# Geometric Progressions

A **geometric progression** (GP) is a sequence of non-zero terms in which each succeeding term is obtained by multiplying its preceding term by a constant. Let's begin by illustrating this with a familiar example:

That's right: the height to which a ball rises in each successive bounce follows a geometric progression! Let's explore the concept of geometric progressions so that we can mathematically consider several such phenomena.

Geometric progressions are very tractable. We can solve for an arbitrary term, a finite or infinite sum, and apply them in various contexts, including some difficult problems.

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## Terminology and Examples

The example below shows a geometric progression with initial term 10 and each succeeding term bearing a ratio of \(3:1\) to its preceding term.

\[\LARGE \color{blue}{10} \underbrace{\quad \quad }_{\times 3} \color{red}{30} \underbrace{\quad \quad }_{\times 3} \color{green}{90} \underbrace{\quad \quad }_{\times 3} \color{cyan}{270} \underbrace{\quad \quad }_{\times 3} \color{orangered}{810} \underbrace{\quad \quad }_{\times 3} \color{grey}{2430} \]

**Initial term:** In a geometric progression, the first number is called the "initial term". In the above example, the initial term is \(10\).

**Common ratio:** The ratio of a term to its preceding term of a geometric progression is called the "common ratio". In the above sequence, the common ratio is \(3\).

**General term:** The general term (\(n^{\text{th}}\) term) of a geometric progression with initial term \( a \) and common ratio \(r\) is represented by \[a_n=a \times r^{n-1}.\]

Note:It is sometimes easier to compute values in a geometric progression based on terms in the middle rather than the initial term. In that case, the \(n^{\text{th}}\) term in the geometric progression is given by\[ a_n = a_k \times r ^ {n-k }. \ _\square\]

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

## 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?

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

\[2r^{4}=32 \Rightarrow r=2 \Rightarrow a=4.\]

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

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

## The product of three consecutive terms in a geometric progression is 216 and their sum is 21. Find these terms.

Let the three consecutive terms be \(\dfrac {a}{r}, a\) and \(ar\) to make it easier to solve further. In this way, when we take the product of the three terms, the common ratio gets cancelled: \(\frac{a}{r}\times a \times ar=a^{3}=216 \Rightarrow a=6.\)

Now, since the sum of the three terms is \(21,\) we have

\[\begin{align} \frac {a}{r}+a+ar&=21\\ \frac {6}{r}+6+6r&=21\\ 6+6r+6r^2&=21r\\ 6r^2-15r+6&=0\\ 2r^2-5r+2&=0\\ (r-2)(2r-1)&=0\\ r&=2 \text{ or } \frac{1}{2}. \end{align}\]

For both values of \(r\), the three terms are \(3, 6\) and \( 12\). \(_\square\)

The following problem is yet another application of geometric progression!

## Sum of a Geometric Progression with Finite Terms

We sometimes want to sum finitely many terms of a geometric progression. We could do that manually but there's actually a general formula for the sum. Let's start with an example.

Find the sum of the first \(10\) terms of the following geometric progression:

\[\begin{array} &3, &15, &75, &375, &1875, &\ldots . \end{array}\]

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

\[ S_n = \begin{cases} a \cdot \left( \frac{ r^n -1 } { r - 1 } \right) & \text{for }r \neq 1 \\ a \cdot n & \text{ for }r = 1. \ _\square\end{cases} \]

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

\[ S_n = a + a \cdot r + a \cdot r^2 + \ldots + a \cdot r^{n-2} + a \cdot r ^ {n-1}. \qquad (1)\]

Multiplying both sides by \(r\) gives

\[ r S_n = a \cdot r + a \cdot r^2 + \ldots + a \cdot r^{n-1} + a \cdot r ^ {n}. \qquad (2)\]

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 \(3?\)

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

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

Let \(S\) and \(P\) be the sum and product of the first \(n\) terms of a geometric progression, respectively. Also, let \(R\) be the sum of the reciprocals of the \(n\) terms. Then prove \(P^2=\left( \dfrac SR \right)^n\).

