Sum of n terms of Geometric Progression: Definition, Formula, Solved Examples & Practice Questions

Sum of n terms of Geometric Progression: Progression is a series of numbers related by a common relation. If the numbers in the series are obtained by adding or subtracting the same number, that series is known as the arithmetic series or arithmetic progression. If the numbers in the series are obtained by multiplying or dividing with the same number, it is said to be in geometric progression.

The geometric progression is the series of numbers in which each number is obtained by dividing or multiplying the previous term by the same number. That same or common number is called the common ratio. In this article, we shall discuss the sum of n terms of geometric progression.

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What is Geometric Progression?

The geometric progression is a series of numbers in which each number is obtained by dividing or multiplying the previous number by the same number. Geometric progression is a special kind of number series in which each term of the series is obtained by multiplying or dividing with the common number except the first term.

The common number that can be multiplied or divided with each term except the first term is called the common ratio. It may be a positive number, negative number, or zero. The geometric progression is generally abbreviated as G.P. Some of the real-life examples of geometric progressions are:

1. Calculating the interest earned by the bank
2. Population growth

Formula of Geometric Progression

The geometric progression is the series of numbers in which each number is obtained by dividing or multiplying with the same number (common ratio). The geometric progressions are generally written as

\(a,ar,a{r^2},a{r^3}, \ldots \ldots \) 

In the above figure, each term of the geometric progression is obtained by multiplying the previous term by the common ratio \((r),\) excluding the first term \((a).\)

Here,

1. \(a-\)First-term
2. \(r-\)Common Ratio

Practice 10th CBSE Exam Questions

The common ratio of the geometric progression is obtained by multiplying its previous term by the common number excluding the first term

Also, the common ratio is obtained by dividing any term by the previous term.

The formula used to calculate the general term or the \({n^{{\rm{th}}}}\) term or the last term of the geometric progression is given below:

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Sum of Terms of a Geometric Progression

The formula for the sum of the geometric progression or series is used to find the total value of the given terms of the given geometrical series. There are two types of geometric series, namely finite geometric series or infinite geometric series. Thus, we have different formulas to calculate the sum of terms in the given series, which are listed below:

Sum of n terms of Geometric Progression

The formula for calculating the sum of \(n\) terms of a geometric progression is given by

\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\) when \(r > 1\)

Derivation: Consider the geometric series \(a,ar,a{r^2},a{r^3}, \ldots .{a_{n – 1}},{a_n}\)

The addition of all the terms of the geometric progression is given by

\({S_n} = a + ar + a{r^2} + a{r^3} + \ldots .. + {a_{n – 1}} + {a_n} \ldots \ldots {\rm{(i)}}\)

When \(r=1,\)

\({S_n} = a + a + a + \ldots \ldots . + a(n\,{\rm{times}}) = na\)

When \(r≠1,\) multiply equation \({\rm{(i)}}\) with \(r,\)

\({S_n} \times r = r \times \left( {a + ar + a{r^2} + a{r^3} + \ldots .. + {a_{n – 1}} + {a_n}} \right)……\left( {{\rm{ii}}} \right)\)

Subtracting the above equations, such as \({\rm{(ii) – (i),}}\)

\( \Rightarrow r{S_n} – {S_n} = r\left( {a + ar + a{r^2} + a{r^3} + \ldots \ldots + {a_{n – 1}} + {a_n}} \right) -\)
\(-\left( {a + ar + a{r^2} + a{r^3} + \ldots \ldots + {a_{n – 1}} + {a_n}} \right)\)

\( \Rightarrow r{S_n} – {S_n} = a{r^n} – a\)

\( \Rightarrow {S_n}(r – 1) = a\left( {{r^n} – 1} \right)\) 

\( \Rightarrow {S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)

The formula for calculating the sum of n terms of a geometric progression is given by

\({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\) when \(r < 1\)

Derivation: Consider the geometric series \(a,ar,a{r^2},a{r^3}, \ldots .{a_{n – 1}},{a_n}.\)

The addition of all the terms of the geometric progression is given by

\({S_n} = a + ar + a{r^2} + a{r^3} + \ldots \ldots + {a_{n – 1}} + {a_n} \ldots \left( {\rm{i}} \right)\)

