Sum of a Finite Geometric Series
Aligned to A-SSE.B.4 — Common Core State Standards for Mathematics.
What this lesson teaches
A geometric series adds terms with a constant ratio r. The sum of n terms starting at a is S = a(1 − r^n)/(1 − r), which comes from subtracting r×S from S to cancel the middle terms.
Worked examples
Example 1
Find 3 + 6 + 12 + 24 + 48. Here a = 3, r = 2, and n = 5 terms. S = 3(1 − 2^5)/(1 − 2) = 3(1 − 32)/(−1) = 3(−31)/(−1) = 93.
Example 2
A savings plan deposits $200 at the start of each year and earns 5% annually. After 10 years the deposits have grown to 200(1.05)¹⁰ + ... + 200(1.05)¹. That is a geometric series with a = 200(1.05) = 210 and r = 1.05, so S = 210(1.05¹⁰ − 1)/(1.05 − 1) = 210(0.6289)/0.05 which is about $2,641.
Example 3
Common mistake: using the sum formula on a sequence that is not geometric. In 2, 4, 6, 8 the terms grow by ADDING 2, so the ratio 4/2 = 2 but 6/4 = 1.5 — it is arithmetic, and needs S = n(first + last)/2 = 4(2 + 8)/2 = 20. Test the ratio between at least two consecutive PAIRS before reaching for a(1 − rⁿ)/(1 − r).
Practice questions
- Find the sum 5 + 15 + 45 + 135.
- Use the formula to sum the first 6 terms of a = 2, r = 3.
- A ball drops and each bounce is 1/2 the height: sum 100 + 50 + 25 + 12.5.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Suma de una Serie Geométrica Finita
Una serie geométrica suma términos con una razón constante r. La suma de n términos que empieza en a es S = a(1 − r^n)/(1 − r), que se obtiene al restar r×S de S para cancelar los términos intermedios.
Ejemplo 1: Encuentra 3 + 6 + 12 + 24 + 48. Aquí a = 3, r = 2, y n = 5 términos. S = 3(1 − 2^5)/(1 − 2) = 3(1 − 32)/(−1) = 3(−31)/(−1) = 93.
Ejemplo 2: Un plan de ahorro deposita $200 al inicio de cada año y gana 5% anual. Tras 10 años los depósitos han crecido a 200(1.05)¹⁰ + ... + 200(1.05)¹. Esa es una serie geométrica con a = 200(1.05) = 210 y r = 1.05, así que S = 210(1.05¹⁰ − 1)/(1.05 − 1) = 210(0.6289)/0.05, unos $2,641.
Ejemplo 3: Error común: usar la fórmula de la suma en una sucesión que no es geométrica. En 2, 4, 6, 8 los términos crecen SUMANDO 2, así que la razón 4/2 = 2 pero 6/4 = 1.5 — es aritmética y necesita S = n(primero + último)/2 = 4(2 + 8)/2 = 20. Comprueba la razón entre al menos dos PARES consecutivos antes de recurrir a a(1 − rⁿ)/(1 − r).
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