Last updated on July 9th, 2025
A geometric progression (GP) is a sequence of a number, where each term is obtained by multiplying the previous term by a constant called the common ratio. The sum of the first n terms is: Sn=a(1-rn)/1-r where r is not equal to 1
In a GP sum, every term is obtained by multiplying the previous one by a fixed number called the common ratio. For a better understanding, consider this example where a student deposits ₹100 in the first week, ₹200 in the second week, ₹400 in the third week, and continues this in sequence till 6 weeks. The total amount in six weeks would be ₹6,300. Calculating step by step is easy. To calculate this for 15 weeks, we will be using the formula,
Sn=a(1-rn) / 1-r When r is not equals to 1
First term a = ₹100
Common ratio r = 2
Number of terms n = 15
Now, putting the numbers in the formula.
S15=100(1-215) / 1-2
Calculating 215= 32,768, we get:
Substituting the values into the formula:
Sn=100 x 1-215 / 1-2
S15=100 x 1-32768 / -1
S15=100 x -32767 / -1
S15=100 x 32767
S15=3,276,700
Therefore, by the end of 15 weeks, the student will have ₹3,276,700.
The sum of an infinite GP is the aggregate of its infinitely many terms. This sum is used only if the common ratio r satisfies ∣r∣ < 1, if the terms get smaller and smaller, approaching zero
For calculating the sum of an infinite GP, use this formula:
S=a / 1-r
a is the first term,
r is the common ratio, and
∣r∣<1 for convergence.
Calculate the series:
₹100 + ₹50 + ₹25 + ₹12.5 + ₹6.25 + …
So a = 100, r = 0.5
Now we will put the numbers in the formula:
S=100 / 1-0.5=100 / 0.5=200
As a result, we get the total sum = 200
There are two main formulas for the sum of GP
1. Sum of the First n Terms of a Finite GP
For a geometric progression, we define the first term a, the common ratio r, and n terms; the formula for the sum is:
Sn=a(1-rn) / 1-r When r 1
Special Case: If r = 1, the sum can be computed in a simple way:
Sn = a×n
2. Sum of an Infinite GP
For an infinite geometric progression, we define the first term a and the common ratio r, the sum exists only if ∣r∣<1. The sum is:
S=a / 1-r (for ∣r∣<1)
If ∣r∣≥1, the series does not converge and has no finite sum.
The sum of the first n terms of a geometric progression (GP) is computed using a formula, depending on the common ratio r. For a GP, we define the first term a, common ratio r, and number of terms n. The formula is:
Sn=a(1-rn) / 1-r When r 1
If r = 1, the sum is:
Sn = a×n
A student is saving money for someone. He saves ₹2 in the first week, ₹4 in the second week, ₹8 in the third week, and goes on till 6 weeks. This doubles his savings each week.
First term a = ₹2
Common ratio r = 2
Number of terms n = 6
Now we will put the numbers in the formula.
S6=2(1-26) / 1-2=2(1-64) / -1=2(-63) / -1=126
So after 6 weeks, the student will have a total amount of ₹126.
Special cases in geometric progressions (GP) happen when the common ratio (r) takes specific values; this modifies the series' behavior and convergence.
Special Cases in Geometric Progression
Finite GP with r = 1:
When the common ratio r is equal to 1, each term will be the same. The sum of the first n terms formula is:
Sn = an Where we consider a as the first term.
Infinite GP with ∣r∣<1:
For an infinite geometric progression to converge, the absolute value of the common ratio must be less than 1. The sum of an infinite GP is:
S=a /1-r
This formula will be valid only when ∣r∣<1; otherwise, the sequence will change.
Geometric progressions (GPs) have many real-life applications where numbers change at a constant rate, making them important in fields like finance, biology, and technology.
Learning and avoiding common mistakes in computing the sum of a geometric progression (GP) is important for accurate mathematical computations.
What is the sum of the first 5 terms of the GP: 1, 2, 4, 8, 16?
The sum of the first 5 terms is 31.
The formula should be used Sn=a(1-rn) / 1-r
Given: a = 1, r = 2, n = 5. Substituting these values into the formula:
Sn=1 x 1-25/1-2=1 x 1-32 / -1=31
Calculate the sum of the infinite GP: 5, 2.5, 1.25, ...
The final calculation for the sum of the infinite GP: 5, 2.5, 1.25, … will be S = 5 / (1 - 0.5) = 5 / 0.5 = 10
For an infinite GP where we use the formula |r| < 1, the sum is given by:
S = a / (1 - r)
So, a = 5 and r = 0.5.
Now we have to substitute the values:
S = 5 / (1 - 0.5) = 10
What is the sum of the infinite GP: 7, 3.5, 1.75, ...?
The answer for the sum of the infinite GP: 7, 3.5, 1.75, ...
S = 7 / (1 - 0.5) = 7 / 0.5 = 14
For an infinite GP with |r| < 1, the sum is given by:
S = a / (1 - r)
So, a = 7 and r = 0.5. Now we have to substitute the values:
S = 7 / (1 - 0.5) = 14
Find the sum of the first 4 terms of the GP: 2, 6, 18, 54.
The final answer for the sum of the first 4 terms of the GP is: 2, 6, 18, 54.
S₄ = 2 × (1 - 3⁴) / (1 - 3) = 2 × (1 - 81) / (-2) = 2 × (-80) / (-2) = 80
So, a = 2, r = 3, and n = 4. Take the sum formula for a finite GP:
Sₙ = a × (1 - rⁿ) / (1 - r). Now we have to substitute the values:
S₄ = 2 × (1 - 3⁴) / (1 - 3) = 80
Find the sum of the first 3 terms of the GP: 4, 12, 36.
The answer for 3 terms of the GP: 4, 12, 36.
S₃ = 4 × (1 - 3³) / (1 - 3) = 4 × (1 - 27) / (-2) = 4 × (-26) / (-2) = 52
So, a = 4, r = 3, and n = 3. Using the formula,
Sₙ = a × (1 - rⁿ) / (1 - r)
Now we have to substitute the values:
S₃ = 4 × (1 - 3³) / (1 - 3) = 52