# Geometric Progression – Meaning, Formulas & Examples

A sequence is a collection of objects (or events) arranged logically in mathematics. A sum of a sequence of terms is referred to as a series. In other words, a series is a collection of numbers connected by addition operations. Many types of sequences (also called progressions) and series are studied in mathematics such as arithmetic progression, geometric progression, harmonic progression, etc.

Let’s understand what is geometric progression and what the different formulas used to solve problems are based on it.

## What is a Geometric Progression?

A geometric progression(abbreviated as GP) is a progression where the successive terms bear a constant ratio known as a common ratio.

It is generally represented in form $a$, $ar$, $ar^2$, … where $a$ is the first term and $r$ is the common ratio of the progression. The common ratio can have both negative as well as positive values.

Note: To get the common ratio $r$ of a geometric progression we divide any succeeding term by its preceding term, i.e., $r = \frac{a_n}{a_{n-1}}$.

For example, $1, 2, 4, 8, 16, 32$, … is a geometric progression as the ratio between every two consecutive terms is the same (i.e., $2$). $\frac{2}{1} = \frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = … = 2$. We can also notice that every term (except the first term) of this GP is obtained by multiplying $2$ by its previous term.

Thus, a geometric progression, in general, can be written as $a$, $ar$, $ar^2$, $ar^3$, …, where $a$ is a first term and $r$ is a common ratio.

### Examples of Geometric Progression

Example 1: The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.

Let’s consider that the initial amount of air in a cylinder = $100$ cubic units

Amount of air removed by vacuum pump after first time = $\frac{1}{4} \times 100 = 25$. Amount of air left in the cylinder = $100 – 25 = 75$ cubic units.

Amount of air removed by vacuum pump after second time = $\frac{1}{4} \times 75 = 18.75$. Amount of air left in the cylinder = $75 – 18.75 = 56.25$ cubic units.

Amount of air removed by vacuum pump after third time = $\frac{1}{4} \times 56.25 = 14.0625$. Amount of air left in the cylinder = $56.25 – 14.0625 = 42.1875$ cubic units.

The amount of air in the cylinder(in cubic units) after each use of the vacuum pump is $100$, $75$, $56.25$, $42.1875$, … which is not an AP.

Ratio between $2^{nd}$ and $1^{st}$ terms is $\frac{75}{100} = \frac{3}{4}$

Ratio between $3^{rd}$ and $2^{nd}$ terms is $\frac{56.25}{75} = \frac{3}{4}$

Ratio between $4^{th}$ and $3^{rd}$ terms is $\frac{42.1875}{56.25} = \frac{3}{4}$

The ratio between consecutive terms of the sequence is the same.

Example 2: The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally, then the number of bacteria in the culture after every hour.

Number of bacteria in $1^{st}$ hour = $30$.

Number of bacteria in $2^{nd}$ hour = $60$.

Number of bacteria in $3^{rd}$ hour = $120$.

Number of bacteria in $4^{th}$ hour = $240$.

Ratio between $2^{nd}$ and $1^{st}$ terms is $\frac{60}{30} = 2$

Ratio between $3^{rd}$ and $2^{nd}$ terms is $\frac{120}{60} = 2$

Ratio between $4^{th}$ and $3^{rd}$ terms is $\frac{240}{120} = 2$

The ratio between consecutive terms of the sequence is the same.

Example 3: What will ₹$500$ amount to in $10$ years after its deposit in a bank which pays an annual interest rate of $10\%$ compounded annually?

Amount of money in the beginning of $1^{st}$ year = ₹$500$

Amount of money in the beginning of $2^{st}$ year = $500\left( 1 + \frac{10}{100}\right) =$ ₹$550$

Amount of money in the beginning of $3^{rd}$ year = $550\left( 1 + \frac{10}{100}\right) =$ ₹$605$

Amount of money in the beginning of $4^{th}$ year = $605\left( 1 + \frac{10}{100}\right) =$ ₹$665.50$

Ratio between $2^{nd}$ and $1^{st}$ terms is $\frac{550}{500} = \frac{11}{10}$

Ratio between $3^{rd}$ and $2^{nd}$ terms is $\frac{605}{550} = \frac{11}{10}$

Ratio between $4^{th}$ and $3^{rd}$ terms is $\frac{665.50}{605} = \frac{11}{10}$

The ratio between consecutive terms of the sequence is the same.

