GP Sum
The sum of a GP is the sum of a few or all terms of a geometric progression. A series of numbers obtained by multiplying or dividing each preceding term, such that there is a common ratio between the terms (that is not equal to 0) is the geometric progression and the sum of all these terms formed so is the sum of geometric progression (GP).
Let us learn more about GP sum formulas (for both finite and infinite series) along with examples.
1.  What is GP Sum? 
2.  Sum of n Terms in GP 
3.  Sum of Infinite GP 
4.  Sum of GP Formulas 
5.  FAQs on GP Sum 
What is GP Sum?
GP sum is the sum of a few or all terms of a geometric progression. Let us start understanding GP sum using an example. Clara saves a few dollars every week in a particular fashion. In week 1 she deposits $2. In week 2  $4, in week 3  $8, in week 4  $16, and so on. How much will she have at the end of 6 weeks in her piggy bank? The amount deposited in the 6 weeks would be $2, $4, $8, $16, $32, and $64. So, at the end of 6 weeks she will have $2+ $4 + $8+ $16+ $32 + $64 = $126 accumulated in her piggy bank. Since we had to calculate the amount only after 6 weeks, the manual calculation was easy. But what if we had to find the amount after so many weeks, say 50 weeks? GP sum formulas make this process easier.
Sum of n Terms in GP
Let's discuss how to calculate the sum of n terms of GP. Consider the sum of the first n terms of a geometric progression (GP) with first term a and common ratio r. Then the first 'n' terms of GP are of the form a, ar, ar^{2}, ... ar^{n1}. Let S be the sum of the geometric progression of n terms. Then:
S_{n} = a + ar + ar^{2 }+ ... + ar^{n1} ... (1)
Multiply both sides by r:
rS_{n} = ar + ar^{2 }+ ... + ar^{n} ... (2)
Subtracting equation (1) from equation (2):
rS_{n}  S_{n} = (ar + ar^{2 }+ ... + ar^{n})  (a + ar + ar^{2 }+ ... + ar^{n1})
S_{n} (r  1) = ar^{n}  a
S_{n} (r  1) = a(r^{n}  1)
S_{n} = a(r^{n}  1) / (r  1)
Note that, here, r ≠ 1. This formula can also be written as:
S_{n} = a(1  r^{n}) / ((1  r)) = a(1  r^{n}) / (1  r).
What happens when r = 1? Then the GP is of form a, a, a, ..., (n terms). Then the sum is, S_{n} = na.
Now, let us work on the example (from the last section) using the sum of n terms of GP formula. The amounts saved by Clara in the order of weeks are, 2, 4, 8, 16, .... Clearly, this is a geometric progression as 4/2 = 8/4 = 16/2 = ... = 2.
Here, the first term is, a = 2, the common ratio is, r = 2, and the number of terms is, n = 6 (as we want the sum of the amounts after 6 weeks).
We have got the same answer using the GP sum formula also.
Sum of Infinite GP
The above formula gives the sum of a finite GP. But what if we have to find the sum of an infinite GP? Let us consider a GP a, ar, ar^{2}, ... (up to an infinite number of terms) such that the absolute value of its common ratio is less than 1. i.e., r < 1. Let us assume the sum of all these infinite number of terms be S_{∞}. Then:
S_{∞} = a + ar + ar^{2} + ... ... (1)
Multiplying both sides by r,
rS_{∞} = ar + ar^{2} + ... ... (2)
Subtracting (2) from (1):
(1  r) S_{∞} = a
S_{∞} = a / (1  r)
This is the formula for the sum of infinite gp. Note that for any infinite geometric series,
 If r < 1, then the geometric series converges and it has a sum.
 If r ≥ 1, then the geometric series diverges and it cannot have a sum.
Example: A square is drawn by joining the midpoints of the sides of the original square. A third square is drawn inside the second square in the same way and this process is continued indefinitely. If a side of the first square is "s" units, determine the sum of areas of all the squares so formed?
Solution:
The areas of squares thus formed are, s, s^{2}/2, s^{2}/4, s^{2}/8, ....
Taking 's^{2}' as common factor, the sum of areas is, s^{2} ( 1 + 1/2 + 1/4 + ... ) ... (1)
Here, 1 + 1/2 + 1/4 + ... is an infinite geometric series with a = 1 and r = 1/2. So its sum is
S_{∞} = a / (1r) = 1 / (1  1/2) = 2.
Substituting this in (1):
The sum of areas = 2s^{2}.
Sum of GP Formulas
Let us summarize all formulas used for finding the sum of a GP.
 To find the sum of finite (n) terms of a GP,
S_{n} = a(r^{n}  1) / (r  1) [OR] S_{n} = a(1  r^{n}) / (1  r), if r ≠ 1.
S_{n} = an, if r = 1.  To find the sum of infinite terms of a GP,
S = a / (1  r), if r < 1 (and in this case, we say that the series converges)
S cannot be found if r ≥ 1 (and in this case, we say that the series diverges)
These GP sum formulas are summarized in the flowchart below.
Important Notes on GP Sum:
 The sum of GP (of n terms) is: S_{n} = a(r^{n}  1) / (r  1) [OR] S_{n} = a(1  r^{n}) / (1  r), if r ≠ 1.
 The sum of GP (of n terms) is: S_{n} = na, when r = 1.
 The sum of GP (of infinite terms) is: S_{∞} = a/(1r), when r < 1.
 The sum of GP (of infinite terms) is: S_{∞} = does not exist, when r ≥ 1.
☛ Related Topics:
GP Sum Examples

