- Written By Keerthi Kulkarni
- Last Modified 25-01-2023
Geometric Progression: It is the sequence or series of numbers such that each number is obtained by multiplying or dividing the previous number with a constant number. The constant number is called the common ratio of the series.
Geometric progression is the series of numbers that are related to each other by a common ratio. The common ratio is a positive or negative integer or fraction. It is finite or an infinite series.
Definition of Geometric Progression
Geometric progression is the special type of sequence in the number series. It is a series of numbers in which each term is obtained by multiplying the previous term by a fixed number, known as the common ratio.
The common ratio of a geometric progression is a positive or negative integer. The geometric progression generally abbreviated as \(G.P.\)
Real-time Examples:
1. Calculating the interest earned by the bank
2. Population growth
Terms Used in Geometric Progression
In geometrical progression \(\left({G.P.} \right),\) we will use certain terms to solve the mathematical problems, which are given below:
1. \(a \)-First term
2. \(r \)-common ratio
3. \({a^n} – {n^{th}}\) term
4. \({s_n}\)-sum of \(n\) terms (Finite)
5. \({S_\infty }\)-sum of terms(infinite)
Formula of Geometric Progression
Geometric progression is a series of numbers in which each term is obtained by multiplying the previous term by a fixed number (common ratio).
The progressions are generally written as:
\(a,ar,a{r^2},a{r^3},….\)
Here, each term is obtained by multiplying the previous term by the common ratio \(\left( r \right)\) except the first term \(\left( {a.} \right)\)
Here, \(a\)-First-term
\(r\)-Common ratio
Geometric Progression – First Term
Each number associated with a series is known as the term. The first term of the \(G.P.\) is known as the first number of the given series. Thus, starting or initial number is known as the first term of the \(G.P.\)
It is generally denoted by \({a_1}\) or \(a.\)
Example:
For series \(6,12,24,48,96……\)
The first term is \(6.\)
Geometric Progression – Common Ratio
Geometric progression is the sequence or series of numbers such that each number is obtained by multiplying or dividing the previous number with a constant number. The constant number is called the common ratio of the series.
The geometric progression, in which each number is obtained by multiplying with the common ration \(r,\) except the first term is given below:
Here, the common ratio is obtained by dividing the term by its preceding term except for the first term, such as: \(r = \frac{{{a_1}r}}{{{a_1}}} = \frac{{{a_1}{r^2}}}{{{a_1}r}} = \frac{{{a_1}{r^3}}}{{{a_1}{r^2}}}\)
The general formula for the common ratio of the geometrical progression is given by:
\(r = \frac{{{a_n}}}{{{a_{n – 1}}}}\)
Example:
In the geometric series \(1,3,9,27,….\)
The common ratio is given by: \(r = \frac{{{2^{nd}}term}}{{{1^{st\,}}term}} = \frac{3}{1} = \frac{9}{3} = \frac{{27}}{9} = 3\)
In the geometric progression \(a,ar,a{r^2},a{r^3},……\) and so on.
The formula gives \({n^{th}}\) term or last term of a geometric progression
\({a_n} = {a_1}{r^{n – 1}}\)
Where, \({a_1}\)- First-term of \(G.P.\)
\(r\)-Common ratio
\(n\)-Number of terms
Sum of Terms of Geometric Progression
The sum of the geometric series formula is used to find the total of all the terms of the given geometrical series. There are two types of geometric progressions, such as infinite or infinite series. So, there are different formulas to calculate the sum of series, which are given below:
Sum of Finite Terms of Geometric Progression
Finite geometric progression is the series of numbers, which has finite numbers. In the finite series, the last term is defined.
The sum of terms of a geometric progression is given by:
\({S_n} = \frac{{a\left({{r^n} – 1} \right)}}{{r – 1}},\) When \(r > 1\) and \({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}},\)
When \(r < 1\)
Derivation of the Sum of Infinite Geometric Progression
Consider the geometric series \(a,ar,a{r^2},a{r^3},…..{a_{n – 1}},{a_{n.}}\)
The sum of all the terms of a geometric progression is given by:
\({S_n} = a + a{r^2} + a{r^3} + …… + {a_{n – 1}} + {a_n} – – – – \left( i \right)\)
When \(r = 1,\)
\({S_n} = a + a + a + …… = a\left({n\,times} \right) = na\)
When \(r \ne 1,\) multiply equation \(\left( i \right)\) with \(r,\)
\({S_n} \times r = r\left({a + ar + a{r^2} + a{r^3} + ……{a_{n – 1}} + {a_n}} \right) – – – – \left({ii} \right)\)
Subtracting the above equations, \(\left({ii} \right) – \left( i \right),\)
\( \Rightarrow r{S_n} – {S_n} = r\left({a + ar + a{r^2} + a{r^3} + ….. + {a_{n – 1}} + {a_n}} \right) – \left({a + ar + a{r^2} + a{r^3} + …..{a_{n – 1}} + {a_n}} \right)\)
\( \Rightarrow r{S_n} – {S_n} = a{r^n} – a\)
\( \Rightarrow {S_n}\left({r – 1} \right) = a\left({{r^n} – 1} \right)\)
\( \Rightarrow {S_n}=\frac{{a\left({{r^n} – 1} \right)}}{{r – 1}}\)
Sum of Infinite Geometric Series
In the geometric progression \(a,ar,a{r^2},a{r^3},…\infty .\)
The sum of terms of the above geometric series is found by using the simple formula, which is given below:
\({S_\infty } = \frac{a}{{1 – r}};1 < r < 1\)
Here,
\({S_\infty }\)-Sum of infinite geometric series
\(a\)-first term of the geometric series
\(r\)-common ratio
\(n\)-number of terms in the geometric progression
Geometric Mean
In mathematics, geometric mean gives the mean of the central tendency of a set of numbers.
