Learning Objectives
Distinguish a sequence term from a partial sum.
Calculate finite arithmetic and geometric sums.
Translate a repeated-addition context into a series with the correct number of terms.
Prerequisites: arithmetic and geometric nth-term formulas. This lesson treats finite sums. Infinite-series convergence is a separate topic and should not be inferred from these formulas.
1. Terms and Sums Answer Different Questions
A sequence lists terms u₁, u₂, … . A series adds them. The notation Sₙ = u₁ + u₂ + … + uₙ means the sum of the first n terms. For 2, 4, 6, the third term is u₃ = 6, while the third partial sum is S₃ = 12.
2. Pair the Ends of an Arithmetic Series
For an arithmetic sequence, the first and last terms sum to u₁ + uₙ. So do the second and second-last terms. Pairing the sequence with a reversed copy gives 2Sₙ = n(u₁ + uₙ), hence Sₙ = n(u₁ + uₙ)/2.
Substitute uₙ = u₁ + (n − 1)d to get the equivalent formula Sₙ = n[2u₁ + (n − 1)d]/2. The first version is convenient when the last term is known; the second is convenient when the common difference is given.
Worked Example: Seats in Rows
A hall has 12 rows. The first row has 18 seats, and each successive row has 2 more seats. Find the total.
The last row has u₁₂ = 18 + 11(2) = 40 seats.
The total is S₁₂ = 12(18 + 40)/2 = 348 seats. The average row size is 29 seats, so 12 × 29 provides a check.
3. Subtract Shifted Geometric Sums
For a geometric series Sₙ = a + ar + … + arⁿ⁻¹, multiplication by r gives rSₙ = ar + ar² + … + arⁿ. Subtracting cancels the middle terms: Sₙ − rSₙ = a − arⁿ. For r ≠ 1, this gives Sₙ = a(1 − rⁿ)/(1 − r).
The equivalent form a(rⁿ − 1)/(r − 1) can be convenient when r > 1. If r = 1, all n terms equal a, so Sₙ = na. Do not divide by zero in the geometric formula.
Worked Example: Five Geometric Terms
Find the sum 3 + 6 + 12 + 24 + 48. Here a = 3, r = 2, and n = 5.
Use S₅ = 3(2⁵ − 1)/(2 − 1) = 3(31) = 93.
Adding the five listed terms also gives 93. Notice that the last term uses r⁴, while the sum formula uses r⁵.
4. Count What Is Actually Included
If a problem asks for u₄ through u₁₀ inclusive, there are 10 − 4 + 1 = 7 terms. You can sum these directly as a shorter series, or calculate S₁₀ − S₃. Subtracting S₄ would incorrectly remove the first requested term.
Common Mistakes
Do not give uₙ when the question asks for the total Sₙ.
Count inclusive endpoints carefully.
Use the finite geometric formula with n terms; do not replace it with an infinite-sum formula.
5. Practice
1. Find the sum of the first 20 terms of 5, 8, 11, … .
2. Find the sum of the first four terms of 80, 40, 20, … .
3. Find the sum of terms u₃ through u₇ for uₙ = 2n + 1.
Worked Solutions
1. u₂₀ = 5 + 19(3) = 62. Thus S₂₀ = 20(5 + 62)/2 = 670.
2. S₄ = 80[1 − (1/2)⁴]/(1 − 1/2) = 150. The terms 80 + 40 + 20 + 10 confirm it.
3. The five terms run from u₃ = 7 to u₇ = 15. Their sum is 5(7 + 15)/2 = 55.
6. Summary
- A partial sum adds terms; it is not itself the nth term.
- Arithmetic sums use the number of terms and their average endpoint value.
- Finite geometric sums come from subtracting a shifted copy.
- Check the first included term, the last included term, and the term count.