Algebra • Sequences & Series Flagship

Arithmetic Sequence Calculator

Solve arithmetic progressions (AP) for any target $n$-th term ($a_n$), common difference ($d$), initial term ($a_1$), or partial series sum ($S_n$). Generates explicit formulas, discrete SVG progression charts, and cumulative sum tables.

Reviewed by Sanjay Samanta & Academic Review Board

Verified Gauss Summation Proofs

Last Updated: September 2026

Sequence & Series Solver

Calculate nth terms, finite partial sums, infinite series convergence, and explicit formulas.

Input Mode:

Quick Presets:

First Term (a₁)

Common Difference (d)

Target Term Index (n)

Arithmetic Progression Solution (AP)

Calculated nth-term value, partial series sum, and closed-form algebraic formulas.

Nth Term (a₁₀) Target

a₁₀ = 3 + 9(4)

Sum of First n Terms (S₁₀) Series Sum

210

S₁₀ = (10/2)(3 + 39)

Common Difference (d) Step

Constant delta

Initial Value (a₁) Base

Starting term at n=1

Explicit Formula

a_n = 3 + 4(n - 1) = 4n - 1

Recursive Formula

a₁ = 3, a_n = a_{n-1} + 4

Discrete Sequence Progression Chart

First 8 Terms

Sequence Points $(k, a_k)$

Linear Progression (+d)

Sequence Terms Table

Cumulative Sum ($S_k$)

n	Term (a_n)	Sum (S_n)

Direct Answer & Overview

Verified Educational Guide

Arithmetic Sequence Definition & Formula

An arithmetic sequence is an ordered progression of numbers where the difference between consecutive terms is constant. This constant is called the common difference (d).

Primary Mathematical Formula Standard Mathematical Model

Standard Equation

ƒ(x)

Q.E.D.

Nth Term: a_n = a_1 + (n − 1)d, Series Sum: S_n = (n / 2)(a_1 + a_n) = (n / 2)[2a_1 + (n − 1)d]

Evaluated with exact mathematical formulation • Rigorously verified

Exact Formula

Input Parameters

Required

First Term (a₁)

Common Difference (d)

Target Term Index (n)

Alternatively: Two known terms (a_p, a_q) or comma-separated list

Expected Outputs

Calculated

Target Nth Term (a_n)

Partial Series Sum (S_n)

Explicit algebraic formula

Recursive algebraic formula

Discrete sequence progression plot

Worked Numerical Example

Instant Verification

Find the 10th term and sum of first 10 terms for sequence 3, 7, 11, 15, ...

→ First term a_1 = 3, difference d = 4. 10th Term a_10 = 3 + (10 - 1)(4) = 3 + 36 = 39. Sum S_10 = (10 / 2)(3 + 39) = 5(42) = 210.

a_10 = 39, S_10 = 210

Arithmetic Progression Foundations & Common Difference

An arithmetic progression (AP) is a sequence of real numbers where each successive term after the first is generated by adding a constant scalar quantity $d$, known as the common difference:

$$d = a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1}$$

Increasing AP (d > 0)

Each term grows larger than the preceding term (e.g. $2, 5, 8, 11\dots$ with $d = +3$).

Decreasing AP (d < 0)

Each term is smaller than the preceding term (e.g. $50, 42, 34, 26\dots$ with $d = -8$).

Constant AP (d = 0)

Every term is identical to the first term ($a_1, a_1, a_1\dots$ with zero slope).

Derivation of the Nth-Term Explicit Formula

Rather than iteratively calculating dozens of intermediate steps to find a distant term, we derive a closed-form formula by inspecting the additive pattern from the initial term $a_1$:

Term 1 ($n=1$): $a_1 = a_1 + 0d$

Term 2 ($n=2$): $a_2 = a_1 + 1d$

Term 3 ($n=3$): $a_3 = a_2 + d = (a_1 + d) + d = a_1 + 2d$

Term 4 ($n=4$): $a_4 = a_3 + d = (a_1 + 2d) + d = a_1 + 3d$

Term $n$: $a_n = a_1 + (n - 1)d$

Notice that to reach the $n$-th position from the 1st position, exactly $n - 1$ steps of magnitude $d$ are traversed.

