Sum of Geometric Series: A geometric series is a series where each subsequent number is obtained by multiplying or dividing the number preceding it. The sum of the geometric series refers to the sum of a finite number of terms of the geometric series. A geometric series can be finite or infinite as there are a countable or uncountable number of terms in the series. The sum of infinite geometric series is greater than the sum of finite geometric series.

Geometric series have several applications in Physics, Engineering, Biology, Economics, Computer Science, Queueing Theory, Finance etc. It also has various applications in the field of Mathematics. In this article, we will provide detailed information on the Sum of Infinite Geometric Series. Scroll down to learn more about Sum of Geometric Series!

Sum of Geometric Series: Overview

In a Geometric Series, each term is obtained by multiplying or dividing the preceding term by a factor, which is a constant.

In general, we write a geometric sequence in the following format: \(a,\,ar,\,a{r^2},\,a{r^3},….,\)
Where \(a \to \) is the first term
\(ar \to \) is the second term.
\(r \to \) common ratio (The factor between the terms)

Note: In case of a geometric progression, \(r \ne 0,\) because when \(r = 0,\) we get the sequence \(\left\{ {a,\,0,\,0,\,…} \right\}\) which is not geometric.

Sum of Infinite Geometric Series Formula

A geometric series is a set of numbers where each term after the first is found by multiplying or dividing the previous term by a fixed number.
The common ratio, abbreviated as \(r,\) is the constant amount.

Let the first, second, third, \(……,\,{n^{{\rm{th}}}}\) term be denoted by \({T_1},\,{T_2},\,{T_3},\,….{T_n},\) then we can write,
\({T_1} = a\)
\({T_2} = ar\)
\({T_3} = a{r^2}\)
\({T_n} = a{r^{n – 1}}\)
\( \Rightarrow r = \frac{{{T_2}}}{{{a_1}}} = \frac{{{T_3}}}{{{T_2}}} = \frac{{{T_4}}}{{{T_3}}}\)
\( \Rightarrow r = \frac{{{T_n}}}{{{T_{n – 1}}}}.\)
Therefore, the common ratio can be determined by dividing each term by its proceeding term.

Example: Find the common ratio of the geometric series \(1,\,2,\,4,\,8,\,16,\,….\)
By using the formula,
\( \Rightarrow r = \frac{{{T_2}}}{{{a_1}}} = \frac{{{T_3}}}{{{T_2}}} = \frac{{{T_4}}}{{{T_3}}}\)
\( \Rightarrow r = \frac{2}{1} = \frac{4}{2} = \frac{8}{4}\)
\( \Rightarrow r = \frac{2}{1} = 2\)
So, the common ratio of the geometric series is \(2.\)

Sum of n Terms of a Geometric Series: Types

Sum of n Terms of a Geometric Series can be classified into two types. Below we have provided the types of Sum of n terms of a Geometric Series:

1. Sum of Finite Geometric Series
2. Sum of Infinite Geometric Series

1. Sum of Finite Geometric Series

Let us consider that the first term of a geometric series is \(“a”,\) and the common ratio is \(r\) and the number of terms is \(n.\)
There are two cases here.
Case-1: When \(r > 1\)
In this case, the sum of all the terms of the geometric series is given by
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}.\)
Case-2: When \(r > 1\)
In this case, the sum of all the terms of the geometric series is given by
\({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}.\)

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Formula for the Sum of a Finite Geometric Series

Let us consider that,
\(n \to \) the number of terms, \(a \to \) first-term
\(r \to \) common ratio, \({S_n} \to \) Sum of first \(n\) terms
Let \({S_n} = a + ar + a{r^2} + ….a{r^{n – 1}}\)…..(i)
Multiply the equation (i) with \(r,\) we get,
\({S_n} \times r = ar + a{r^2} + a{r^3} + …a{r^{n – 1}} – a{r^n}\)…… (ii)
Subtract the equation (i) from (ii), we get,
\({S_n}r – {S_n} = a{r^n} – a\)
\( \Rightarrow {S_n}\left( {r – 1} \right) = a\left( {{r^n} – 1} \right)\)
\( \Rightarrow {S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{\left( {r – 1} \right)}}\)
Hence, the formula to find the sum of the geometric series is
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{\left( {r – 1} \right)}}.\)

