“Three consecutive integers add to 48 — find them.” This Consecutive Integers Calculator solves the classic algebra setup instantly: choose plain consecutive, consecutive even, or consecutive odd integers, say how many and what they must sum to, and it returns the sequence — or explains exactly why no such sequence exists.
Worked Example: Three Consecutive Integers Summing to 48
The calculator returns 15, 16, 17 — and 15 + 16 + 17 = 48, as required. The algebra behind it: call the first integer x, so the sequence is x, x + 1, x + 2 and the sum is 3x + 3 = 48, giving x = 15. Notice the shortcut hiding there: the sum of a consecutive run is the count times the middle value (for odd counts), 3 × 16 = 48 — the middle term is the average, which is why dividing the target by the count is always the first move. For even counts there is no single middle term, but the sum is still the count times the average of the two middle terms, and the calculator’s solve handles both shapes through the same formula.
The General Formula
For n consecutive integers starting at x with step d (1 for plain, 2 for even/odd runs), the sum is S = n·x + d·n(n−1)/2, so the first term is x = (S − d·n(n−1)/2) / n. A valid sequence exists exactly when that division comes out whole and, for the even and odd modes, when x has the required parity. Those two conditions explain the classic impossibility results: three consecutive integers can never sum to 50 (x would be 47/3), and two consecutive even integers can never sum to an odd number (their sum is always even — indeed always divisible by 2 but never odd). Sums of plain consecutive runs connect to triangular numbers: 1 + 2 + … + n = n(n+1)/2, and any later run is a difference of two triangular numbers. The formula also settles which representations exist: for 48, testing each run length k in x = (48 − k(k−1)/2)/k gives a whole positive x only for k = 3 (x = 15) — lengths 2, 4, 5 and 6 all strand x on a fraction. So 15 + 16 + 17 is the unique representation of 48 as a sum of consecutive positive integers (besides the trivial one-term “run”), a neat demonstration of how restrictive the divisibility condition really is.
Word Problems This Solves
Consecutive-integer setups are the training ground of algebra word problems: page numbers on facing pages (sum given, find both), ages in sequence, seats in a row, odd-numbered houses on one side of a street (step 2, odd mode), even-numbered jersey numbers, and calendar runs. The pattern to recognize is always “numbers equally spaced, total known” — once you name the first term x and the step d, the formula does the rest. The same machinery answers feasibility questions contest writers love: is a target achievable at all, and if so, by how many different run lengths?
Common Mistakes
Modeling even or odd runs with step 1: consecutive even integers differ by 2, so the run is x, x + 2, x + 4 — forgetting the step doubles back as a wrong answer that often still “looks” reasonable. Assuming every target works: divisibility and parity quietly veto many sums, and a non-integer x is the algebra telling you so. Using the middle-term shortcut for even counts without adjusting: with n even, the “middle” is the average of the two central terms, a half-integer for plain runs, so x = S/n − (n−1)/2 lands a half-step off if you round carelessly. And reporting the sequence without re-adding it — ten seconds of checking catches every one of these slips.
Even and Odd Runs, Worked
Two examples show the step-2 machinery behaving. Four consecutive even integers summing to 68: the formula gives x = (68 − 2 × 4 × 3/2) / 4 = (68 − 12)/4 = 14, so the run is 14, 16, 18, 20 — adding to 68 exactly. Five consecutive odd integers summing to 75: the center is the average, 75 / 5 = 15, and the run is built symmetrically around it in steps of 2: 11, 13, 15, 17, 19 — again 75 on re-addition. Note how the average shortcut is exact for odd counts in every mode, because the run is symmetric about its middle term; for even counts the average falls between the two central terms, and the full formula earns its place.
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Frequently Asked Questions
What three consecutive integers sum to 48?
15, 16 and 17. The middle term is the average, 48 ÷ 3 = 16, and the run is built symmetrically around it.
How do I write consecutive even integers algebraically?
As x, x + 2, x + 4, … — the step is 2, and x itself must be even. Odd runs use the same step with x odd.
Why can some sums not be formed at all?
The first term x = (S − d·n(n−1)/2)/n must come out a whole number with the right parity. When it does not, no sequence exists — for instance, no three consecutive integers sum to 50.
What is the fastest way to find a consecutive run for a sum?
Divide the sum by the count to find the center, then build outward. For odd counts the center is the middle term itself; the calculator applies the exact formula for every case.
Can consecutive integers be negative?
Yes — runs like −2, −1, 0, 1 are perfectly consecutive, and the formula handles negative targets and terms without modification.
How many ways can 48 be written as a sum of consecutive positive integers?
Only one way: 15 + 16 + 17. Each candidate run length imposes a divisibility condition, and for 48 only length 3 satisfies it with positive terms.