Consecutive Integers Formula n, n+1, n+2 Examples

Key Points

• The consecutive integers formula writes a run of integers as $n$, $n+1$, $n+2$, each one more than the last.

• Consecutive even or odd integers step by 2 instead, so they are written $n$, $n+2$, $n+4$.

• The sum of $k$ consecutive integers starting at $n$ is $kn + \frac{k(k-1)}{2}$, which follows from adding the step amounts.

• For an odd count of consecutive integers, the sum equals the count times the middle value, which turns many problems into one division.

• Only the first term needs a variable, since every other term is fixed relative to it.

What Is the Consecutive Integers Formula?

The consecutive integers formula is a way of setting up a problem rather than a value to compute. Three consecutive integers are written:

$$n, \quad n+1, \quad n+2$$

For consecutive even or consecutive odd integers the gap is 2 rather than 1:

$$n, \quad n+2, \quad n+4$$

Why Only the First Term Needs a Variable

A run of consecutive integers has no freedom beyond its starting point. Once the first value is fixed, every later one is determined, so a single unknown describes the whole run.

Using two variables would overstate the freedom and leave the problem underdetermined. Writing three integers as $a$, $b$, $c$ loses the very relationship the word "consecutive" supplies.

That is why the setup carries the work. Translating "three consecutive integers sum to 42" into $n + (n+1) + (n+2) = 42$ reduces a sentence to a one-variable equation, and the rest is routine.

Variable Key

Examples of the Consecutive Integers Formula

Example 1

Three consecutive integers sum to 42. Find them.

$$n + (n+1) + (n+2) = 42 \implies 3n + 3 = 42 \implies n = 13$$

Final answer: 13, 14 and 15

Example 2

Three consecutive even integers sum to 48. Find them, using $n$, $n+1$, $n+2$.

Setting them up as $n$, $n+1$, $n+2$ gives $3n+3 = 48$, so $n = 15$ and the integers would be 15, 16 and 17.

But 15 and 17 are odd, so that run is not three even integers at all. The setup allowed a step of 1, while consecutive even numbers are 2 apart.

Using the correct step:

$$n + (n+2) + (n+4) = 48 \implies 3n + 6 = 48 \implies n = 14$$

Final answer: 14, 16 and 18. The step size is part of the problem statement, not a detail.

Example 3

Five consecutive integers sum to 90. Find them using the middle-term shortcut.

With an odd count, the sum is the count times the middle value:

$$90 \div 5 = 18$$

Final answer: 16, 17, 18, 19, 20, with 18 as the middle term.

The shortcut works because the terms either side of the middle cancel: 16 is 2 below and 20 is 2 above.

Example 4

The sum of four consecutive integers is 70. Find them.

With an even count there is no middle term, so use the setup:

$$4n + 6 = 70 \implies n = 16$$

Final answer: 16, 17, 18 and 19

Note that 70 divided by 4 is 17.5, which is the average of the run but not a member of it. That is the structural difference between odd and even counts.

Example 5

Two consecutive odd integers have a product of 143. Find them.

$$n(n+2) = 143 \implies n^2 + 2n - 143 = 0 \implies (n+13)(n-11) = 0$$

Final answer: $n = 11$, giving 11 and 13. The root $n = -13$ gives $-13$ and $-11$, which also multiply to 143 and is a valid second answer.

Three Setups Compared

The even and odd rows share a setup, which surprises people. Nothing in $n, n+2, n+4$ forces evenness, so the parity comes from the solution rather than from the algebra.

Consecutive Integers, Where Things Go Sideways

Mistake 1: Using a step of 1 for even or odd runs

Where it slips in: The phrase "consecutive" suggests a gap of one, and the word "even" gets treated as decoration.

Don't do this: $n, n+1, n+2$ for consecutive even integers.

The correct way: $n, n+2, n+4$. Check the answer's parity: if a run of supposedly even integers contains an odd number, the step was wrong.

Students who write out a concrete example of the run before introducing algebra catch this immediately, because writing 14, 16, 18 makes the gap of 2 obvious. Going straight to symbols is what hides the step size.

Mistake 2: Introducing a variable for every term

Where it slips in: Three unknown numbers seem to need three unknowns.

Don't do this: Let the integers be $a$, $b$ and $c$.

The correct way: Use $n$, $n+1$, $n+2$. The word consecutive supplies the relationships, and discarding them leaves an equation that cannot be solved.

Mistake 3: Using the middle-term shortcut on an even count

Where it slips in: The shortcut is quick and feels general.

Don't do this: Divide 70 by 4 and expect a member of the run.

The correct way: The shortcut needs an odd count, where a middle term exists. For an even count the division gives the average, which falls between two terms, as 17.5 does in Example 4.

Bottom Line

• The consecutive integers formula is the setup $n$, $n+1$, $n+2$, using one variable for the whole run.

• Even and odd runs step by 2 and share the setup $n$, $n+2$, $n+4$.

• For an odd count, the sum divided by the count gives the middle term directly.

• An even count has no middle term, so the division returns an average that is not in the run.

• To work through word problems with a teacher, explore a 7th grade math tutor or math tutoring.

Where to Go From Here

Find four consecutive integers summing to 102. Then find three consecutive odd integers summing to 81, and check the parity of your answer before accepting it.

Want a live Bhanzu trainer to walk through more algebra problems? Book a free demo class.

Read More

• What Is an Integer

• What Is a Natural Number

• Sum of Natural Numbers

• Integers Formula

• Greatest Integer Function