Let the initial term and common ratio of the geometric progression be \(a\) and \(r,\) respectively. Then

\[\begin{align} S&=a+ar+ar^2+\cdots+ar^{n-1}\\ &=\dfrac{a(r^n-1)}{r-1} &\qquad (1)\\ P&=a \cdot ar \cdot ar^2 \cdots ar^{n-1}\\ &=a^n \cdot r^{1+2+3+\cdots+(n-1)}\\ &=a^n \cdot r^{\frac{n(n-1)}{2}}. &\qquad (2) \end{align}\]

Now, the sum of the reciprocals of the \(n\) terms is

\[\begin{align} R &=\dfrac{1}{a} +\dfrac{1}{ar}+\dfrac{1}{ar^2}+\cdots +\dfrac{1}{ar^{n-1}} \\ &=\dfrac 1a \left[ 1+\left( \dfrac 1r \right)+\left( \dfrac 1r \right)^2+\cdots +\left( \dfrac 1r \right)^{n-1} \right] \\ &=\dfrac{1}{a} \cdot \dfrac{1-\left( \dfrac 1r \right)^n}{1-\left( \dfrac 1r \right)} \\ &=\dfrac 1a \cdot \dfrac{(r^n-1)}{(r-1)} \cdot \dfrac{r}{r^n} \\ &=\dfrac 1a \cdot \dfrac{(r^n-1)}{(r-1)} \cdot \dfrac{1}{r^{n-1}}. \qquad (3) \end{align}\]

Taking \((1)\div (3)\) gives

\[\begin{align} \dfrac SR &= \dfrac{a(r^n-1)}{(r-1)} \cdot \dfrac{a(r-1)r^{n-1}}{(r^n-1)}\\ &=a^2 \cdot r^{n-1}\\ \Rightarrow \left( \dfrac SR \right)^n &= \left( a^2 \cdot r^{n-1} \right)^n\\ &=\left[ a^n \cdot r^{\frac{n(n-1)}{2}} \right]^2\\ &=P^2. \qquad (\text{from }(2) \text{ above}) \end{align}\]

Hence, we proved that \(P^2=\left( \dfrac SR \right)^n. \ _\square\)

## Sum of a Geometric Progression with Infinite Terms

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:

\[5+ \dfrac 53 +\dfrac 59 +\dfrac{5}{27}+\cdots.\]

Let the given sum be \(S,\) then

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

Multiplying \(S\) by \(\dfrac 13\), 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

\[ S_{\infty} = \frac{ a } { 1-r } . \ _\square \]

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 :

\(\hspace{1cm}\)

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

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

Now we can return to bouncing balls:

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 \(100 \text{ m}?\)

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

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

Since we are given \(h=100\) and \(e=2/3,\)

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

Find the sum of the geometric series

\[5 - \frac{10}{3} + \frac{20}{9} - \frac{40}{27} +\ldots .\]

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 \(S = \dfrac{a}{1-r}\) to evaluate the value of the given series. Hence

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

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

If the first three terms of a geometric progression is given to be \( \sqrt2+1,1,\sqrt2-1 \). Find the sum to infinity of all of its terms.

If the answer is in the form of \( \dfrac{a+b\sqrt c}d \) for positive integers \(a,b,c\) and \(d\) with \(c\) square-free, find the minimum value of \(a+b+c+d\).

## Applications

In this section we will work out some examples and problems based on applications of GP:

**1. Recognizing a geometric progression**

Convert the recurring decimal \(2. \overline{357}\) into the fractional form.

Let the given recurring decimal be \(\lambda,\) then

\[\begin{align} \lambda&=2.\overline{357}\\ &=2.357357357 \ldots\\ &= 2+0.357+0.000357+0.000000357+\cdots \\ &=2 +\left(\dfrac{357}{10^3}+\dfrac{357}{10^6}+\dfrac{357}{10^9}+\cdots\right). \end{align}\]

Since the geometric series in the parenthesis is the sum of a geometric progression with initial term \(\dfrac{357}{10^3} \) and common ratio \(\dfrac{1}{10^3},\) we have

\[\begin{align} \lambda =2+\dfrac{\dfrac{357}{10^3}}{1-\dfrac{1}{10^3}} =2+ \dfrac{357}{999}

=\dfrac{2355}{999}. \ _\square \end{align}\]

You and your best friend are playing a game with a fair die. If your friend rolls the die first and whoever gets a six first wins, what are the chances that you win the game?

