Formula Derivation And Solved Examples Of Sum To N Terms Of A GP
The concept of sum to n terms of a GP is essential in mathematics and helps in solving real-world and exam-level problems efficiently. It is commonly encountered in sequence and series questions across school, board, and competitive exams. Understanding this concept ensures you can quickly calculate the total of a geometric series with any number of terms.
Understanding Sum to n Terms of a GP
A sum to n terms of a GP (Geometric Progression) refers to the process of adding up the first n terms of a geometric sequence. A geometric progression is a sequence in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio (r). This concept is widely used in compound interest calculations, population growth models, and physics problems involving repeated multiplications.
Formula Used in Sum to n Terms of a GP
The standard formula is: \( S_n = a\dfrac{(r^n - 1)}{(r-1)} \) if \( r \neq 1 \), where:
r = common ratio
n = number of terms
Here’s a helpful table to understand sum to n terms of a GP more clearly:
Sum to n Terms of a GP Table
| Type | Formula | Condition |
|---|---|---|
| General Case | \( S_n = a\dfrac{(r^n - 1)}{r - 1} \) | \( r \neq 1 \) |
| When r = 1 | \( S_n = n \times a \) | \( r = 1 \) |
| Alternative Form | \( S_n = a\dfrac{(1-r^n)}{1-r} \) | \( r \neq 1 \) |
This table shows how the pattern of sum to n terms of a GP changes based on the value of the common ratio.
Stepwise Proof of the Formula
1. Let the GP be \( a, ar, ar^2, \ldots, ar^{n-1} \), and let their sum be \( S_n \).
2. So, \( S_n = a + ar + ar^2 + \ldots + ar^{n-1} \).
3. Multiply both sides by r: \( rS_n = ar + ar^2 + ar^3 + \ldots + ar^n \).
4. Subtract the second equation from the first:
\( S_n - rS_n = a - ar^n \)
5. Factor out \( S_n \): \( S_n(1 - r) = a(1 - r^n) \).
6. Therefore, \( S_n = \dfrac{a(1 - r^n)}{1 - r} \), or equivalently
\( S_n = \dfrac{a(r^n - 1)}{r - 1} \) (by multiplying numerator and denominator by -1).
Worked Example – Solving a Problem
Example: Find the sum to first 5 terms of a GP with a = 2, r = 3.
1. Write the formula: \( S_n = a\dfrac{(r^n - 1)}{r-1} \).
2. Substitute the values: \( a = 2, r = 3, n = 5 \):
\( S_5 = 2\dfrac{(3^5 - 1)}{3-1} \)
3. Calculate \( 3^5 \): \( 3^5 = 243 \).
4. Compute numerator: \( 243 - 1 = 242 \).
\( S_5 = 2\dfrac{242}{2} \)
5. Simplify denominator: \( 3-1 = 2 \).
6. Calculate: \( S_5 = 2 \times 121 = 242 \).
Final Answer: The sum of the first 5 terms is 242.
Practice Problems
- Find the sum to first 7 terms of a GP where a = 4, r = 2.
- If the sum to n terms of a GP is 315, a = 5 and r = 2, find n.
- Calculate the sum of the first 6 terms for the GP: 3, 6, 12, ...
- A GP has a = 2, r = 0.5. Find the sum to the first 4 terms.
Common Mistakes to Avoid
- Using the AP (arithmetic progression) formula instead of the GP formula.
- Forgetting to check if r = 1, in which case use \( S_n = n \times a \).
- Not applying brackets properly for negative or fractional r.
Real-World Applications
The concept of sum to n terms of a geometric progression appears in areas such as calculating compound interest, spreading information through social networks, and modeling exponential growth and decay. Vedantu helps students see how these series form the foundation for advanced scientific and economic calculations.
Quick Comparison: AP vs GP vs HP
| Type | Sum to n Terms Formula |
|---|---|
| Arithmetic Progression (AP) | \( S_n = \frac{n}{2}[2a + (n-1)d] \) |
| Geometric Progression (GP) | \( S_n = a\frac{(r^n-1)}{r-1} \) |
| Harmonic Progression (HP) | Sum is taken by converting to AP of reciprocals |
We explored the idea of sum to n terms of a GP, how to apply it, how to solve problems step by step, and why it matters in real life. Practice more such questions on Vedantu and reinforce your understanding of geometric progressions and other sequences.
