Arithmetic 3, 7, 11, … to 10 terms
- Sequence
- Arithmétique
- First term a₁
- 3
- Common difference d
- 4
- Term number n
- 10
- nth term
- 39
- Sum of the first n terms
- 210
Source de vérification : Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39
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.
Mis à jour Exemples vérifiés : 7
Starting at 3 and adding 4 each time, term 10 is 39 and the first 10 terms add up to 210.
| k | Term aₖ | Running sum Sₖ |
|---|---|---|
| 1 | 3 | 3 |
| 2 | 7 | 10 |
| 3 | 11 | 21 |
| 4 | 15 | 36 |
| 5 | 19 | 55 |
| 6 | 23 | 78 |
| 7 | 27 | 105 |
| 8 | 31 | 136 |
| 9 | 35 | 171 |
| 10 | 39 | 210 |
Pair the first and last terms: each pair has the same total.
An arithmetic series with a non-zero term or step has no finite sum to infinity.
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.
Source de vérification : Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39
Source de vérification : Python 3.8 decimal loop over 20 terms (forkA/verify_seq.py)
Source de vérification : Python 3.8: 2·3⁵ = 486, Σ 2·3^k (k < 6) = 728
Source de vérification : Python fractions: Σ (1/2)^k for k < 10 = 1023/512; sum to infinity 1/(1 − 1/2) = 2
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.
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.
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.
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.
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.
La précision dépend de vos données et des hypothèses de la méthode. Le calcul décimal utilise 50 chiffres significatifs, mais les estimations, méthodes numériques et données sources peuvent être moins précises ; l’arrondi affiché ne supprime pas ces limites. Exemples résolus vérifiés à partir de sources indépendantes : 7. Par exemple, « Arithmetic 3, 7, 11, … to 10 terms » est vérifié à l’aide de Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39.
Abramowitz & Stegun, Handbook of Mathematical Functions, §3.6 (series); OEIS A000045 — Fibonacci numbers; Wikipedia — Geometric series.
Ce calculateur comprend 7 exemples résolus dont les réponses proviennent de sources indépendantes. Ils font partie de la suite de tests et peuvent aussi être exécutés ici.
Raise a number to any exponent, including fractional and negative ones, or take its square, cube or nth root, with exact simplified radicals such as 5√2.
The logarithm of a number to any base, including ln and log₁₀, with change-of-base steps, or the exponent x that solves bˣ = y, exact when rational.