Let \(P\) be the probability that you win the game when your friend starts it.

When a die is rolled, the probability of getting a six is \(P=\dfrac 16\) and the probability of not getting a six is \(1-P=\dfrac 56\). Therefore, using Multiplication theorem in probability, we get

\[P= \dfrac 56 \cdot \dfrac 16 +\dfrac 56 \cdot \dfrac 56 \cdot\dfrac 56 \cdot \dfrac 16 +\dfrac 56 \cdot \dfrac 56 \cdot\dfrac 56 \cdot \dfrac 56 \cdot \dfrac 56 \cdot \dfrac 16 +\cdots,\]

which is the sum of an infinite geometric progression with initial term \(\dfrac{5}{36}\) and common ratio \(\dfrac{25}{36}\). Applying the formula for sum of infinite geometric progression, we get

\[P=\dfrac{\dfrac{5}{36}}{1-\dfrac{25}{36}}= \dfrac{5}{11}. \ _\square\]

Find the value of the following series up to 20 terms:

\[2+22+222+2222+\cdots.\]

Let the required sum be \(S\), then

\[\begin{align} S&=2+22+222+\cdots +\underbrace{222 \cdots 22}_{\text{ 20 times 2's}}\\ &=2 \left(1+11+111+ \cdots +\underbrace{111 \cdots 11}_{\text{ 20 times 1's}} \right)\\ &=\dfrac 29 \left( 9+99+999+\cdots+\underbrace{999 \cdots 99}_{\text{ 20 times 9's}} \right)\\ &=\dfrac 29 \left[ (10-1) +\left(10^2-1\right)+\left(10^3-1\right)+ \cdots +\left(10^{20}-1\right) \right]\\ &=\dfrac 29 \left[10+10^2+10^3+\cdots +10^{20} -20 \right] \\ &=\dfrac 29 \left[ \dfrac{10\left(10^{20}-1\right)}{10-1}-20 \right] \\ &=\dfrac 29 \left[ \dfrac{10\left(10^{20}-1\right)}{9}-20 \right]\\ &=\frac{2\left(10^{21}-190\right)}{81}. \ _\square \end{align}\]

Simplify the following series:

\[\left( x+\dfrac 1x \right)^2+\left(x^2+\dfrac{1}{x^2} \right)^2+\cdots+ \left(x^n+\dfrac{1}{x^n} \right)^2.\]

Let the given sum be \(T,\) then the given expression can be rewritten as

\[\begin{align} T&= \left( x^2+\dfrac{1}{x^2}+2 \right) + \left( x^4+\dfrac{1}{x^4}+2 \right) + \cdots + \left( x^{2n}+\dfrac{1}{x^{2n}}+2 \right) \\ &=\left( x^2+x^4+ \cdots +x^{2n} \right) +\left( \dfrac{1}{x^2}+\dfrac{1}{x^4}+\cdots+\dfrac{1}{x^{2n}} \right) + 2n\\ &=\dfrac{x^2(x^{2n}-1)}{x^2-1}+\dfrac{\dfrac{1}{x^2} \left[ 1- \left( \dfrac{1}{x^2} \right)^n \right] }{1-\dfrac{1}{x^2}}+2n\\ &=\dfrac{x^2(x^{2n}-1)}{x^2-1}+\dfrac{x^{2n}-1 }{x^{2n} (x^2-1)}+2n. \ _\square \end{align}\]

Here is a problem for you to work on your own.

In your class test, there were \(\xi \) questions. During the test, for each \(p\) satisfying \(1 \leq p \leq \xi \), \(2^{\xi -1}\) students give **unexpected** answers to at least \(p\) questions. If the total number of **unexpected** answers given is equal to \(2047,\) then what was the total number of questions in the test?

**2. Miscellaneous**

Here you will find miscellaneous examples as well as problems having geometric progression as the main theme.

