Skip to main content
🔢 Series

Sum of Consecutive Integers Calculator

Add every whole number from one value to another. Get the total, the count of terms, and the average — with the arithmetic-series formula shown.

✓ Sum a to b
✓ 1 to n shortcut
✓ Term count
✓ Average
✓ 100% Free
🔢 Open All Math Calculators 📖 Read the Guide

Sum of integers — Quick answer

Multiply the number of terms by the average of the first and last.

sum = n × (a + b) ÷ 2  where  n = b − a + 1  ·  1…n = n(n+1)/2

Worked example: 1 to 100 → 100 × 101 ÷ 2 = 5050.

Examples

RangeTermsSum
1 to 101055
5 to 1511110
1 to 10001000500500

The average is just the midpoint (a + b) ÷ 2.

🔢 Sum of Consecutive Integers

Enter the first and last whole number of the range.

Sum
—
Number of terms
—
Average
—
Range
—

ℹ️ Both endpoints are included. If you enter them in reverse order, they're swapped automatically.

Standards & method

✓ Independently verified 12 July 2026
Basis
First principles
Method
Gauss’s summation formula for an arithmetic series.
Core formula
Σ(1..n) = n(n+1)/2  ·  general: S = n(a₁ + aₙ)/2
Why this matters
Closed form — no loop needed. The sum of the first n integers grows with n², not n, which is why naive O(n²) algorithms blow up.
Independently verified
12 July 2026 — Formula re-derived from first principles and verified numerically against hand-computed reference cases, including edge cases and unit handling.

Results are for guidance. Verify against the current edition of the governing standard and have a qualified professional review before use in practice.

The sum of consecutive integers from a to b is an arithmetic series: it equals the number of terms times the average of the first and last value, sum = n × (a + b) ÷ 2. For the special case 1 to n this becomes n(n + 1)/2. This calculator returns the total, the term count, and the average.

Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: arithmetic series formula, recomputed in code.

The formula

Arithmetic series
sum = n × (a + b) ÷ 2,   n = b − a + 1   (and 1…n = n(n+1)/2)

The idea is Gauss's pairing trick: line up the range and add the smallest to the largest, the next-smallest to the next-largest, and so on. Every pair sums to a + b, and there are n/2 pairs, giving n(a + b)/2. Because the average of an evenly spaced list is just the midpoint of its ends, the sum is simply count × average.

Worked examples

1 to 100:

Gauss's sum
n = 100 · (1 + 100) ÷ 2 = 50.5 · 100 × 50.5 = 5050

5 to 15:

11 terms
n = 15 − 5 + 1 = 11 · 11 × (20) ÷ 2 = 110

−3 to 3 (symmetric):

Cancels to 0
7 terms · midpoint 0 · 7 × 0 = 0

The average is always the midpoint of the two ends, so for 1 to 100 it's 50.5 and for 5 to 15 it's 10. A symmetric range around zero, like −3 to 3, sums to 0 because the positives and negatives cancel.

Frequently Asked Questions

How do I sum integers from a to b?⌄

sum = n × (a + b) ÷ 2, with n = b − a + 1. 5 to 15: 11 × 20 ÷ 2 = 110.

What is the sum of 1 to 100?⌄

5050, from 100 × 101 ÷ 2 — Gauss's famous result.

What is the 1 to n formula?⌄

n(n+1)/2. For n = 10: 10 × 11 ÷ 2 = 55.

Why does the pairing trick work?⌄

Each first+last pair sums to a + b, and there are n/2 pairs → n(a+b)/2.

Can it include negatives?⌄

Yes. −3 to 3 has 7 terms summing to 0, with average 0.

Need more math tools?

Explore arithmetic and geometric sequences, midrange, quotient & remainder and more across the AI Calculator math suite.

🔢 Open Math Calculators — Free

No registration required · 350+ calculators · PDF report export