- Written By Keerthi Kulkarni
- Last Modified 26-01-2023

**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.

Learn the Concepts of Geometric Progression

## 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:

- Calculating the interest earned by the bank
- 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,

- \(a-\)First-term
- \(r-\)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:

**Attempt 10th CBSE Exam Mock Tests**

### 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 **\({12^{{\rm{th}}}}\)

**\(12\)**

*generation kid. As she thinks about the*

*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** \(1+3+9+…\)

**\(121\)**

*are there, whose sum is given by*

*?***The given series is \(1+3+9+…\)**

*Ans:*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** \(3.\)

**\(6\)**

*What would be the total count of the virus after*

*hours?***Given that the count of the virus gets doubled in every hour,**

*Ans:*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 **\(5\)

**\({\rm{4}}\,{\rm{a}}{\rm{.m}}{\rm{.}}\)**

*of her friends at***\(5\)**

*Each of her friends shared the same information with***\({\rm{5}}\,{\rm{a}}{\rm{.m}}{\rm{.}}\)**

*unique friends, at***\(5\)**

*And, again, each of these five people shared the same information with***\({\rm{6}}\,{\rm{a}}{\rm{.m}}{\rm{.}}\)**

*unique people at***\({\rm{12}}\,{\rm{p}}{\rm{.m}}{\rm{.}}\)**

*In this sequence, how many would get the same information by*

*?***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.**

*Ans:*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.***Given geometric series is \(4, -12, 36, -108….\)**

*Ans:*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.

Learn the Concepts on Sequences and Series

### 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.

*We hope this detailed article on the sum of n terms of geometric progression helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel free to ask us in the comment section and we will be more than happy to assist you. Happy learning!*

## FAQs

### How do you find the sum of n terms in a geometric progression? ›

The sum of the first n terms of a geometric sequence, given by the formula: **Sn=a1(1−rn)1−r**, r≠1. An infinite geometric series where |r|<1 whose sum is given by the formula: S∞=a11−r.

**What is the formula of SN in geometric progression? ›**

To find the sum of a finite geometric series, use the formula, **Sn=a1(1−rn)1−r**,r≠1 , where n is the number of terms, a1 is the first term and r is the common ratio .

**What are examples on geometric progression? ›**

For example, the sequence **2, 6, 18, 54, ...** is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2.

**Which term of the sequence √ 3 3 3 √ 3 is 729? ›**

Thus, the 12th term of the given sequence is **729**.

**How do you solve geometric progression? ›**

The formula for the nth term of a geometric progression whose first term is a and common ratio is r is: **a _{n}=ar^{n}^{-}^{1}**. The sum of n terms in GP whose first term is a and the common ratio is r can be calculated using the formula: S

_{n}= [a(1-r

^{n})] / (1-r).

**How do we find the nth term of a geometric sequence *? ›**

How do you find the nth term of a geometric progression with two terms? First, calculate the common ratio r by dividing the second term by the first term. Then use the first term a and the common ratio r to calculate the nth term by using the formula **an=arn−1 a n = a r n − 1** .

**What are three 3 examples of geometric shapes? ›**

There are many shapes in geometry based on their dimensions. Circle, Triangle, Square, Rectangle, Kite, Trapezium, Parallelogram, Rhombus and different types of polygons are the 2-d shapes. **Cube, Cuboid, Sphere, Cone and Cylinder** are the basic three-dimensional shapes.

**What is an example of a geometric term? ›**

**Point, line, line segment, ray, right angle, acute angle, obtuse angle, and straight angle** are common geometric terms.

**Which term of the geometric sequence 2 √ 3 6 6 √ 3 is 1458? ›**

Find the geometric progression with fourth term = 54 **seventh term** = 1458.

**Which term of the sequence root3 3 3 root3 is 243 ?!? ›**

Answer : **Tenth term** of the G.P. is 243.

### Which term of the geometric sequence 2 8 32 is 131072? ›

Hence, 131072 is the **9th term** in the given G.P.

**How do you solve geometric probability problems? ›**

To find the geometric probability, **divide the area of the circle, which is the desired area, by the area of the whole target**. The area of the circle is πr2 or (π)(22), which is (3.14)(4) or 12.56 ft^{2}. The area of the whole target is 5 × 5 = 25 ft^{2}. The geometric probability is or 0.5024.

