SSC CGL Maths · Averages

Averages of Consecutive Numbers

Numbers that go up in equal steps average to their middle term, which is also the mean of the first and the last.

Why this matters

Twelve questions in the bank, eight EASY and four MODERATE: consecutive or evenly spaced numbers, multiples of a number in a range, and the first n natural numbers or their squares. Each falls to one formula, with no adding up at all.

Concept 1 of 3: Average of numbers in equal steps

In 3, 5, 7, 9, 11 the first and last pair up to 14, and so do the second and fourth. Every pair averages 7, the middle term, so the whole list does too.

Definition

Numbers that rise by the same step form an arithmetic progression (AP): consecutive numbers, consecutive even or odd numbers, ages of children born at equal gaps.
Average of an AP $=$ middle term $= \frac{\text{first} + \text{last}}{2}$ .
Sum $=$ count $\times$ middle term. Given the sum, the middle term is sum $\div$ count, then step out to the ends.
An even count has two middle terms and the average lies halfway between them. So six consecutive integers always average to a number ending in .5, and four consecutive odd numbers average to an even number (5, 7, 9, 11 average 8).
For an even count, write the terms as $a, a + d, a + 2d, \ldots$ and solve for $a$ .

Average of an arithmetic progression

\text{Average} = \frac{\text{first} + \text{last}}{2} \qquad \text{Sum} = \text{count} \times \text{middle term}

Worked example

The sum of seven consecutive odd numbers is 147. What is the largest of them?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q64Easy

Example 1 · Averages · Averages of Consecutive Numbers

The average of a group of 8 consecutive even numbers is 41. What is the sum of the smallest and largest numbers in the group?

Expecting a whole-number average from an even count of consecutive integers

Six consecutive integers such as 4 to 9 average 6.5: an even count of consecutive integers always averages to a number ending in .5.

Taking a middle term that does not exist

With an even count there is no single middle term. Six consecutive even numbers averaging 25 have no term equal to 25; the two middle terms are 24 and 26.

Drill 4 more on average of numbers in equal steps

Concept 2 of 3: Average of the multiples of a number in a range

The multiples of 13 go up in steps of 13, so they form an arithmetic progression. Find the first and the last multiple in the range and take their mean: you never need to list or count them.

Definition

First multiple: divide the lower end by $k$ and round up. Last multiple: divide the upper end by $k$ and round down.
Average $= \frac{\text{first multiple} + \text{last multiple}}{2}$ , or $k \times$ the mean of the two multipliers.
'Between' normally leaves the ends out. If both ends are themselves multiples of $k$ , adding or dropping both leaves the average unchanged.
The count is needed only for a sum: count $= \frac{\text{last} - \text{first}}{k} + 1$ .

Multiples of k in a range

\text{Average} = \frac{\text{first multiple} + \text{last multiple}}{2} \qquad \text{Count} = \frac{\text{last} - \text{first}}{k} + 1

Worked example

Find the average of all numbers between 200 and 400 that are divisible by 17.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 12 Sep 2025, Shift 1 · Q57Easy

Example 2 · Averages · Averages of Consecutive Numbers

What is the average of all integers between 100 and 250 that are exactly divisible by 11?

Averaging the ends of the range instead of the multiples

Between 50 and 100 the multiples of 8 run from 56 to 96, so their average is 76, not 75. Use the first and last multiples, not the ends of the range.

Drill 3 more on average of the multiples of a number in a range

Concept 3 of 3: Averages of the first n natural numbers and their squares

The numbers 1 to n are an arithmetic progression, so their average is the middle,

\frac{n + 1}{2}

. Squares do not rise in equal steps, so for them use the standard sum formula and divide by n.

Definition

$1 + 2 + \cdots + n = \frac{n(n + 1)}{2}$ , so the average is $\frac{n + 1}{2}$ .
$1^2 + 2^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ , so the average of the squares is $\frac{(n + 1)(2n + 1)}{6}$ . Cancel before multiplying.
$1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2$ , so the average of the cubes is $\frac{n(n + 1)^2}{4}$ .
The first $n$ even numbers average $n + 1$ ; the first $n$ odd numbers average $n$ .

Sums and averages of the first n natural numbers

\frac{1 + 2 + \cdots + n}{n} = \frac{n + 1}{2} \qquad \frac{1^2 + 2^2 + \cdots + n^2}{n} = \frac{(n + 1)(2n + 1)}{6} \qquad \frac{1^3 + 2^3 + \cdots + n^3}{n} = \frac{n(n + 1)^2}{4}

Worked example

Find the average of the squares of the first 20 natural numbers.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 17 Sep 2024, Shift 1 · Q55Easy

Example 3 · Averages · Averages of Consecutive Numbers

The average of the squares of the first 47 natural numbers is

Using n ÷ 2 for the average of 1 to n

The average of 1 to n is (n + 1) ÷ 2. The numbers 1 to 9 average 5, not 4.5.

Giving the sum of squares as the average

n(n + 1)(2n + 1) ÷ 6 is the SUM of the first n squares. Divide by n to get their average, which is (n + 1)(2n + 1) ÷ 6.

Drill 2 more on averages of the first n natural numbers and their squares

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 (3)

Average of numbers in equal steps
Average of an arithmetic progression
$\text{Average} = \frac{\text{first} + \text{last}}{2} \qquad \text{Sum} = \text{count} \times \text{middle term}$
Average of the multiples of a number in a range
Multiples of k in a range
$\text{Average} = \frac{\text{first multiple} + \text{last multiple}}{2} \qquad \text{Count} = \frac{\text{last} - \text{first}}{k} + 1$
Averages of the first n natural numbers and their squares
Sums and averages of the first n natural numbers
$\frac{1 + 2 + \cdots + n}{n} = \frac{n + 1}{2} \qquad \frac{1^2 + 2^2 + \cdots + n^2}{n} = \frac{(n + 1)(2n + 1)}{6} \qquad \frac{1^3 + 2^3 + \cdots + n^3}{n} = \frac{n(n + 1)^2}{4}$

Watch out for (5)

Expecting a whole-number average from an even count of consecutive integers
→ Average of numbers in equal steps
Taking a middle term that does not exist
→ Average of numbers in equal steps
Averaging the ends of the range instead of the multiples
→ Average of the multiples of a number in a range
Using n ÷ 2 for the average of 1 to n
→ Averages of the first n natural numbers and their squares
Giving the sum of squares as the average
→ Averages of the first n natural numbers and their squares

Test yourself on Averages

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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