2. Infinite Sum of Geometric Series

In the case where there are an infinite number of terms in the geometric series, meaning in the situation where \(n \to \infty ,\) the sum of infinite geometric series is given by
\({S_n} = \frac{a}{{1 – r}},\) where \(a\) is the first term, \(r\) is the common ratio and \(\left| r \right| < 1.\)

Applications of Geometric Series

The geometric series has many applications. Some of them are listed below:
1. Geometric series have huge applications in physics, engineering, biology, economics, computer science, queueing theory, finance etc. They are utilised across mathematics.
2. To calculate the area encompassed by a parabola and a straight line, Archimedes utilised the sum of a geometric series.
3. The Koch snowflake’s interior is made up of an unlimited number of triangles. Geometric series frequently appear in the study of fractals as the perimeter, area, or volume of a self-similar figure.
4. The knowledge of infinite series makes us solve ancient problems like zeno’s paradoxes.

Solved Examples on Sum of Geometric Series

Q.1: Find the common ratio of the geometric series \(3,\,6,\,12,\,….\)
Ans: By using the formula, we get,
\(r = \frac{{{T_2}}}{{{a_1}}} = \frac{{{T_3}}}{{{T_2}}}\)
\( \Rightarrow r = \frac{6}{3} = \frac{{12}}{6}\)
\( \Rightarrow r = \frac{2}{1} = 2\)
Therefore, the common ratio of the geometric series is \(2.\)

Q.2: Find the common ratio of the geometric series: \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + ….\)
Ans: By using the formula, we get,
\(r = \frac{{{T_2}}}{{{a_1}}}\)
\( \Rightarrow r = \frac{{\frac{1}{4}}}{{\frac{1}{2}}} = \frac{1}{2}\)
\( \Rightarrow r = \frac{1}{2}\)
Therefore, the common ratio of the given geometric series is \(\frac{1}{2}.\)

Q.3: Find the sum of the infinite geometric sequence \(27,\,18,\,12,\,8,\,….\)
Ans: From the given geometric series, \(27,\,18,\,12,\,8,\,….,\) we get,
\(r = \frac{{{a_2}}}{{{a_1}}} = \frac{{18}}{{27}} = \frac{2}{3}\)
Now, the sum of the infinite geometric series is given by
\(S = \frac{{{a_1}}}{{1 – r}}\)
\( \Rightarrow S = \frac{{27}}{{1 – \frac{2}{3}}}\)
\( \Rightarrow S = \frac{{27}}{{\frac{{3 – 2}}{3}}} = \frac{{27}}{{\frac{1}{3}}} = 27 \times 3 = 81\)
Hence, the obtained sum is \(81.\)

Q.4: Find the \({8^{{\rm{th}}}}\) term of the following geometric series: \(4,\,12,\,36,\,108,\,….\)
Ans: From the given geometric series, we get, \(a = 4,\,r = \frac{{12}}{4} = 3,\,n = 8\)
\({T_n} = a{r^{n – 1}}\)
\( \Rightarrow {T_8} = 4 \times {3^{8 – 1}}\)
\( \Rightarrow {T_8} = 4 \times {3^7}\)
\( \Rightarrow {T_8} = 4 \times 2187\)
\( \Rightarrow {T_8} = 8748\)
Therefore, the \({8^{{\rm{th}}}}\) term of the given geometric series is \(8748.\)

Q.5: Find the sum of first \(5\) terms of the geometric series \(10,\,30,\,90,\,270,\,810,\,2430,\,….\)
Ans: From the given geometric series, we get, \(a = 10,\,r = \frac{{30}}{{10}} = 3\)
In this case \(r > 1.\)
So, the formula to find the sum of the first \(5\) terms of the geometric series is,
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
\( \Rightarrow {S_5} = \frac{{10\left( {{3^5} – 1} \right)}}{{3 – 1}}\)
\( \Rightarrow {S_5} = \frac{{10\left( {243 – 1} \right)}}{{3 – 1}}\)
\( \Rightarrow {S_5} = \frac{{10\left( {242} \right)}}{2}\)
\( \Rightarrow {S_5} = \frac{{2420}}{2}\)
\( \Rightarrow {S_5} = 1210\)
Therefore, the required sum is \(1210.\)

