AFCAT Numerical Ability · Average

Consecutive Numbers: the Middle Term Does the Work

Consecutive integers, even numbers, odd numbers and multiples are evenly spaced, so their average is the middle term and every other term is a fixed number of steps away from it.

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Why this matters

About one question in eight here is HARD, and nearly all of them use one idea in different directions. Marks go when the count is even, because then no single number sits in the middle, and when a step of 2 (even or odd numbers) is treated as a step of 1.

Concept 1 of 4: An odd count: the average is the middle number

Five consecutive numbers balance on the third one. So the average is the middle number, and the others step away from it on both sides.

Definition

Consecutive integers step by 1; consecutive even or odd numbers step by 2; consecutive multiples of $k$ step by $k$ .
With an odd count $n$ , the middle term equals the average $A$ . The largest is $A + \tfrac{n - 1}{2}d$ and the smallest is $A - \tfrac{n - 1}{2}d$ , where $d$ is the step.
Given the sum instead of the average, divide by the count first.

Ends of an odd-length run

\text{largest} = A + \dfrac{n - 1}{2}\,d, \qquad \text{smallest} = A - \dfrac{n - 1}{2}\,d

Worked example

The average of 5 consecutive multiples of 4 is 48. Find the smallest and the largest.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Average · Consecutive Numbers

The average of

9

consecutive odd numbers is

15

. Find the smallest number:

Use the right step

Five consecutive odd numbers averaging 31 run from 27 to 35, not from 29 to 33. Odd and even numbers step by 2.

Drill 29 more on an odd count: the average is the middle number

Concept 2 of 4: An even count: halfway between the two middle numbers

Six consecutive numbers have no middle number. The average falls halfway between the third and the fourth, so for consecutive integers it ends in .5.

Definition

With an even count $n$ , the two middle terms are $A - \tfrac d2$ and $A + \tfrac d2$ .
The same formula still gives the ends: largest $= A + \tfrac{n - 1}{2}d$ , where $\tfrac{n - 1}{2}$ is now a half-number such as 2.5.
Parity check: an even count of consecutive odd numbers always averages an even number, and an even count of consecutive even numbers averages an odd number.
Or name the first term $a$ and solve $a + \tfrac{n - 1}{2}d = A$ .

Ends of an even-length run

\text{largest} = A + \dfrac{n - 1}{2}\,d \quad (n \text{ even, so } \tfrac{n - 1}{2} \text{ ends in } .5)

Worked example

The average of 6 consecutive odd numbers is 40. Find the largest.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Average · Consecutive Numbers

If the average of

8

consecutive even numbers is

19

, what is the largest number?

Check the parity

Six consecutive odd numbers can never average 25: an even count of odd numbers averages an even number. A question like that cannot be solved as printed.

Drill 24 more on an even count: halfway between the two middle numbers

Concept 3 of 4: Pairs from the two ends

In an evenly spaced run, the first and the last add up to twice the average, and so do the second and the second-last. Pairs taken from the two ends always balance.

Definition

First + last = 2 × average, whatever the count.
Any two terms the same distance from the two ends also add up to 2 × average.
Largest − smallest = (n − 1) × step, whatever the average.
Sum = average × count, so the sum of 5 consecutive numbers is 5 times the middle one.

End pairs and spread

\text{first} + \text{last} = 2\bar x, \qquad \text{last} - \text{first} = (n - 1)\,d

Worked example

The average of 11 consecutive odd numbers is 63. Find the sum of the largest and the smallest, and their difference.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Average · Consecutive Numbers

The sum of first and last of

9

consecutive numbers is

56

. What is their average?

The spread does not depend on the average

Any 12 consecutive integers span 11 from the smallest to the largest, not 12: there are 11 gaps between 12 numbers.

Drill 14 more on pairs from the two ends

Concept 4 of 4: Letters, products, and a run that changes

When the average is a letter, write the numbers around it:

x - 2, x - 1, x, x + 1, x + 2

. When a term is added or removed, go back to totals.

Definition

Centre the run on its middle term: five consecutive numbers averaging $x$ are $x - 2$ to $x + 2$ . A run that starts at $m$ with 11 terms averages $m + 5$ .
Products: consecutive numbers whose product is given sit near its root. For three of them try near the cube root: $11 \times 12 \times 13 = 1716$ , and $\sqrt[3]{1716}$ is just under 12. For two, try near the square root.
Adding or removing a term: total = average × count, add or remove the term, divide by the new count.

A run centred on its average

x - 2,\; x - 1,\; x,\; x + 1,\; x + 2 \quad (\text{five terms averaging } x)

Worked example

The average of 11 consecutive integers starting from

m

is 40. Find

m

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Average · Consecutive Numbers

The product of

3

consecutive numbers is

504

. What is their average?

Centre the run, do not start it

Writing five consecutive numbers as

x, x + 1, \dots, x + 4

makes

x

the smallest, not the average. Centring them on

x

keeps the average equal to

x

Drill 13 more on letters, products, and a run that changes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

An odd count: the average is the middle number
Ends of an odd-length run
$\text{largest} = A + \dfrac{n - 1}{2}\,d, \qquad \text{smallest} = A - \dfrac{n - 1}{2}\,d$
An even count: halfway between the two middle numbers
Ends of an even-length run
$\text{largest} = A + \dfrac{n - 1}{2}\,d \quad (n \text{ even, so } \tfrac{n - 1}{2} \text{ ends in } .5)$
Pairs from the two ends
End pairs and spread
$\text{first} + \text{last} = 2\bar x, \qquad \text{last} - \text{first} = (n - 1)\,d$
Letters, products, and a run that changes
A run centred on its average
$x - 2,\; x - 1,\; x,\; x + 1,\; x + 2 \quad (\text{five terms averaging } x)$

Watch out for (4)

Use the right step
→ An odd count: the average is the middle number
Check the parity
→ An even count: halfway between the two middle numbers
The spread does not depend on the average
→ Pairs from the two ends
Centre the run, do not start it
→ Letters, products, and a run that changes

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