Statistics

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Subtopic 5 of 5·← Range, variance and standard deviation

Averages of special series

Closed-form sums give instant averages for standard series (first $n$ of each):

series	sum	mean
naturals $1..n$	$\frac{n(n+1)}{2}$	$\frac{n+1}{2}$
squares	$\frac{n(n+1)(2n+1)}{6}$	$\frac{(n+1)(2n+1)}{6}$
cubes	$\left[\frac{n(n+1)}{2}\right]^2$	$\frac{n(n+1)^2}{4}$
AP (first $a$ , last $l$ )	$\frac{n(a+l)}{2}$	$\frac{a+l}{2}$

For consecutive equally-spaced numbers (AP), the mean is the middle term — with an even count, halfway between the two middles.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 14 practice Q

Detailed notes

The closed-form sum table

series	sum	mean
naturals $1..n$	$\frac{n(n+1)}{2}$	$\frac{n+1}{2}$
squares	$\frac{n(n+1)(2n+1)}{6}$	$\frac{(n+1)(2n+1)}{6}$
cubes	$\left[\frac{n(n+1)}{2}\right]^2$	$\frac{n(n+1)^2}{4}$
AP (first $a$ , last $l$ )	$\frac{n(a+l)}{2}$	$\frac{a+l}{2}$

Divide any sum by $n$ for the mean — the mean column is pre-divided for speed. The $n = 10$ anchors (mean of squares 38.5, sum of cubes 3025) make instant sanity checks.

Consecutive numbers: the mean is the middle

Any AP's mean is $\frac{\text{first} + \text{last}}{2}$ — the middle term. Five consecutive even numbers averaging 16 are $12, 14, 16, 18, 20$ : the largest is $16 + 2 \times 2$ . General form for odd counts: with average $m$ and step $d$ over $k$ terms, the largest is $m + \frac{(k-1)}{2}d$ .

Working backwards

"The average of squares of the first n naturals is 38.5" → $\frac{(n+1)(2n+1)}{6} = 38.5$ → $(n+1)(2n+1) = 231$ → $n = 10$ . Reverse-engineering n from a mean is a favourite: build the equation and test the factor pair.

Sum-of-cubes ≡ square-of-sum

$\sum k^3 = \left(\sum k\right)^2$ — the same number wearing two hats. "The sum of the cubes of the first n naturals is 441" instantly gives $\sum k = 21$ , so $n = 6$ via $\frac{n(n+1)}{2} = 21$ . Questions exploit this identity in both directions.

Weighted AP averages

The mean of consecutive odd numbers $1, 3, ..., (2n-1)$ is $n$ (mean of first n odds = n); of consecutive multiples of $k$ up to $kn$ : $\frac{k(n+1)}{2}$ . These sub-cases save the last minute of the exam.

Worked examples at speed

Mean of the first 20 odd numbers: 20 (mean of the first $n$ odds $= n$ ).
Mean of the cubes of the first 5 naturals: $\frac{5 \times 36}{4} = 45$ (sum 225 ÷ 5).
Mean of all two-digit numbers (10 to 99): $\frac{10 + 99}{2} = 54.5$ .
Mean of the multiples of 7 between 1 and 100 (7 to 98, 14 terms): $\frac{7 + 98}{2} = 52.5$ .
Mean of the squares of the first 5 naturals: $\frac{6 \times 11}{6} = 11$ (sum 55 ÷ 5). Each is one line once the series type is named — naming it is the real skill. Always count the terms first: the term count is where most slips happen.

Quick revision

Sums: $\frac{n(n+1)}{2}$ , $\frac{n(n+1)(2n+1)}{6}$ , $\left[\frac{n(n+1)}{2}\right]^2$ .
AP mean = middle term $= \frac{a + l}{2}$ .
First n odds average n; first n evens average $n+1$ .
Cubes sum = (naturals sum)²; exploit both ways.
Reversing: set the mean formula equal to the given and solve for n.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Mean of squares / cubes of first nvery common3 practice Q

How to spot it:

'The mean of the squares (cubes) of the first n natural numbers is...' — n given or asked.