Q.6: Find the sum of the first \(7\) terms of the geometric series if \(a = 1,\,r = 2.\)
Ans: From the given geometric series, we get, \(a = 1,\,r = 2.\)
The formula to find the sum of the first \(7\) terms of the geometric series is,
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
\( \Rightarrow {S_7} = \frac{{1\left( {{2^7} – 1} \right)}}{{2 – 1}}\)
\( \Rightarrow {S_7} = \frac{{1\left( {128 – 1} \right)}}{1}\)
\( \Rightarrow {S_7} = \frac{{1\left( {127} \right)}}{1}\)
\( \Rightarrow {S_7} = 127\)
Therefore, the sum of the first \(7\) terms of the geometric series is \(127.\)

Q.7: Find the sum, if it exists for the geometric series: \(10 + 9 + 8 + 7 + ….\)
Ans: In this case, we observe that the given series \(10 + 9 + 8 + 7 + ….\) is not a geometric series because the ratio between the consecutive terms is not constant.
It is an arithmetic progression. Finding the sum of an infinite arithmetic series is not possible.

Q.8: Find the sum of first \(5\) terms of the geometric series: \(1,\,2,\,4,\,8,\,16,\,32,\,….\)
Ans: From the given geometric series, we get, \(a = 1,\,r = \frac{2}{1} = 2\)
The formula to find the sum of first \(5\) terms of the geometric series is,
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
So, \({S_5} = \frac{{1\left( {{2^5} – 1} \right)}}{{2 – 1}}\)
\({S_5} = \frac{{1\left( {32 – 1} \right)}}{1}\)
\({S_5} = \frac{{32}}{1}\)
\({S_5} = 32\)
Therefore, the required sum is \(32.\)

Important Practice Questions on Sum of Geometric Series

Below we have provided some of the important practice questions on the sum of geometric series:

1. Find the equivalent fraction of the recurring decimal \(0.797970…..\)
2. Find out the 12th term of the sequence \(3, -6, 9, -12,….?\)
3. Write the first five terms of a GP whose first term is 5 and the common ratio is 6.
4. Find the sum, if it exists for the geometric series: \(20 + 19 + 18 + 17 + ….\)
5. Find the sum of the first \(9\) terms of the geometric series if \(a = 3,\,r = 6.\)

Summary

The geometric progression is a set of integers generated by multiplying or dividing each preceding term in such a way that there is a common ratio between the terms (that is not equal to \(0\)), and the sum of all these terms is the sum of the geometric progression. The article includes the definition of geometric series, formula to find the sum of n terms of finite and infinite geometric series. We also discussed the derivation of the formula. We mentioned the various applications of the geometric series that helps in understanding its importance. The learning outcome from the topic is that it will help in understanding the method in finding the sum of the different sum of geometric series.

FAQs on Sum of Geometric Series

The following are the most frequently asked questions on the sum of a Geometric Series:

Q.1: Explain the Sum of Geometric Series with an example?
Ans: A geometric series is a series where each term is obtained by multiplying or dividing the previous term by a constant number, called the common ratio. And, the sum of the geometric series means the sum of a finite number of terms of the geometric series.
Example: Let us consider the series \(27,\,18,\,12,\,…\)
Here, we observe that \(\frac{{18}}{{27}} = \frac{2}{3}\) and \(\frac{{12}}{{18}} = \frac{2}{3}.\)
So, the ratio of the consecutive terms, in this case, is constant. Hence, the above series can be called a geometric series.

Q.2: How do you find the sum of a geometric series?
Ans: Let us consider that the first term of a geometric series is \(“a”,\) and the common ratio is \(r\) and the number of terms be \(n.\)
To find the sum of a finite geometric series, we use the formula,
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\) for \(r > 1\)
\({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\) for \(r < 1.\)

Q.3: What is the sum of \(n\) terms of a Geometric series?
Ans: The formula of the sum of geometric series is given by,
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\) for \(r > 1\)
\({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\) for \(r < 1\)
where \(n \to \)the number of terms, \({a_1} \to \)first term and \(r \to \)common ratio.

Q.4: What is the common ratio in geometric series?
Ans: The common ratio in the geometric progression is the ratio of any one term in the series divided by the previous term. The letter \(“r”\) is usually used to symbolise it.

Q.5: What are the two types of geometric series?
Ans: The two types of geometric series are listed below:
1. Finite geometric series
2. Infinite geometric series

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