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.

1, 3, 9, 27, 81, 243, 729...

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, ar2, ar3 ... arn−2, arn−1

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

Sn = a + ar + ar2 + ar3 + ... + arn−2 + arn−1

Multiply both sides of this equation by r.

rSn = ar + ar2 + ar3 + ar4 + ... + arn−1 + arn

Subtract the left and right sides of these two equations.

  Sn = a + ar + ar2 + ... + arn−2 + arn−1    
rSn =     ar + ar2 + ... + arn−2 + arn−1 + arn
  Sn − rSn = a − arn

The left side becomes Sn − rSn, and since all the parts in between disappear on the right side, it becomes a − arn.

Sn − rSn = a − arn

Solve this for Sn.

Sn(1 − r) = a(1 − rn)

Sn = a(1 − rn)(1 − r)

Therefore

Sn= a(1 − rn)(1 − r) (r ≠ 1)

Sum of the first to nth terms = First term × (1 − Common ration)(1 − Common ratio)

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

Sn= a + a + a + ... + a + a = na

Therefore

Sn = na (r = 1)
Sum of the first to nth terms = n × First term

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.

  Sm−n = arn−1 + arn + arn+1 + ... + arm−2 + arm−1    
rSm−n =     arn + arn+1 + ... + arm−2 + arm−1 + arm
  Sm−n − rSm−n = arn−1 − arm

The left side becomes rSm−n − Sm−n, and since all the parts in between disappear on the right side, it becomes arn−1 − arm.

rSm−n − Sm−n = arn−1 − arm

Solve this for Sm−n.

Sm−n(1 − r) = a(rn−1 − rm)

Sm−n= a(rn−1 − rm)(1 − r)

Therefore

Sm−n= a(rn−1 − rm)(1 − r) (r ≠ 1)
Sum of nth through mth term = First term × (Common ration−1 − Common ratiom)(1 − Common ratio)

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

Sm−n= a + a + a + ... + a + a = (m − n + 1)a

Therefore

Sm−n = (m − n + 1)a (r = 1)
Sum of the first to nth terms = (m − n + 1) × First term

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