1
One mean is the one in the middle.
arithmetic mean
a + b2
the ordinary average
geometric mean
√a · b
multiply, then root
4 and 20
(4 + 20) ÷ 2 = 12
4, 12, 20 — differences of 8 ✓
3 and 27
√3 × 27 = √81 = 9
3, 9, 27 — ratios of 3 ✓
The word “mean” just names the term that sits between two others in a sequence.
Which kind of sequence decides which formula.
With x in it, square the root away. If x is the geometric mean of x − 2 and x + 4, then x² = (x − 2)(x + 4) = x² + 2x − 8, so 2x = 8 and x = 4. Check: the numbers are 2 and 8, and √16 = 4 ✓.
2
Count the gaps, not the blanks.
Three means between two numbers gives five terms — and four jumps.
| Means inserted | Terms in total | Gaps |
|---|
| 1 | 3 | 2 |
|---|
| 2 | 4 | 3 |
|---|
| 3 | 5 | 4 |
|---|
| k | k + 2 | k + 1 |
|---|
Dividing by 3 is the trap
Three blanks, so three shares — except the distance from 4 to 20 is crossed in
four steps, not three. Every wrong answer in this topic comes from that
off-by-one.
3
Arithmetic: solve for d.
aₙ = a₁ + (n − 1)d
the sequence formula
20 = 4 + 4d
a₅, so n − 1 = 4
16 = 4d, d = 4
the common difference
means: 8, 12, 16
add 4 each time
Shortcut worth knowing: d is simply the total distance shared between the gaps —
(20 − 4) ÷ 4. The formula and the shortcut are the same arithmetic.
4
Geometric: solve for r.
Two means, so four terms and three jumps — each one a multiplication.
aₙ = a₁ · rn−1
the geometric formula
24 = 3r³
a₄, so the power is 3
means: 6, 12
3, 6, 12, 24
4, 8, 12, 16, 20 · 3, 6, 12, 24
5
An even root brings a second answer.
one geometric mean between 2 and 8
three terms, two jumps
8 = 2r², r² = 4
a square, so ±
r = 2 → 2, 4, 8
the usual answer
r = −2 → 2, −4, 8
also a genuine sequence
Odd powers have one root, even powers have two
r³ = 8 gives only r = 2. r² = 4 gives 2 and −2. If the question says the terms
are positive, take the positive root — otherwise both belong in the answer.
And a geometric mean needs a and b to have the same sign: √a·b does not exist in
the reals when the product is negative.
6
The sum of all the means.
The means pair off from the outside in, and every pair averages the same as the two ends.
average of the means = 5 + 352 = 20
same centre as the ends
k means add to 20k
k terms averaging 20
Check: five means give six gaps, so d = 30 ÷ 6 = 5 and the means 10, 15, 20, 25, 30 add to 100 ✓.
The ends are not means
The two given numbers are not part of the count. Adding them in, or counting them as means, is the off-by-two the wrong choices are built on.
7
The arithmetic mean is never smaller.
4 and 20
AM = 12 · GM = √80 ≈ 8.94
AM > GM
9 and 9
AM = 9 · GM = 9
equal only here
AM ≥ GM for any two positive numbers, with equality only when they are the same.
A quick sanity check: a geometric mean that came out larger than the average is
wrong.
8
Backwards: the two numbers from their means.
The AM gives the sum and the GM gives the product. A sum and a product make a quadratic.
a + b2 = 15, so a + b = 30
double the AM
√ab = 9, so ab = 81
square the GM
t² − 30t + 81 = 0
t² − (sum)t + (product) = 0
(t − 3)(t − 27) = 0
3 + 27 = 30 and 3 × 27 = 81
Check: (3 + 27) ÷ 2 = 15 and √81 = 9 ✓. Then give what is asked — the larger number, the difference 24, or something else.
9
Variations: going down, and means that are fractions.
A negative, fraction d. Insert two arithmetic means between 10 and 3.
3 = 10 + 3d
two means, so three gaps
d = −73
the sequence goes down
means: 233, 163
163 − 73 = 93 = 3 ✓
What is different: the last term is smaller, so d is negative, and the gap 7 does not share evenly into 3. Keep the thirds; do not round.
A fraction r. Insert two geometric means between 250 and 2.
2 = 250 r3
four terms, power 3
r3 = 2250 = 1125
divide by the first term
r = 15
cube root of top and bottom
means: 50, 10
250, 50, 10, 2
What is different: the terms shrink, so r is a fraction. The equation is the same; divide by the first term, not the last.
Dividing the wrong way
Writing r3 = 250 ÷ 2 = 125 gives r = 5 and means 1250 and 6250, which run away from 2. The first term goes on the bottom.
10
AM ≥ GM finds a least value.
When two positive parts have a fixed product, their sum is smallest when the parts are equal.
x · 16x = 16
the product is fixed
x + 16x2 ≥ √16 = 4
AM ≥ GM
x + 16x ≥ 8
double both sides
equal when x = 16x, that is x = 4
the two parts match
Test it: x = 2 gives 2 + 8 = 10, x = 8 gives 8 + 2 = 10, and x = 4 gives 4 + 4 = 8, the least.
On the exam
The two means usually come together: AM and GM given with the numbers wanted (sum and product, then a quadratic), means hidden inside a sum, or AM ≥ GM used for a least value. Always turn “k means” into k + 1 gaps. Ten-second check: the AM can never be smaller than the GM — if yours is, a slip happened.
✓
Check: walk the sequence back.
differences
8−4, 12−8, 16−12, 20−16
4, 4, 4, 4 ✓
ratios
6÷3, 12÷6, 24÷12
2, 2, 2 ✓
endpoints
first is 4, last is 20
the given numbers survived ✓
Remember the flow
k means → k + 2 terms → k + 1 gaps · Solve for d or r · Fill in
Add for arithmetic, multiply for geometric.
