Sequences of numbers are having some rules and patterns. This pattern may be of any routine kind, applicable from one term to another term. A sequence where multiplication is needed to get the next term is the geometric sequence. Terms of the sequence can be accumulated with plus or minus symbols, to get its sum. This will give the geometric series. For example one geometric series is 1 + 2 + 4 + 8 + 16 , … This article will explain about the geometric series and geometric series formula with examples. Let us learn the interesting concept!

## W**hat is Geometric Series?**

A geometric series is also termed as the geometric progression. It is a series formed by multiplying the first term by a fixed value to get the second term. This process is continued until we get a required number of terms in the series. Such a progression increases in a specific way and hence giving a geometric progression.

The geometric series formula will refer to determine the general term as well as the sum of all the terms in it. For example, in the above series, if we multiply by 2 to the first number we will get the second number and so on.

As such series behave according to a simple rule of multiplying a constant number to one term to get to another. Therefore, we can generate any number of the term of such series. So we can examine such series to know about the fixed numbers for multiplication, which is called the common ratios. It is denoted as a symbol ‘r’.

**The Formula for Geometric Series**

General Generic Sequence is: \(a_{1} , a_{2}, a_{3} …\)

Take \(a_{1}\) as the first term of the sequence.. The common ratio ‘r’ has the formula for its computation is follows:

\(r = \frac {a_{n}}{a_{n-1}}\)

where n is a positive integer and n >1.

The formula for the general term for any geometric series will be: \(a_{n} = a_{1} r^{n-1}\)

Now formula to add finite n number of terms of the series will be:

\(S_{n} = \frac {a_{1} (1- r^{n})}{1-r} , for, r < 1 \\\)

And, \(S_{n} = \frac {a_{1} (r^{n}-1)}{r-1} , for, r >1\)

For the infinite number of terms, we have much more compact formula as:

\(S_{\infty } = \frac {a_{1}}{1-r} for, r < 1 \\\)

And \(S_{\infty } = \frac {a_{1}}{r-1} for, r > 1 \\\)

\(a_{1}\) | Frist term |

\(a_{n}\) | General nth term |

n | Number of terms |

r | Common ratio |

\(S_{n}\) | Sum of n terms |

\(S_{\infty }\) | Sum of infinity terms |

**Solved Examples for Geometric Series Formula**

Q.1: Add the infinite sum 27 + 18 + 12 + …

Solution: It is a geometric sequence. So using Geometric Series Formula

Here , \(a_{1} = 27\)

\(r = \frac{a_{2}}{a{1}}\)

\(r = \frac {18}{27} = \frac{2}{3}\)

Sum of the infinity terms will be:

\(S_{\infty } = \frac {a_{1}}{1-r}\\\)

\(S_{\infty } = \frac {27}{1-\frac{2}{3}}\\ = 81\)

Thus sum of given infinity series will be 81.

Q.2: Find the sum of the first 10 terms of the given sequence: 3 + 6 + 12 + ….

Solution: The given series is a geometric series, due to a fixed common ratio.

Also its first term is , \(a_{1} = 3\)

n = 10

Common ratio, \(r = \frac {6}{3}\)

i.e. r = 2

since r is greater than 1. So formula for sum of finite terms will be,

\(S_{n} = \frac {a_{1} (r^{n}-1)}{r-1} , for, r >1\)

Thus substituting values , we have

\(S_{10} = \frac {3 (2^{10}-1)}{2-1} \\ = 3 \times (1024-1) \\\)

\(S_{10} = 3069\)

Thus sum will be 3069.

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26

Hi

Same

yes