Example 1: In a GP, the sum of the first three terms is 16, and the sum of the next three terms is 128. Find the sum of the first n terms of the GP.
Solution:
Let 'a' and 'r' be the first term and the common ratio of the given GP respectively. Then:
a + ar + ar^{2} = 16
ar^{3} + ar^{4} + ar^{5} = 128We can rewrite these equations as:
a (1 + r + r^{2}) = 16
ar^{3} (1 + r + r^{2}) = 128Dividing the second equation by first,
[ar^{3} (1 + r + r^{2})] / [a (1 + r + r^{2})] = 128/16
r^{3} = 8
r = 2
Substituting this in the first equation:
a (1 + 2 + 2^{2}) = 16
7a = 16
a = 16/7
The sum of n terms in GP is:
S_{n} = a(r^{n}  1) / (r  1) = [ (16/7) (2^{n}  1) ] / (2  1) = 16 (2^{n}  1) / 7
Answer: S_{n} = 16 (2^{n}  1) / 7.

Example 2: Sara shared a message with 5 unique people at 1 am. At 2 am, each of her friends shared it with 5 unique people. Then at 3 pm each of their friends shared with 5 unique people. In this sequence, how many unique people would have received the message by 8 am?
Solution:
The number of unique people who received messages after each hour starting at 1 AM are 5, 25, 125, ...
Clearly, this is a GP with a = 5 and r = 5. We have to find the GP sum for 8 terms (8 PM).
Using the sum of GP formula,
S_{n} = a(r^{n}  1) / (r  1)
S_{8} = 5(5^{8}  1) / (5  1) = 488,280
Answer: The total number of unique people = 488,280.

Example 3: The distance traveled by a ball dropped from a height (in inches) are 128/9, 32/3, 8, 6... What could be the distance traveled by the ball before coming to rest?
Solution:
The distances traveled by the ball = 128/9, 32/3, 8, 6....
This is a GP because (32/3) / (128/9) = 8 / (32/3) = 6/8 = ... = 3/4.
So the first term is, a = 128/9; and the common ratio is, r = 3/4.
The total distance traveled by the ball will be the sum of this infinite GP.
S_{∞} = a/(1r)
= (128/9) / (1  (3/4))
= (128/9) / (1/4)
= 128/9 × 4/1
= 56.89 inches
Answer: The total distance traveled by the ball = 56.89 inches.
FAQs on GP Sum
What is the Formula of Finding GP Sum for Finite Terms?
The GP sum formula used to find the sum of n terms in GP is, S_{n} = a(r^{n}  1) / (r  1), r ≠ 1 where:
 a = the first term of GP
 r = the common ratio of GP
If r = 1, then S_{n} = na.
When Does the Sum of an Infinite GP Diverge?
The sum of an infinite GP diverges when the absolute value of the common ratio of GP is either equal to 1 or greater than 1. i.e., when r ≥ 1.
How to Find the Sum of n Terms in GP?
The sum of n terms in GP with 'a' to be its first term and 'r' to be its common ratio can be found using one of the formulas:
 S_{n} = a(r^{n}  1) / (r  1)
 S_{n} = a(1  r^{n}) / (1  r)
These two formulas would work when r ≠ 1. If r = 1, then the sum of n terms is S_{n} = na.
What is the Sum of Geometric Progression Formula for Infinite Terms?
If GP is infinite, then we can find its sum only when r < 1 and we use the formula S_{∞} = a/(1r) in this case. In case, r ≥ 1 for an infinite GP, then it diverges and hence we cannot find its sum.
What is the Sum of GP with r = 1?
If r = 1, then the GP is of form a, a, a, ... The sum of finite (n) terms of such GP = a + a + a + ... (n times) = na. We cannot find the sum of infinite terms of such GP.
When Does the Sum of an Infinite GP Converge?
The sum of an infinite GP converges when the absolute value of the common ratio of GP is less than 1. i.e., when r < 1.
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