Let us consider three numbers, \(a,b,c\) are In geometric progression, then the geometric mean is given by \({b^2} = ac\) or \(b = \sqrt {ac} .\)
Formulas of Geometric Progression
For the given geometric progression, \(a,ar,a{r^2},a{r^3},….a{r^n}.\) The various formulas used to calculate the problems on geometric progression are listed as below:
Description | Formula |
Common ratio | \(\frac{{{a_n}}}{{{a_{n – 1}}}}\) |
\({n^{th}}\) term | \({a_n} = {a_1}{r^{n – 1}}\) |
sum of \(n\) terms \(\left( {{S_n}} \right),\) when \(r > 1\) | \({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\) |
sum of \(n\) terms \(\left( {{S_n}} \right),\) when \(r < 1\) | \({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\) |
sum of infinite terms | \({S_\infty } = \frac{a}{{1 – r}}\) |
Geometric mean of \(a,b,c\) | \(\sqrt {a \times c} \) |
Properties of Geometric Progression
There are some unknown properties of the geometric progression, which help to solve the mathematical problems easily. Some of them are listed below:
1. When we multiply each term of the geometric series with the non-zero constant number, the new series also forms geometric series with the same common ratio.
2. The new series also forms geometric series with the same common ratio when we divide each term of the geometric series with the non-zero constant number.
3. The series of reciprocal of the terms of geometric progression also forms geometric series.
4. When each term of the series changes to the terms’ square, the new series also forms geometric series.
5. When the terms of a geometric progression are selected at the intervals, then the new series is also geometric.
6. The logarithm of each term of geometric progression (non-zero, non-negative series) changes to arithmetic series.
Tricks for Geometric Progression
For solving, different types of mathematical problems on geometrical progression, follow some tricks, which help to solve the problems easily:
1. When the product of three terms of the geometric progression is given, consider the numbers are \(\frac{a}{r},a,ar,\) where \(r\) is the common ratio.
2. When the product of four terms of the geometric progression is given, consider the numbers are \(\frac{a}{{{r^3}}},\frac{a}{r}a,a{r^3},\) where \(r\) is the common ratio.
3. When the product of five terms of the geometric progression is given, consider the numbers are \(\frac{a}{{{r^2}}},\frac{a}{r}a,ar,a{r^2},\) where \(r\) is the common ratio.
Solved Examples – Geometric Progression
Q.1. Write the next three terms of the given geometric progression: \(1,2,4,8,16, \ldots \)
Ans: Given geometric series is \(1,2,4,8,16, \ldots \)
We know that geometric progression is a series of numbers in which each term is obtained by multiplying the previous term by a fixed number, known as the common ratio.
The common ratio of the given geometric series is given by:
\(r = \frac{2}{1} = \frac{4}{2} = 2\)
The next number after \(16\) is obtained by multiplying \(16\) with the common ratio \(\left({r = 2} \right).\)
\(16 \times 2 = 32\)
The next number after \(32\) is obtained by multiplying \(32\) with the common ratio \(\left({r = 2} \right).\)
\(32 \times 2 = 64\)
The next number after \(16\) is obtained by multiplying \(16\) with the common ratio \(\left({r = 2} \right).\)
\(64 \times 2 = 128\)
The next three terms of the given series are \(32,64,128.\)
Q.2. Find the \({10^{th}}\) term of the given geometric series \({\text{4,12,36,108,}}….\)
Ans: Given series is \({\text{4,12,36,108,}}….\)
From the above geometric series, \(a = 4\) and the common ratio \(r=\frac{{12}}{4} = 3\)
The \(10\,th\) term of the geometric progression is found by using the formula: \({a_n} = a{r^{n – 1}}.\)
\( \Rightarrow {a_{10}} = \left( 4 \right) \times {\left( 3 \right)^{10 – 1}}\)
\( \Rightarrow {a_{10}} = 4 \times {3^9}\)
\( \Rightarrow {a_{10}} = 4 \times 19683\)
\( \Rightarrow {a_{10}} = 78,732\)
Hence, the tenth term of the given series is \(78,732.\)
Q.3. In a certain scenario, the count of the virus gets doubled after every hour. Initially, the count of the virus is \(3.\) What would be the total count of the virus after \(6\) hours?