Can \(10, 11\) and \(12\) be any three terms that appear in the same geometric progression?

Let \(10, 11\) and \(12\) be the \(p^{\text{th}},\) \(q^{\text{th}}\) and \(k^{\text{th}}\) terms of a geometric progression, respectively, with initial term \(a\) and common ratio \(r\). Then we have

\[\begin{align} 10&=ar^{p-1} &\qquad (1)\\ 11&=ar^{q-1} &\qquad (2)\\ 12&= ar^{k-1}. &\qquad (3) \end{align}\]

Dividing \((2)\) by \((1)\) and \((3)\) by \((2),\) respectively, we get

\[\begin{align} \dfrac{11}{10}&=r^{q-p} &\qquad (4)\\ \dfrac{12}{11}&=r^{k-q} .&\qquad (5) \end{align}\]

From \((4),\) we have \(\left( \dfrac{11}{10} \right)^{k-q}=\left(r^{q-p}\right)^{k-q}=r^{(q-p)(k-q)}. \qquad (6)\)

From \((5),\) we have \(\left( \dfrac{12}{11} \right)^{q-p}=\left(r^{k-q}\right)^{q-p}=r^{(k-q)(q-p)}. \qquad (7)\)

Equating \((6)\) and \((7)\) gives

\[\begin{align} \left( \dfrac{11}{10} \right)^{k-q}&=\left( \dfrac{12}{11} \right)^{q-p}\\ (11)^{k-q+q-p}&=(10)^{k-q} (12)^{q-p} \\ (11)^{k-p}&= 5^{k-q} \cdot 2^{k-q} \cdot 2^{2(q-p)} \cdot 3^{q-p} \\ (11)^{k-p}&= 5^{k-q} \cdot 2^{k+q-2p} \cdot 3^{q-p}, \end{align} \]

which is possible only when

\[\begin{array} &k-p=0, &k-q=0, &k+q-2p=0, &q-p=0. \end{array}\]

This implies \(k=p=q\), which is not possible as \(p,q,k\) are distinct.

Hence \(10, 11\) and \(12\) can never be the terms appearing in the same geometric progression. \(_\square\)

If you borrow Rs. \(12000\) from a merchant, then you will be charged an interest of \(10\)% compounded annually. Find the total interest charged to you during the \(5^{\text{th}}\) year.

Observe that the amount of interest charged after every year follows the geometric progression. Also, let \(P\) be the principal, \(r\) the rate of interest, and \(I\) the interest charged during the \(5^{\text{th}}\) year.

The main concept here is\[n^{\text{th}} \text{ term of a geometric progression }= \text{(sum of first } n \text{ terms) } - \text{ (sum of first }n-1 \text{ terms)} .\]

The amount payable in \(5\) years (in Rs.) is

\[P\left( 1+\dfrac{r}{100} \right)^5=12000 \left( 1+\dfrac{10}{100} \right)^5=12000 \times \left( \dfrac{11}{10} \right)^5. \]

The amount payable in \(4\) years (in Rs.) is

\[P\left( 1+\dfrac{r}{100} \right)^4=12000 \left( 1+\dfrac{10}{100} \right)^4=12000 \times \left( \dfrac{11}{10} \right)^4 .\]

Hence, the interest charged for the \(5^{\text{th}}\) year (in Rs.) is

\[I= 12000 \times \left( \dfrac{11}{10} \right)^5 -12000 \times \left( \dfrac{11}{10} \right)^4=1756.92 . \ _\square\]

Accomplish the objectives of this page by solving the following problems:

Cody has started running in a well organized manner. He runs \(100 \text{ m}\) east, then turns left and runs another \(10 \text{ m}\) north, turns left and runs \(1 \text{ m},\) again turns left and runs \(0.1 \text{ m},\) and on next turn \(0.01 \text{ m}\) and so on. Assuming that Cody can run in this pattern infinitely, then the displacement from his initial position can be written as \(\frac{a}{\sqrt{b}}\) with \(a\) and \(b\) as positive integers and \(b\) square-free.

What is the value of \( a \times b?\)

## See also

**Cite as:**Geometric Progressions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/geometric-progressions/