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# Geometric Progression – Meaning, Formulas & Examples

March 15, 2023

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.

Letâ€™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 $a$, $ar$, $ar^2$, â€¦ 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., $2$). $\frac{2}{1} = \frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = â€¦ = 2$. 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 $a$, $ar$, $ar^2$, $ar^3$, â€¦, 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.

Letâ€™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 $100$, $75$, $56.25$, $42.1875$, â€¦ 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 â‚¹$500$ 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 = â‚¹$500$

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

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

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

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 $3, 9, 27, 81, 243,$ â€¦ 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 $a_n = ar^{n â€“ 1}$.

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

### Examples on the nth Term of Geometric Progression

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

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 $a_n = ar^{n â€“ 1}$.

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

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

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

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 $a_n = ar^{n â€“ 1}$.

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

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