Ans: Given, every hour, the count of the virus gets doubled.
Here, the count of the virus forms a geometric progression with the first term \(\left({a = 3} \right)\) and the common ratio \(\left({r = 2} \right).\)
So, the total count of the virus after \(6\) hours is found by using the sum of the first 6 terms of \(G.P.\)
\({S_n} = \frac{{a\left({{r^n} – 1} \right)}}{{r – 1}}\)
\({S_6} = \frac{{3\left({{2^6} – 1} \right)}}{{\left({2 – 1} \right)}}\)
\( \Rightarrow {S_6} = 3\left({64 – 1} \right)\)
\( \Rightarrow {S_6} = 3 \times 63\)
\( \Rightarrow {S_6} = 189\)
Hence, the total count of the virus after \(6\) hours is \(189.\)
Q.4. Find the following sum of the terms of the given geometrical series \(\frac{1}{3},\frac{1}{9},\frac{1}{{27}},….\infty.\)
Ans: Given series is \(\frac{1}{3},\frac{1}{9},\frac{1}{{27}},….\infty .\)
Here, the first term of the series is \(a = \frac{1}{3}\) and the common ratio is \(r = \frac{{\frac{1}{9}}}{{\frac{1}{9}}} = \frac{1}{3}\)
The given series is infinite geometric series. The sum of the infinite series is \({S_\infty } = \frac{a}{{1 – r}}\)
\( \Rightarrow {S_\infty } = \frac{{\frac{1}{3}}}{{1 – \frac{1}{3}}} = \frac{{\frac{1}{3}}}{{\frac{2}{3}}} = \frac{1}{2}\)
Hence, the sum of the given series is \(\frac{1}{2}.\)
Q.5. How many terms of the geometric progression \(1 + 3 + 9 + …..\) are there, whose sum is given by \(121\)? Ans: The given series is \(1 + 3 + 9 + …..\)
The first term of the series is \(a = 1,\) and the common ratio of the given series is \(r = \frac{3}{1} = 3.\)
Let the number of terms in the given series be \(n.\)
Given, the sum of terms of the given series is \({S_n} = 121.\)
By using the formula, \({S_n} = \frac{{a\left({{r^n} – 1} \right)}}{{r – 1}}\)
\( \Rightarrow 121 = \frac{{1\left({{3^n} – 1}\right)}}{{\left({3 – 1} \right)}}\)
\( \Rightarrow 121 = \frac{{\left({{3^n} – 1} \right)}}{2}\)
\( \Rightarrow {3^n} – 1 = 121 \times 2 = 242\)
\( \Rightarrow {3^n} = 242 + 1 = 243\)
\( \Rightarrow {3^n} = {3^5}\)
Equating the powers of the above exponents with the same base.
\(n = 5\)
Therefore, the number of terms in the given series is \(5.\)
Summary
In this article, we studied the definition of geometric progression, which is the series of numbers related to each other by a common ratio. We have discussed various terms of the geometric progression, which includes first term, common ratio, \({n^{th}}\) term of the geometric progression along with the formulas.
We have studied the finite and infinite series and the formulas to be used to find the sum of terms of the geometric progression. This article discussed the properties of the geometrical progression and the tricks to be followed for some type of problems. Some of the concepts are explained by using solved examples.
Frequently Asked Questions(FAQ) – Geometric Progression
Q.1. What is the formula of the sum of \(G.P.\)?
Ans: The formula of the sum of \(G.P\) is \({S_n} = \frac{{a\left({{r^n} – 1} \right)}}{{r – 1}},r > 1.\)
Q.2. What is the sum of infinite geometric progression?
Ans: For the geometric progression \(a,ar,a{r^2},a{r^3},…\infty \)
\({S_\infty } = \frac{a}{{1 – r}}; – 1 < r < 1\).
Q.3. What is geometric progression?
Ans: Geometric progression is the series of numbers that are related to each other by a common ratio.
Q.4. What is the geometric progression formula?
Ans: The geometric formula is given by \({a_n} = {a_1}{r^{n – 1}}.\)
Q.5. How do you find the common ratio?
Ans: The common ratio of the series can be found by dividing the number by its previous number.
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