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210 LearnersLast updated on October 18, 2025

ChatGPT said: A Geometric Progression (GP) is a sequence where each term is obtained by multiplying the previous term by a fixed common ratio. The previous term can be found by dividing by this ratio. For example, 3, 6, 12, 24,… is a GP with a ratio of 2. GPs can have finite or infinite terms, and this article covers their meaning, formulas, and types.
The sequence in which each term is obtained by multiplying the previous term by a fixed number (common ratio) is known as a geometric progression. It is usually expressed as: \(a, ar, ar^2, ar^3…\), where ‘a’ represents the first term and ‘r’ represents the common ratio. The common ratio can be positive or negative. Any term in a GP can be determined using the first term and the common ratio.
Geometric progressions are mainly classified into two types based on their length.
The different types of geometric progressions are:
We will now learn about each type in detail:
Finite geometric progression: A finite geometric progression has a limited number of terms, and the last term is known. For example: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), …, \(\frac{1}{32768}\) is a finite geometric progression. Here, \(\frac{1}{32768}\) is the last term.
Infinite geometric progression: An infinite geometric progression has an endless number of terms. Since there is no fixed number of terms, the last term cannot be specified. For example, the infinite series 3, -6, 12, -24, … does not have a definite end term.
To help you identify the sequence effectively, we will now look at the key differences between GP and AP.
| Geometric Progression (GP) | Arithmetic Progression (AP) |
| Each term is obtained by multiplying the previous term by a fixed common ratio 𝑟. |
Each term is obtained by adding a fixed common difference 𝑑 to the previous term. |
| No common difference between the terms. | There is no fixed ratio between the terms |
| For example: 2, 4, 8, 16,...(r = 2) | For example: 3, 6, 9, 12,...(d = 3) |
| Such series can converge or diverge depending on r. | The series is always divergent unless the common difference is zero. |
| Formula for n-th term is \(a_n = a_1 \cdot r^{\,n-1} \). | Formula for n-th term is \(a_n = a_1 + (n-1)d \). |
Understanding the unique features of a progression helps us identify it more easily. Here are a few properties that geometric progressions (GP) follow.
In a GP, the sum of the terms can be calculated using the following formulas:
For a \(GP: a, ar, ar^2, ar^3\), …
The sum does not exist when \(|r| ≥ 1\).
Learn how to quickly identify, analyze, and apply geometric progressions in problems and real-life scenarios.
Geometric progression is a simple mathematical concept, but many students struggle with its problems. Here are a few common mistakes and tips to avoid them:
Geometric progression has a vital role in various real-life situations. Let's explore how this concept applies in real-life scenarios.
Find the 5ᵗʰ term of a GP Given: First term (a) = 3 Common ratio (r) = 2
a5 = 48
First, apply the formula for the nᵗʰ term:
\(a_n = a \cdot r^{\,n-1} \)
Substituting the values into the formula:
\(a_5 = 3 \times 2^{\,5-1} = 3 \times 2^4 \)
Here, we get:
a5 = 3 × 16 = 48
Find the sum to infinity of a GP Given: a = 8, r = 1/2
S∞ = 16
Let’s first check if |r| < 1
It holds true for the infinite sum since |1/2| < 1.
Using the formula:
S∞ = a / (1 – r)
Substituting the values into the formula:
S∞ = 8 / (1 – 1/2) = 8 / (1/2)
So,
S∞ = 8 × 2 = 16
Find the sum of the first 6 terms of a GP Given: a = 5, r = 3, n = 6
S6 = 1820
Here, we use the formula for the sum of the first n terms
\(S_n = a(1 − r^n)/(1 − r)\)
Let’s substitute the values:
S6 = 5(36 – 1) / (3 – 1)
We now calculate powers and simplify:
36 = 729
S6 = 5 (729 – 1) / 2 = (5 × 728) / 2
So,
S6 = 3640 / 2 = 1820
Find the 8ᵗʰ term of the GP 5, 10, 20, 40,... Given: a = 5, r = 2, n = 8
a8 = 640
Here, we apply the formula for the nᵗʰ term:
\(a_n = a \times r^{\,n-1} \)
Substituting the values into the formula:
a8 = 5 × 28 - 1 = 5 × 27
So,
a8 = 5 × 128 = 640
Find how many terms of the GP 3, 6, 12, 24,... are needed to make the sum 93 Given: a = 3, r = 2, Sₙ = 93
n = 5
Using the formula:
\(S_n = a(1 − r^n)/(1 − r)\)
Substituting the given values:
93 = 3(2n – 1) / (2 – 1)
Now, simplify to get the result:
93 = 3(2n – 1)
93 ÷ 3 = 2n – 1
31 = 2n – 1
2n = 31 + 1
2n = 32
Here, n is the exponent to which 2 needs to be raised to obtain 32.
Since 25 = 32
→ n = 5
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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