**What is a practical example of GP? ›**

GP occurs in real life when each actor in a system behaves independently and is fixed. Examples include: **If each person decides not to have another child depending on the current population, then annual population increase is geometric**.

**What is an example of a geometric series sum? ›**

A geometric series is the sum of the first few terms of a geometric sequence. For example, 1, 2, 4, 8,... is a geometric sequence, and **1+2+4+8+...** is a geometric series.

**How do you find the nth term of an example? ›**

Finding the nth Term of an Arithmetic Sequence

Given an arithmetic sequence with the first term a1 and the common difference d , the nth (or general) term is given by **an=a1+(n−1)d** . Example 1: Find the 27th term of the arithmetic sequence 5,8,11,54,... . a8=60 and a12=48 .

**Can you give at least 3 example of geometric sequence? ›**

**{2,6,18,54,162,486,1458,...}** is a geometric sequence where each term is 3 times the previous term.

**What is geometry explained simply? ›**

Geometry is defined as “**a branch of mathematics that deals with the measurement, properties, and relationships of points, lines, angles, surfaces, and solids**.” Put even more simply, geometry is a type of math that deals with points, lines, shapes, and surfaces.

**What are examples of geometric forms quizlet? ›**

Geometric forms are used in construction, for organization, and as parts in machines; **cube, pyramid**.

**What is an example of geometric problem? ›**

Example: **A triangle has a perimeter of 50.** **If 2 of its sides are equal and the third side is 5 more than the equal sides, what is the length of the third side?**

**How many terms are there in a geometric progression 3 6 12 24 384? ›**

∴ Number of terms =**8**. Was this answer helpful?

### What is the 8th term of the geometric sequence 3 6 12 24? ›

Hence, the 8th term from the end = **96**.

**What is the geometric of 6 and 24? ›**

=√6×24=**√144**=12.

**What is the 10th term of the sequence √ 3 √ 12 √ 27? ›**

The 10th term of the sequence 3 ,12 ,27 ..... is **103** .

**Which term of the progression 18 12 8 is 729 512? ›**

Thus, the **9th term** of the given G.P. is 512729.

**What is the square root of 243 in radical form? ›**

The Square root of 243 = **√243** where '√' is the radical, and 243 is the radicand. Exponential Form of Square root of 243 = 243. Solution for √243 = 9√3.

**How many terms are there in the geometric sequence 32 256 2048 16384 2⁵⁰? ›**

In the given Geometric progression find the number of terms. 32, 256, 2048, 16384,.........,2^{50}. Explanation: n^{th} term = first term(ratio^{n} ^{–} ^{1})., 2^{50} = 2^{5}(2^{3}^{(}^{n}^{-}^{1}^{)}), n=15. This implies **16 ^{th} term**.

**Which term of 0.0004 0.02 0.1 is 12? ›**

Thus, **6th term** of the given G.P. is 12.5.

**How many terms of the G.P. 3 3 by 2 3 by 4 are needed to give the sum 3069 512? ›**

Solution : Here, `a = 3 and r = 1/2`<br> `S_n = (a(1-r^n))/(1-r)`<br> `3069/512 = (3(1-(1/2)^n))/(1-(1/2))`<br> `1023/1024= 1-(1/2)^n`<br> `(1/2)^n = 1/1024=(1/2)^10`<br> So, `n = 10` is the required answer. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

**What is the formula of sum of n terms? ›**

**S _{n} = n(n+1)/2**

Hence, this is the formula to calculate sum of 'n' natural numbers.

**How do you find the missing n in a geometric sequence? ›**

Step 1: Find the common ratio of each pair of consecutive terms in the sequence by dividing each term by the term that came before it. Step 2: Multiply the common ratio with the number prior to the first missing number in the sequence. Step 3: Repeat Step 2 for any other missing numbers.

### How do you find the n term formula? ›

Solution: To find a specific term of an arithmetic sequence, we use the formula for finding the nth term. Step 1: The nth term of an arithmetic sequence is given by **an = a + (n – 1)d**. So, to find the nth term, substitute the given values a = 2 and d = 3 into the formula.