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331 LearnersLast updated on October 18, 2025


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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