Divide the closed-form sum by n: squares mean $\frac{(n+1)(2n+1)}{6}$; cubes mean $\frac{n(n+1)^2}{4}$.

Write the sum formula; divide by n mentally.
Substitute n; keep fractions exact.
Reversed: set the mean equal to the given and solve for n.

Why: the formulas do the work; the only skill is not confusing sum with mean.

Example: The mean of the squares of the first 10 natural numbers is:

$\frac{(11)(21)}{6} = 38.5$ .

Type 2: AP / consecutive numbers averagevery common4 practice Q

How to spot it:

Consecutive integers, evens, odds or multiples; the average (or an end term) asked.

Mean = middle term $= \frac{a+l}{2}$; odd counts land exactly on a term, even counts between the two middles.

The average pins the middle of the run.
Largest = average + $\frac{(k-1)}{2}d$ ; smallest = average − same.
Sum = average × count when a total is asked.

Why: symmetry — deviations from the middle cancel pairwise to zero.

Example: The average of 5 consecutive even numbers is 16. The largest of them is:

Middle (3rd) = 16: numbers 12–20; largest $= 16 + 4 = 20$ .

Type 3: Cubes-sum = square-of-sum identitycommon2 practice Q

How to spot it:

'The sum of cubes of the first n naturals is K — find n' or a combined mean question exploiting the identity.

$\sum k^3 = \left[\frac{n(n+1)}{2}\right]^2$; a given cube-sum is a perfect square — its root is the naturals-sum.

Root the given sum to get $\frac{n(n+1)}{2}$ .
Solve the triangular-number equation for n.
Mean of cubes = that root squared ÷ n.

Why: the identity links two sums the exam alternates between — one conversion covers both.

Example: The sum of the cubes of the first n natural numbers is 441. The value of n is:

$\sqrt{441} = 21 = \frac{n(n+1)}{2} \Rightarrow n(n+1) = 42 \Rightarrow n = 6$ .

Type 4: First n odds / evens / multiplesoccasional2 practice Q

How to spot it:

Averages of the first n odd numbers, even numbers, or multiples of k.

First n odds average n; first n evens average $n+1$; multiples of k up to kn average $\frac{k(n+1)}{2}$.

Recognise the special series.
Quote the mean directly (or via $\frac{a+l}{2}$ ).
For 'how many terms', divide the last term appropriately.

Why: these are AP means with canned answers — instant marks.

Example: The average of the first 9 odd numbers is:

9 (mean of first n odds = n).

Formulas

First n naturals

\sum_{1}^{n}k=\frac{n(n+1)}{2},\quad \text{mean}=\frac{n+1}{2}

Squares

\sum_{1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}

Cubes

\sum_{1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2

AP mean

\text{mean of AP}=\frac{\text{first}+\text{last}}{2}

Shortcut tricks

⚡ Consecutive numbers: average = middle

For any run of consecutive terms (odd, even, naturals), the average equals the median — the middle term. No summation needed.

Example: The average of 5 consecutive even numbers is 16. Find the largest.

Middle (3rd) $=16$ ; largest $=16+2\times2=20$ .

⚡ Squares and cubes via the sums

Divide the closed-form sum by $n$ ; memorise the $n=10$ anchor (mean of squares $=38.5$ ) as a sanity check.

Example: Find the mean of the cubes of the first 6 natural numbers.

Sum $=\left(\frac{6\times7}{2}\right)^2=441$ ; mean $=\frac{441}{6}=73.5$ .

Where students lose marks

Using $\frac{(n+1)(2n+1)}{6}$ (the mean of squares) as if it were a sum.
Off-by-one on 'consecutive': between the middle two for even counts, exactly the middle term for odd.
Cubes sum confused with square of sum of naturals (they are equal, but the mean needs dividing by $n$ ).

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.