Explore Related Concepts
- Arithmetic Progression
- Sequence and Series
- Nth Term of AP
- Harmonic Progression
- Formula Sheet for Class 11
FAQs on Sum To N Terms Of A Geometric Progression Explained
1. What is the sum to n terms of a GP?
The sum to n terms of a GP is the total obtained by adding the first n terms of a geometric progression. If the first term is a and the common ratio is r, then:
Sn = a(1 − rⁿ) / (1 − r), for r ≠ 1.
This formula helps calculate the sum quickly without adding each term individually.
2. What is the formula for the sum of n terms of a geometric progression?
The formula for the sum of n terms of a geometric progression (GP) is Sn = a(1 − rⁿ) / (1 − r) when r ≠ 1.
- a = first term
- r = common ratio
- n = number of terms
- Sn = sum of first n terms
If r = 1, then the sum becomes Sn = na.
3. How do you find the sum of n terms of a GP step by step?
To find the sum of n terms of a GP, use the formula Sn = a(1 − rⁿ)/(1 − r).
- Step 1: Identify the first term (a).
- Step 2: Find the common ratio (r).
- Step 3: Determine the number of terms (n).
- Step 4: Substitute into Sn = a(1 − rⁿ)/(1 − r).
Example: For 2, 6, 18, 54 (first 4 terms):
a = 2, r = 3, n = 4
Sn = 2(1 − 3⁴)/(1 − 3) = 2(1 − 81)/(-2) = 80.
4. What is the sum to n terms of a GP when r is less than 1?
When the common ratio r is less than 1, the sum to n terms of a GP is still given by Sn = a(1 − rⁿ)/(1 − r).
Since rⁿ becomes smaller as n increases, the sum approaches a fixed value if |r| < 1.
This property is important when studying converging geometric series.
5. What happens to the sum of a GP if r = 1?
If the common ratio r = 1, then the sum of n terms of a GP is Sn = na.
This is because every term in the GP is equal to the first term a, so:
- a, a, a, ..., a (n times)
Adding n identical terms gives na.
6. What is the difference between the sum of an AP and the sum of a GP?
The main difference is that an AP uses a constant difference, while a GP uses a constant ratio.
- Sum of AP: Sn = n/2 [2a + (n − 1)d]
- Sum of GP: Sn = a(1 − rⁿ)/(1 − r)
In an arithmetic progression (AP), terms increase by addition, whereas in a geometric progression (GP), terms increase by multiplication.
7. Can you give an example of finding the sum to n terms of a GP?
Yes, the sum to n terms of a GP can be calculated using Sn = a(1 − rⁿ)/(1 − r).
Example: Find the sum of first 5 terms of 3, 6, 12, 24, 48.
- a = 3
- r = 2
- n = 5
Sn = 3(1 − 2⁵)/(1 − 2) = 3(1 − 32)/(-1) = 93.
8. What is the sum to infinity of a geometric progression?
The sum to infinity of a geometric progression is S∞ = a/(1 − r) when |r| < 1.
This formula applies only when the common ratio satisfies:
- |r| < 1 (the series converges)
If |r| ≥ 1, the GP does not have a finite sum to infinity.
9. Why does the formula Sn = a(1 − rⁿ)/(1 − r) work?
The formula Sn = a(1 − rⁿ)/(1 − r) works because subtracting rSn from Sn eliminates most terms of the geometric progression.
- Sn = a + ar + ar² + ... + arⁿ⁻¹
- rSn = ar + ar² + ... + arⁿ
Subtracting gives:
Sn − rSn = a − arⁿ
Sn(1 − r) = a(1 − rⁿ)
Dividing by (1 − r) gives the formula.
10. What are common mistakes when finding the sum to n terms of a GP?
Common mistakes when finding the sum to n terms of a GP usually involve incorrect substitution or sign errors.
- Using the wrong formula (confusing AP and GP).
- Forgetting to use rⁿ instead of r.
- Making sign errors when r is negative.
- Not checking if r = 1.
Always identify a, r, and n correctly before applying Sn = a(1 − rⁿ)/(1 − r).
