Geometric progression is calculated from the first term and common ratio, the sum of the progression of the specified range is displayed.

- Front page
- Calculate progression
- Sum of eometric progression

## Sum of eometric progression

## About the calculation of Sum of eometric progression

Enter the first term, the common ratio, and the range of the progression you want to calculate and click the button "Calculate sum of geometric progression". The sum of the geometric progression in the specified range is calculated.

It also shows how to calculate the sum of a geometric sequence.

Please enter up to 15 digits for First term and Common ratio, and up to 10,000 for nth term.

## What is Geometric progression?

A geometric sequence is a sequence in which the ratio of each adjacent term is the same.

The first term in a sequence is called the first term, and the ratio of each adjacent term is called the common ratio.

For example, the following sequence has a first term of 1 and a common ratio of 3.

Since the common ratio is 3, the ratio of all adjacent terms is 1:3.

## How to calculate Sum of eometric progression

### Sum of the first to nth terms

Calculates the sum from the first term to the nth term.

If the first term is a and the common ratio is r, the sequence is as follows.

a, ar, ar^{2}, ar^{3} ... ar^{n−2}, ar^{n−1}

If the sum of the first through nth terms of this sequence is Sn, then Sn can be expressed as follows:

S_{n} = a + ar + ar^{2} + ar^{3} + ... + ar^{n−2} + ar^{n−1}

Multiply both sides of this equation by r.

rS_{n} = ar + ar^{2} + ar^{3} + ar^{4} + ... + ar^{n−1} + ar^{n}

Subtract the left and right sides of these two equations.

S_{n} |
= | a | + | ar | + | ar^{2} |
+ ... + | ar^{n−2} |
+ | ar^{n−1} |
|||

− | rS_{n} |
= | ar | + | ar^{2} |
+ ... + | ar^{n−2} |
+ | ar^{n−1} |
+ | ar^{n} |
||

S_{n} − rS_{n} |
= | a − ar^{n} |

The left side becomes S_{n} − rS_{n}, and since all the parts in between disappear on the right side, it becomes a − ar^{n}.

S_{n} − rS_{n} = a − ar^{n}

Solve this for Sn.

S_{n}(1 − r) = a(1 − r^{n})

S_{n} = a(1 − r^{n})(1 − r)

Therefore

_{n}= a(1 − r

^{n})(1 − r) (r ≠ 1)

Sum of the first to nth terms = First term × (1 − Common ratio^{n})(1 − Common ratio)

When r = 1, all the numbers in a geometric progression are equal to the first term, so we get:

S_{n}= a + a + a + ... + a + a = na

Therefore

_{n}= na (r = 1)

### Sum of nth through mth term

To calculate the sum from the nth term to the mth term, you similarly reverse the sequence and add both sides.

S_{m−n} |
= | ar^{n−1} |
+ | ar^{n} |
+ | ar^{n+1} |
+ ... + | ar^{m−2} |
+ | ar^{m−1} |
|||

− | rS_{m−n} |
= | ar^{n} |
+ | ar^{n+1} |
+ ... + | ar^{m−2} |
+ | ar^{m−1} |
+ | ar^{m} |
||

S_{m−n} − rS_{m−n} |
= | ar^{n−1} − ar^{m} |

The left side becomes rS_{m−n} − S_{m−n}, and since all the parts in between disappear on the right side, it becomes ar^{n−1} − ar^{m}.

rS_{m−n} − S_{m−n} = ar^{n−1} − ar^{m}

Solve this for S_{m−n}.

S_{m−n}(1 − r) = a(r^{n−1} − r^{m})

S_{m−n}= a(r^{n−1} − r^{m})(1 − r)

Therefore

_{m−n}= a(r

^{n−1}− r

^{m})(1 − r) (r ≠ 1)

^{n−1}− Common ratio

^{m})(1 − Common ratio)

When r = 1, all the numbers in a geometric progression are equal to the first term, so we get:

S_{m−n}= a + a + a + ... + a + a = (m − n + 1)a

Therefore

_{m−n}= (m − n + 1)a (r = 1)