When \(r=1,\)

\({S_n} = a + a + a + \ldots \ldots . + a(n\,{\rm{times}}) = na\)

When \(r≠1,\) multiply equation \(\left( {\rm{i}} \right)\) with \(r,\)

\({S_n} \times r = r \times \left( {a + ar + a{r^2} + a{r^3} + \ldots \ldots + {a_{n – 1}} + {a_n}} \right) \ldots {\rm{(ii)}}\)

Subtracting the above equations, such as \({\rm{(i) – (ii),}}\)

\( \Rightarrow {S_n} – r{S_n} = \left( {a + ar + a{r^2} + a{r^3} + \ldots \ldots + {a_{n – 1}} + {a_n}} \right) \)
\(-r\left( {a + ar + a{r^2} + a{r^3} + \ldots \ldots + {a_{n – 1}} + {a_n}} \right)\)

\( \Rightarrow {S_n} – r{S_n} = a – a{r^n}\)

\( \Rightarrow {S_n}(1 – r) = a\left( {1 – {r^n}} \right)\)

\( \Rightarrow {S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\)

Sum of Infinite Geometric Series: In an infinite geometric progression \(a,ar,a{r^2},a{r^3}, \ldots \ldots \infty \), the formula used to find the sum of terms in an infinite geometric series is given below:

\({S_\infty } = \frac{a}{{1 – r}}; – 1 < r < 1\)

Here,

\({S_\infty } – \)Sum of the terms in an infinite geometric progression

\(a-\)first term of the geometric progression

\(r-\)common ratio of the geometric progression

Solved Examples- Sum of n terms of Geometric Progression

Q.1. Keerthi is the \({12^{{\rm{th}}}}\) generation kid. As she thinks about the \(12\) generations of her family, she is thinking of the parents only. Now, help Keerthi that how many persons are there in her family.
Ans: Given, Keerthi is the \({12^{{\rm{th}}}}\) generation kid. We have to find the total number of persons in the twelve generations of Keerthi’s family. As she thinks about only parents, as shown below:

So, the number of people in each generation, start from Keerthi, is given below:
\(1, 2, 4, 8,…..\)
The above-given series is the geometric series, in which each term is obtained by multiplying each term by constant term except the first term.
First-term\(=1\)
Common ratio\( = \frac{2}{1} = 2\)
Total generations \((n) = 12\)
The total number of people in the family can be found by using the sum formula of geometric progression: \({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
\( \Rightarrow {S_{12}} = \frac{{1\left( {{2^{12}} – 1} \right)}}{{2 – 1}} = {2^{12}} – 1 = 4096 – 1 = 4095\)
Therefore, the total number of people in the twelve generations of Keerthi is \(4095.\)

Q.2. 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 given geometric series is \(a=1,\) and the common ratio of the given geometric series is \(r = \frac{3}{1} = 3.\)
Let the number of terms in the given geometric series be n.
Given, the sum of terms of the given geometric series is \({S_n} = 121\)
By using the formula, \({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
\( \Rightarrow 121 = \frac{{1\left( {{3^n} – 1} \right)}}{{(3 – 1)}}\)
\( \Rightarrow 121 = \frac{{{3^n} – 1}}{2}\)
\( \Rightarrow {3^n} – 1 = 121 \times 2 = 242\)
\( \Rightarrow {3^n} = 242 + 1 = 243\)
\( \Rightarrow {3^n} = {3^5}\)
We know that exponents have the same base, then their powers are equal.
Equating the powers of the above exponents with the same base.
\(n=5\)
Therefore, the number of terms in the given geometric series is \(5.\)

Q.3. In a specific scenario, the count of the bacteria 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 that the count of the virus gets doubled in every hour,
Here, the count of the virus forms a geometric series as the count increased by multiplying every time.
The first term of the series \((a=3)\) and the common ratio of the series \((r=2).\)
So, the total count of the virus after \(6\) hours is found by using the sum of the first six terms of G.P.
The formula used to find the sum of \(n\) terms is given by \({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
Here, \(n=6\)
\({S_6} = \frac{{3\left( {{2^6} – 1} \right)}}{{(2 – 1)}}\)
\( \Rightarrow {S_6} = 3(64 – 1)\)
\( \Rightarrow {S_6} = 3 \times 63\)
\( \Rightarrow {S_6} = 189\)
Hence, the total count of the virus after \(6\) hours is \(189.\)

