$

Taking $(1)-(2)$ gives

$$\begin{array} {rlllllllll} S&=5+ \dfrac 53& +\dfrac 59& +\dfrac{5}{27}&+\cdots \\ \dfrac 13 S&=0+ \dfrac 53& +\dfrac 59& +\dfrac{5}{27}&+\dfrac{5}{81}&+\cdots \\ \hline S\left(1-\dfrac 13 \right)& =5+0&+0&+0&+0&+ \cdots \\ S \cdot \dfrac 23&=5\\ S&={\dfrac{15}{2}}. \ _\square \end{array}$$

Note that we're using the same trick of multiplying by the common ratio and subtracting! In fact, this trick can be used to find a general formula for the sum of the infinite terms of a geometric progression. Here we go:

For a geometric progression with initial term $a$ and common ratio $r$ satisfying $|r| < 1 ,$ the sum of the infinite terms of the geometric progression is

$$S_{\infty} = \frac{ a } { 1-r }.$$

When $-1 < r < 1$, as $n$ becomes arbitrarily large, $r^n$ tends to zero. Hence, taking the limit of the sequence, we get

$$S_\infty = \lim_{n \rightarrow \infty } S_n = \lim_{n \rightarrow \infty} \frac{ a ( 1 - r^n ) } { 1-r } = \frac{ a} { 1-r }. \ _\square$$

Geometrical Proof:

We can also think of this formula visually. If $S$ is the sum of the series and the initial term is $a$, we can construct a square and a triangle as follows:

We can see that the large triangle and the inverted triangle on the left side of the square are similar. Therefore by similarity,

$$\frac{S}{a} = \frac{a}{a-ar}.$$

Solving for $S$, we get

$$S = \frac{a}{1-r}. \ _\square$$

After striking the floor, your tennis ball bounces to two-thirds of the height from which it has fallen. What is the total vertical distance it travels before coming to rest when it is dropped from a height of $100 \text{ m}?$

Reveal Answer

Let $h$ be the height (in meters) from which the ball is dropped, and $e$ a number such that $0<e<1.$ Also, let $S$ be the total vertical distance covered before the ball comes to rest. Then

$$\begin{aligned} S&=h+2(eh)+2\big(e^2h\big)+2\big(e^3h\big)+2\big(e^4h\big)+\cdots \\ &=h +2eh\big(1+e+e^2+e^3+\cdots\big) \\ &=h+2eh \times \dfrac{1}{1-e} \qquad \qquad \qquad \qquad (\text{since } e<1) \\ &=\left( \dfrac{1+e}{1-e} \right) h. \end{aligned}$$

Since we are given $h=100$ and $e=\frac23,$

$$S=\left( \dfrac{1+\frac 23}{1-\frac 23} \right) 100=500 \text{ (m)}. \ _\square$$

Find the sum of the geometric series

$$5 - \frac{10}{3} + \frac{20}{9} - \frac{40}{27} +\cdots .$$

Reveal Answer

Observe that the given series is a geometric progression with initial term $a= 5$ and common ratio $r=\frac{-2}{3}.$ Then since $-1<r<0,$ the series is convergent and we will use the formula for infinite sum $S = \frac{a}{1-r}$ to evaluate the value of the given series. Hence

$$S=\dfrac{5}{1-\left( \frac{-2}{3} \right) } = 3. \ _\square$$

As we are now familiar with the above concepts, let's try our hand at solving some problems below:

• Arithmetic Progressions
• Arithmetic and Geometric Progressions Problem Solving