حاسبة المتتاليات الحسابية والهندسية وفيبوناتشي

Q: 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.

Q: 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.

Q: 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.

Q: 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.

Q: 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.

Q: ما مدى دقة «⁨حاسبة المتتاليات الحسابية والهندسية وفيبوناتشي⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 7. مثلًا، يجري التحقق من «⁨Arithmetic 3, 7, 11, … to 10 terms⁩» بالرجوع إلى ⁨Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39⁩.

Q: ما مصدر هذه الطريقة؟

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

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.

آخر تحديث 28 سبتمبر 2026أمثلة تم التحقق منها: 7

nth term

الحد الأقصى للمنازل العشرية: 10؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر

Sum of the first n terms: 210

Starting at 3 and adding 4 each time, term 10 is 39 and the first 10 terms add up to 210.

Terms and running sum

Term aₖRunning sum

الشروط الصفوف: 10

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

طريقة الحساب S

nth term
$a_n = a_1 + (n - 1)d = 3 + (10 - 1) \times 4 = 39$
Sum of the first n terms
$S_n = \frac{n}{2}(a_1 + a_n) = \frac{10}{2}\left(3 + 39\right) = 210$
Pair the first and last terms: each pair has the same total.
Sum to infinity
An arithmetic series with a non-zero term or step has no finite sum to infinity.

حول حاسبة المتتاليات الحسابية والهندسية وفيبوناتشي

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.

أمثلة محلولة

Arithmetic 3, 7, 11, … to 10 terms

Sequence: الحساب
First term a₁: 3
Common difference d: 4
Term number n: 10

nth term: 39
Sum of the first n terms: 210

مصدر التحقق: ⁨Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39⁩

Arithmetic with a negative step

Sequence: الحساب
First term a₁: 100
Common difference d: -7.5
Term number n: 20

nth term: -42.5
Sum of the first n terms: 575

مصدر التحقق: ⁨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

مصدر التحقق: ⁨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

مصدر التحقق: ⁨Python fractions: Σ (1/2)^k for k < 10 = 1023/512; sum to infinity 1/(1 − 1/2) = 2⁩

الأسئلة

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.

ما مدى دقة «⁨حاسبة المتتاليات الحسابية والهندسية وفيبوناتشي⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 7. مثلًا، يجري التحقق من «⁨Arithmetic 3, 7, 11, … to 10 terms⁩» بالرجوع إلى ⁨Python 3.8: sum(3 + 4k for k in range(10)) = 210, last term 39⁩.

ما مصدر هذه الطريقة؟

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

حول هذه الحاسبة

\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}

المصادر

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

تم التحقق بالرجوع إلى المصادر

تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 7. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.