Tenth Grade Mathematics: GP: n-th Term and Sum of Finite Terms
Just as summing an arithmetic sequence requires a dedicated formula, calculating the cumulative sum of a geometric progression requires an algebraic summation strategy. Because terms in a geometric progression compound exponentially, attempting to compute large sums through manual addition is impossible. By employing the subtraction technique of multiplying the entire sum by the common ratio, we derive a compact closed-form expression for the sum of the first n terms of any geometric sequence.
Derivation of the Sum of First n Terms of a GP
Let a Geometric Progression have first term a and common ratio r (with r != 1). The sum of the first n terms is:
Sn = a + a*r + a*r^2 + a*r^3 + ... + a*r^(n - 1) --- (Equation 1)
Multiply both sides of Equation 1 by the common ratio r:
r*Sn = a*r + a*r^2 + a*r^3 + ... + a*r^(n - 1) + a*r^n --- (Equation 2)
Now, subtract Equation 2 from Equation 1. Observe the massive internal cancellation (telescoping sum):
Sn - r*Sn = a + (a*r - a*r) + (a*r^2 - a*r^2) + ... + (a*r^(n-1) - a*r^(n-1)) - a*r^n
Sn * (1 - r) = a - a*r^n
Sn * (1 - r) = a * (1 - r^n)
Provided that r != 1, divide both sides by (1 - r):
Theorem (Finite GP Sum Formula):
a * (1 - r^n) a * (r^n - 1)
Sn = --------------- or ---------------
1 - r r - 1
Both algebraic forms are mathematically identical. Customarily, we use the first form when |r| < 1 and the second form when |r| > 1 to keep both numerator and denominator positive.
Special Case: r = 1
If r = 1, every term equals a:
Sn = a + a + a + ... + a (n terms) = n * a
Alternative Form Involving the Last Term
If the last term l = an = a * r^(n - 1) is known, we can rewrite a * r^n as (a * r^(n - 1)) * r = l * r:
Theorem (Sum in terms of Last Term):
a - l * r
Sn = -----------, where r != 1
1 - r
Concept of Infinite Geometric Series (|r| < 1)
Although Tenth Grade mathematics primarily focuses on finite progressions, understanding what happens to a decaying geometric progression (where |r| < 1) provides profound insight into recurring decimals.
Behavior of r^n as n grows infinitely large:
When |r| < 1 (for example r = 1/2), r^n approaches 0:
(1/2)^1 = 0.5, (1/2)^2 = 0.25, (1/2)^10 = 0.00097, (1/2)^n ---> 0
Substituting r^n = 0 into the sum formula yields:
a
S_inf = -----
1 - r
This explains why an infinite sum of positive quantities can converge to a finite, exact rational number!
Detailed Worked Example
Find the sum of the first 8 terms of the GP: 1, 3, 9, 27, ...
Step 1: Identify parameters:
First term a = 1
Common ratio r = 3 / 1 = 3 (r > 1)
Number of terms n = 8
Step 2: Apply the sum formula:
S8 = a * (r^n - 1) / (r - 1)
S8 = 1 * (3^8 - 1) / (3 - 1)
Step 3: Evaluate power and simplify:
3^8 = 6561
S8 = (6561 - 1) / 2 = 6560 / 2 = 3280.
The sum of the first 8 terms is 3280.
Chapter Exercises
Exercise 1: Find the sum of the first 7 terms of the GP: 1, -1/2, 1/4, -1/8, ...
Exercise 2: How many terms of the GP: 3, 3/2, 3/4, ... are needed to give the sum 3069 / 512?
Exercise 3: The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the three terms.
Solutions and Explanations
Solution 1: For the GP 1, -1/2, 1/4, -1/8, ..., first term a = 1, common ratio r = -1/2. Number of terms n = 7. Since |r| < 1, use Sn = a * [1 - r^n] / [1 - r]: S7 = 1 * [1 - (-1/2)^7] / [1 - (-1/2)] = [1 - (-1/128)] / [1 + 1/2] = [1 + 1/128] / [3/2] = (129 / 128) * (2 / 3) = (129 * 2) / (128 * 3) = (43 * 1) / (64 * 1) = 43 / 64. Thus, the sum of the first 7 terms is 43/64.
Solution 2: For the GP 3, 3/2, 3/4, ..., first term a = 3, common ratio r = (3/2)/3 = 1/2. We are given Sn = 3069/512. Using the sum formula: 3 * [1 - (1/2)^n] / [1 - 1/2] = 3069/512 ==> 3 * [1 - (1/2)^n] / (1/2) = 3069/512 ==> 6 * [1 - (1/2)^n] = 3069/512 ==> 1 - (1/2)^n = 3069 / (512 * 6) = 3069 / 3072. Simplify the fraction by dividing numerator and denominator by 3: 3069/3 = 1023, and 3072/3 = 1024. Thus: 1 - (1/2)^n = 1023 / 1024 ==> (1/2)^n = 1 - 1023/1024 = 1 / 1024. Since 1024 = 2^10, (1/2)^n = (1/2)^10 ==> n = 10. Therefore, exactly 10 terms are needed.
Solution 3: Let the three terms of the GP be a/r, a, and ar. Their product is given as -1: (a/r) * a * (ar) = -1 ==> a^3 = -1 ==> a = -1. Now use the sum condition: a/r + a + ar = 13/12. Substitute a = -1: -1/r - 1 - r = 13/12 ==> -(1/r + r) = 13/12 + 1 = 25/12 ==> (r^2 + 1)/r = -25/12. Cross-multiply: 12(r^2 + 1) = -25r ==> 12r^2 + 25r + 12 = 0. Factor this quadratic: product is 12 * 12 = 144, sum is 25; factors are 16 and 9. 12r^2 + 16r + 9r + 12 = 0 ==> 4r(3r + 4) + 3(3r + 4) = 0 ==> (3r + 4)(4r + 3) = 0. Thus r = -4/3 or r = -3/4. If r = -4/3: terms are (-1)/(-4/3), -1, (-1)*(-4/3) ===> 3/4, -1, 4/3. If r = -3/4: terms are 4/3, -1, 3/4. In either case, the three numbers are 3/4, -1, and 4/3.