Q.4. Chinmayi shared one crucial piece of information about the ITR, filling with \(5\) of her friends at \({\rm{4}}\,{\rm{a}}{\rm{.m}}{\rm{.}}\) Each of her friends shared the same information with \(5\) unique friends, at \({\rm{5}}\,{\rm{a}}{\rm{.m}}{\rm{.}}\) And, again, each of these five people shared the same information with \(5\) unique people at \({\rm{6}}\,{\rm{a}}{\rm{.m}}{\rm{.}}\) In this sequence, how many would get the same information by \({\rm{12}}\,{\rm{p}}{\rm{.m}}{\rm{.}}\)?
Ans: Given that, at the start, Chinmayi shared information with five people, and next, those people shared this information with \(5\) more people and so on every hour.
Clearly, the above sequence forms the geometric progression, such that the first term \((a)=5\) and the common ratio \((r)=5.\)
From \({\rm{4}}\,{\rm{a}}{\rm{.m}}\) to \({\rm{12}}\,{\rm{p}}{\rm{.m}}\) there is a total of \(8\) hours.
Thus, \(n=8\)
The total number of people who get the information is found by using the sum of the \(n\) terms of the geometric progression as follows:
\({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
\( \Rightarrow {S_8} = \frac{{5\left( {{5^8} – 1} \right)}}{{(5 – 1)}}\)
\( \Rightarrow {S_8} = \frac{{5(390624)}}{4}\)
\( \Rightarrow {S_8} = 488,280\)
Therefore, by \({\rm{12}}\,{\rm{p}}{\rm{.m}}{\rm{,}}\) there is a total of \(488,280\) people get the same information as shared by the Chinamyi at the starting.

Q.5. Find the sum of geometric series given below:
\(4,-12, 36, -108,…\) up to \(10\) terms.
Ans: Given geometric series is \(4, -12, 36, -108….\)
The first term of given geometric series \((a)=4\)
And, the common ratio of the given geometric series \((r) = \,- \frac{{12}}{4} = \,- 3\)
We need to find the sum of the first \(10\) terms of the given geometric series.
The formula used to find the sum of first \(n\) terms of the geometric progression is given by, when \(r = – 3 < 1\)
\({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\)
\( \Rightarrow {S_{10}} = \frac{{4\left( {1 – {{( – 3)}^{10}}} \right)}}{{1 – ( – 3)}}\)
\( \Rightarrow {S_{10}} = \frac{{4(1 – 59049)}}{4}\)
\( \Rightarrow {S_{10}} =\, – 59048\)
Hence, the sum of the first \(10\) terms of the given geometric series is \(-59048.\)

Summary

In this article, we have discussed the definitions of geometric series and the terms of geometric progression. Here, we have studied the formulas of geometric progressions, such as the common ratio and the general or \({n^{{\rm{th}}}}\) term of the series etc.

This article gives the sum of first \(n\) terms of the given geometric series when the common ratio is negative and also positive. Here we have studied the sum of infinite terms of the geometric series by using solved examples that will help understand the concept easily.

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Frequently Asked Questions (FAQs)

Q.1. What is the sum of infinite geometric progression?
Ans: For the geometric progression \(a,ar,a{r^2},a{r^3}, \ldots \ldots \infty \)
\({S_\infty } = \frac{a}{{1 – r}}; – 1 < r < 1\)

Q.2. What is the formula of the sum of G.P.?
Ans: The formula of the sum of G.P is
1. \({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}},r > 1\)
2. \({S_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}},r < 1\)

Q.3. What is geometric progression?
Ans: Geometric progression is a special kind of number series in which each term of the geometric series is obtained by multiplying or dividing with the common number except the first term.

Q.4. How do you find the sum of the first 10 terms of a geometric progression?
Ans: The sum of the first ten terms of a geometric progression is found by substituting the values of the first term \((a),\) and the common ratio \((r)\) and \(n=10\) in the formula : \({S_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)

Q.5. What is the common ratio in the geometric progression?
Ans: It is the constant number that is used to multiply or divide each term of the geometric progression, excluding the first term.

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