# Kalkulator barisan aritmetika, geometri, dan Fibonacci

> The nth term and the sum of the first n terms of an arithmetic or geometric sequence, the sum to infinity, and exact Fibonacci numbers, with a table.

Versi interaktif: https://www.calcopenly.com/id/math/sequence-calculator
Topik: Kalkulator matematika

An arithmetic sequence adds a fixed difference d at each step, so aₙ = a₁ + (n − 1)d and the first n terms sum to Sₙ = n(a₁ + aₙ)/2. A geometric sequence multiplies by a fixed ratio r, so aₙ = a₁rⁿ⁻¹ and Sₙ = a₁(1 − rⁿ)/(1 − r); when |r| < 1 the sum to infinity is a₁/(1 − r). Fibonacci numbers start from F₀ = 0 and F₁ = 1, add the previous two, and are computed exactly to every digit.

Savings that grow by a fixed amount, compound growth and exam questions all use these formulas. The default, 3, 7, 11, …, reaches 39 at term 10, and those 10 terms sum to 210; the halving series 1 + 1/2 + 1/4 + … totals 1.998046875 after 10 terms and approaches 2.

A geometric series with |r| ≥ 1 has no finite sum to infinity, and an arithmetic series has one only when every term is 0. The Fibonacci sum F₁ + … + Fₙ equals Fₙ₊₂ − 1.

## Masukan

- **Sequence** (pilihan: Aritmetika, Geometric, Fibonacci)
- **First term a₁**
- **Common difference d**
- **Common ratio r**
- **Term number n**: Fibonacci counts from F₀ = 0, F₁ = 1; the others from a₁.

## Hasil

- nth term — hasil utama
- Sum of the first n terms
- Sum to infinity
- nth term, all digits
- Digits in the term

## Rumus

$$
\begin{gathered} a_n = a_1 + (n-1)d \\ S_n = \tfrac{n}{2}(a_1 + a_n) \\[10pt] a_n = a_1 r^{n-1} \\ S_n = a_1\frac{1 - r^n}{1 - r},\quad S_\infty = \frac{a_1}{1 - r} \\[10pt] F_n = F_{n-1} + F_{n-2} \end{gathered}
$$

## Contoh penyelesaian

### Arithmetic 3, 7, 11, … to 10 terms

- Sequence: Aritmetika
- First term a₁: 3
- Common difference d: 4
- Term number n: 10
- **nth term: 39**
- **Sum of the first n terms: 210**
- Sumber pemeriksaan: Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39

### Arithmetic with a negative step

- Sequence: Aritmetika
- First term a₁: 100
- Common difference d: -7.5
- Term number n: 20
- **nth term: -42.5**
- **Sum of the first n terms: 575**
- Sumber pemeriksaan: Python 3.8 decimal loop over 20 terms (forkA/verify_seq.py)

### Geometric 2, 6, 18, … to 6 terms

- Sequence: Geometric
- First term a₁: 2
- Common ratio r: 3
- Term number n: 6
- **nth term: 486**
- **Sum of the first n terms: 728**
- Sumber pemeriksaan: Python 3.8: 2·3⁵ = 486, Σ 2·3^k (k < 6) = 728

### Halving series 1 + 1/2 + 1/4 + …

- Sequence: Geometric
- First term a₁: 1
- Common ratio r: 0.5
- Term number n: 10
- **nth term: 0.001953125**
- **Sum of the first n terms: 1.998046875**
- **Sum to infinity: 2**
- Sumber pemeriksaan: Python fractions: Σ (1/2)^k for k < 10 = 1023/512; sum to infinity 1/(1 − 1/2) = 2

### Ratio 1 (edge)

- Sequence: Geometric
- First term a₁: 5
- Common ratio r: 1
- Term number n: 4
- **nth term: 5**
- **Sum of the first n terms: 20**
- Sumber pemeriksaan: With r = 1 every term is a₁, so Sₙ = n·a₁ = 20

### Fibonacci F₁₀₀

- Sequence: Fibonacci
- Term number n: 100
- **nth term: 3.54225 × 10²⁰**
- **nth term, all digits: 354224848179261915075**
- **Digits in the term: 21**
- **Sum of the first n terms: 9.27373 × 10²⁰**
- Sumber pemeriksaan: OEIS A000045 (F₁₀₀); sum F₁₀₂ − 1 from a Python 3.8 loop

## Pertanyaan

### How do you find the nth term of an arithmetic sequence?

Use aₙ = a₁ + (n − 1)d: start from the first term and add the common difference n − 1 times. For 3, 7, 11, … the difference is 4, so the 10th term is 3 + 9 × 4 = 39 and the 50th is 3 + 49 × 4 = 199. To find d from two terms, divide their difference by the gap in positions: (39 − 3)/(10 − 1) = 4.

### What is the formula for the sum of an arithmetic series?

Sₙ = n(a₁ + aₙ)/2, the number of terms times the average of the first and last term. For 3 + 7 + … + 39, which has 10 terms, S = 10 × (3 + 39)/2 = 210. The pairing idea behind it is often credited to the young Gauss, who summed 1 to 100 as 50 pairs of 101 to get 5,050.

### How do you find the sum of a geometric series?

For n terms, Sₙ = a₁(1 − rⁿ)/(1 − r) whenever r ≠ 1. For 2 + 6 + 18 + … with 6 terms, S = 2 × (1 − 3⁶)/(1 − 3) = 2 × (−728)/(−2) = 728. When r = 1 every term equals a₁, so Sₙ = n × a₁ and 5 + 5 + 5 + 5 = 20.

### When does a geometric series have a sum to infinity?

Only when the common ratio is strictly between −1 and 1. Then rⁿ shrinks toward 0 and the sum approaches a₁/(1 − r): 1 + 1/2 + 1/4 + … = 1/(1 − 1/2) = 2, and 0.9 + 0.09 + 0.009 + … = 0.9/(1 − 0.1) = 1, which is why 0.999… equals 1. With |r| ≥ 1 the terms do not shrink, so the series grows without bound or oscillates.

### What is the 100th Fibonacci number?

F₁₀₀ = 354,224,848,179,261,915,075, a 21-digit number, counting from F₀ = 0 and F₁ = 1 as in OEIS A000045. Consecutive Fibonacci numbers grow by a factor approaching the golden ratio φ ≈ 1.618034, so Fₙ is close to φⁿ/√5 and each term adds about 0.209 digits.

### Seberapa akurat “Kalkulator barisan aritmetika, geometri, dan Fibonacci”?

Akurasi bergantung pada masukan dan asumsi metode. Aritmetika desimal memakai 50 digit signifikan, tetapi perkiraan, metode numerik dan data sumber bisa kurang presisi; pembulatan yang ditampilkan tidak menghilangkan batasan itu. Contoh penyelesaian yang diperiksa dengan sumber independen: 7. Misalnya, “Arithmetic 3, 7, 11, … to 10 terms” diperiksa dengan Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39.

### Dari mana metode ini berasal?

Abramowitz & Stegun, Handbook of Mathematical Functions, §3.6 (series); OEIS A000045 — Fibonacci numbers; Wikipedia — Geometric series.

## Sumber

- Abramowitz & Stegun, Handbook of Mathematical Functions, §3.6 (series)
- [OEIS A000045 — Fibonacci numbers](https://oeis.org/A000045)
- [Wikipedia — Geometric series](https://en.wikipedia.org/wiki/Geometric_series)