Note:

• A GP is an increasing GP when the common ratio $r \gt 1$
• A GP is a decreasing GP when the common ratio $r \lt 1$
• A GP is a constant GP when the common ratio $r = 1$

## nth Term of Geometric Progression

The general term (or) $n^{th}$ term of a GP whose first term is $a$ and the common ratio is $r$ is given by the formula $a_n = ar^{n-1}$.

For example, to find the general term (or $n^{th}$) term of the progression $3, 9, 27, 81, 243,$ … we substitute the first term, $a = 3$, and the common ratio, $r = 3$ in the formula for the $n^{th}$ term formula.

Using the formula we get, $a_n = ar^{n – 1} = 3 \times 3^{n – 1} = 3^n$. Thus, the general term (or) $n^{th}$ term of this GP is $a_n = 3^n$.

### Examples on General Term of Geometric Progression

Example 1: Find the general term of a GP whose first term is $\frac{5}{2}$ and the common ratio is $\frac{1}{2}$.

The first term $a = \frac{5}{2}$ and the common ratio $r = \frac{1}{2}$.

Using the formula we get, $a_n = ar^{n – 1} = \frac{5}{2} \times \left(\frac{1}{2} \right)^{n – 1} = \frac{5^{n – 1}}{2^n}$

Example 2: Find the general term of a GP whose first term is $x^3$ and the common ratio is $x^2$.

The first term $a = x^3$ and the common ratio $r = x^2$.

Using the formula we get, $a_n = ar^{n – 1} = x^3 \times \left(x^2 \right)^{n – 1} = x^3 \times x^{2n – 2} = x^{3 + 2n – 2} = x^{2n + 1}$.

Example 3: Find the general term of a GP whose first term is $1$ and the common ratio is $-a$.

The first term $a = 1$ and the common ratio $r = -a$.

Using the formula we get, $a_n = ar^{n – 1} = 1 \times (-a)^{n – 1} = (-1)^{n – 1}a^{n – 1}$.

## Derivation of the nth Term of Geometric Progression

Let $a_1$, $a_2$, $a_3$, . . . be an GP whose first term $a_1$ is $a$ and the common difference is $r$.

Then, the second term $a_2 = ar = ar^{2 – 1}$

The third term $a_3 = a_2 \times r = ar \times r = ar^2 = ar^{3 – 1}$

The fourth term $a_4 = a_3 \times r = ar^2 \times r = ar^3 = ar^{4 – 1}$

Looking at the pattern, we can say that the $n^{th}$ term $a_n = ar^{n – 1}$.

So, the $n^{th}$ term an of the GP with first term $a$ and common ratio $r$ is given by $a_n = ar^{n – 1}$.

### Examples on the nth Term of Geometric Progression

Example 1: Find the $11^{th}$ term of a GP $1, 3, 9, 27, …$.

The first term $a = 1$

The common ratio $r = \frac{3}{1} = 3$

$n = 11$

Formula for $n^{th}$ term an of the GP is $a_n = ar^{n – 1}$.

Substituting the values we get $a_n = ar^{n – 1} = 1 \times 3^{11 – 1} = 3^10 = 59049$

Therefore, the $11^{th}$ term of a GP $1, 3, 9, 27, …$ is $59049$.

Example 2: Find the $20^{th}$ and $n^{th}$ terms of the G.P. $\frac{5}{2}$, $\frac{5}{4}$, $\frac{5}{8}$, ….

The first term $a = \frac{5}{2}$

The common ratio $r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{1}{2}$

$n = 20$

Formula for $n^{th}$ term an of the GP is $a_n = ar^{n – 1}$.

Substituting the values we get $a_n = ar^{n – 1} = \frac{5}{2} \times \left(\frac{1}{2} \right)^{20 – 1} = \frac{5}{2} \times \left( \frac{1}{2}\right)^{19} = \frac{5}{2^{20}} = \frac{5}{1048576}$

And the $n^{th}$ term is $\frac{5}{2} \times \left(\frac{1}{2} \right)^{n – 1} = \frac{5}{2^n}$.

Example 3: The $5^{th}$, $8^{th}$, and $11^{th}$ terms of a G.P. are $p$, $q$ and $s$, respectively. Show that $q^2 = ps$.

Let the first term and the common ratio of the GP be $a$ and $r$ respectively.

Therefore, $a \times r^{5 – 1} = p => ar^4 = p$

$a \times r^{8 – 1} = p => ar^7 = q$

And $a \times r^{11 – 1} = p => ar^{10} = s$

Therefore, $ps = ar^4 \times ar{10} = a^2r^{14} = \left(ar^7 \right)^2 = q^2$