Arithmetic Series Sum & Gauss Pairing Proof

The sum of the first $n$ terms of an arithmetic progression is known as an arithmetic series ($S_n$). The legendary mathematician Carl Friedrich Gauss demonstrated an elegant pairing proof for this sum:

Gauss's Two-Way Addition Proof

Write the sum $S_n$ forwards:
$S_n = a_1 + (a_1 + d) + (a_1 + 2d) + \dots + (a_n - d) + a_n$
Write the identical sum $S_n$ backwards:
$S_n = a_n + (a_n - d) + (a_n - 2d) + \dots + (a_1 + d) + a_1$
Add the two equations vertically term-by-term. In each column, the $\pm k d$ offsets cancel:
$2S_n = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + \dots + (a_1 + a_n) = n(a_1 + a_n)$
Divide by 2 to isolate $S_n$:
$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}[2a_1 + (n - 1)d]$

Solving from Known Parameters (2-Term Reconstruction)

Case 1 • Two Arbitrary Terms Given

Index Step Division

Given $a_p$ at position $p$ and $a_q$ at position $q$:

$d = \frac{a_q - a_p}{q - p}$

$a_1 = a_p - (p - 1)d$

Case 2 • Sum and Term Count Given

Average Back-Calculation

Given sum $S_n$ and term count $n$:

$\text{Average} = \frac{S_n}{n} = \frac{a_1 + a_n}{2}$

$a_n = \frac{2S_n}{n} - a_1$

Real-World Applications (Depreciation, Physics & Finance)

Straight-Line Asset Depreciation

Corporate accounting depreciates capital machinery by equal yearly write-downs ($d = -\text{depreciation}$). Book value at year $t$ forms a decreasing arithmetic progression.

Stadium & Theater Seating Layout

Concert halls expand radially, adding a constant number of additional seats in each successive row ($d = +4$ seats/row). Total capacity equals the arithmetic series sum $S_n$.

Kinematics & Linear Acceleration

Under constant acceleration $a$, velocities sampled at equal time intervals $\Delta t$ form an arithmetic sequence with step $d = a\Delta t$.

Graded Step-by-Step Numerical Solutions

Example 1 • Standard Calculation Basic Tier

Find the 20th term and sum of first 20 terms for $a_1 = 5, d = 3$.

1. Compute $a_{20}$: $a_{20} = 5 + (20 - 1)(3) = 5 + 19(3) = 5 + 57 = 62$.

2. Compute $S_{20}$: $S_{20} = \frac{20}{2}(5 + 62) = 10(67) = 670$.

Example 2 • Two-Term Reconstruction Intermediate Tier

In an arithmetic sequence, $a_4 = 17$ and $a_9 = 42$. Find $a_1$ and $a_{15}$.

1. Find difference $d$: $d = \frac{42 - 17}{9 - 4} = \frac{25}{5} = 5$.

2. Find first term $a_1$: $a_1 = 17 - (4 - 1)(5) = 17 - 15 = 2$.

3. Find 15th term $a_{15}$: $a_{15} = 2 + (15 - 1)(5) = 2 + 70 = 72$.

Common Calculation Pitfalls & Mistake Avoidance

Pitfall 1: Off-by-One Exponent/Multiplier Error

Never multiply $d$ by $n$ directly. The first term $a_1$ already includes 0 additions of $d$, so the $n$-th term requires exactly $(n - 1)d$. Multiplying by $n$ mistakenly evaluates the $(n+1)$-th term.

Pitfall 2: Forgetting to Retain Negative Signs in Decreasing Sequences

When $d < 0$, ensure the negative sign is preserved throughout Gauss’s summation formula. A negative difference causes the series sum to eventually peak and decrease as terms become negative.

Fact-Checked & Verified • Computational Accuracy Standards

Updated July 2026 • Editorial Policy

Authored By

Sanjay Samanta

Lead Developer & Founder of Basic Math Tools. Specializes in browser-native computational algorithms and applied mathematics.

Reviewed & Verified By

Academic Review Board

Mathematics & curriculum specialists. Audited against standard algebraic and arithmetic principles.

Found an error or have an improvement suggestion? Report a calculation issue

Frequently Asked Questions

What is the formula for the nth term of an arithmetic sequence?

The explicit formula is a_n = a_1 + (n − 1)d, where a_1 is the first term, d is the common difference, and n is the position index of the term.

How do you find the sum of an arithmetic series (S_n)?

Use Gauss’s arithmetic series sum formula: S_n = (n / 2)(a_1 + a_n). Alternatively, substituting a_n yields S_n = (n / 2)[2a_1 + (n − 1)d].

How do you find the common difference (d) given any two arbitrary terms?

If you are given term a_p at index p and term a_q at index q, the common difference is calculated as d = (a_q − a_p) / (q − p). The first term is then a_1 = a_p − (p − 1)d.

Can the common difference of an arithmetic sequence be negative or fractional?

Yes. A negative common difference (d < 0) generates a strictly decreasing sequence (e.g. 100, 93, 86, 79 with d = −7). A fractional difference creates sequences with non-integer steps (e.g. 1, 1.5, 2, 2.5 with d = 0.5).

What is the difference between an explicit and recursive arithmetic formula?

An explicit formula (a_n = a_1 + (n − 1)d) allows calculating any term directly from index n without knowing previous values. A recursive formula (a_1 = given, a_n = a_{n-1} + d) defines each term by adding d to the immediately preceding term.