Average is the balancing point — total spread evenly. The one identity to internalize: Sum = Average × Count. Every "change in average" problem is a story about how the total changed.
Average = (Sum of Observations) / (Number of Observations)
The inverse is more useful: Sum = Average × n
Example: average age of 3 people is 20 → sum = 20 × 3 = 60. If you know any two
of average/sum/n, you know the third.
New person brings a value; average shifts.
Value of new person = Old Average ± (Change × New Total Count) — plus for
increase, minus for decrease.
2×5=10 beyond the
old average — that extra 10 came from the newcomer.20 + 2×5 =30. Check: (80+30)/5=22 ✓. If average
fell to 18 (−2), new person = 20 −2×5=10.
Logic reverses:
150−112=38. Average fell, so leaver was above average (38
>30) ✓.
New person = Removed person ± (Change in average × Total number) — plus for
increase, minus for decrease.
30 + 2×5 =40. Check: old sum 100, new sum 110,
removed 30 → new =140−100=40? Wait: new sum =22×5=110, old sum 100, so new =removed + change×n
=30+10=40 ✓ (total +10).
Combining two groups with different averages — not a simple mean of averages (that ignores size).
Group 1: n₁ items, average A₁; Group 2: n₂,
A₂:
Combined Average = (n₁A₁ + n₂A₂) / (n₁ + n₂)
(20×60+30×70)/50 = (1200+2100)/50=66. Simple mean (60+70)/2=65 would be wrong —
weights matter; larger class pulls average toward 70.
Average speed is not the arithmetic mean of speeds. It is
Total Distance / Total Time.
Special case — equal distances go and return at speeds x and
y:
Average Speed = 2xy / (x + y) — the harmonic mean of the two speeds.
2×60×40/100=48 (closer to 40).
Derivation: total 2D, time = D/x + D/y → avg = 2/(1/x+1/y).For numbers in AP (consecutive, even, odd, multiples):
Average = (First + Last) / 2
Example: 2,4,6,8,10 → average 6 (middle). Proof: AP is symmetric, so pairing first+last,
second+second-last all give first+last — same Gauss pairing as in Progressions
(Module 19).
Linked trick: Sum of AP = average × count — so
(first+last)/2 × n recovers the AP sum formula. Averages and progressions are the
same idea.
A number was wrongly recorded (e.g., 80 written as 50):
Correct Sum = Wrong Sum − Wrong Value + Correct Value
Correct Average = Correct Sum / n
Shortcut: error = Correct − Wrong (signed). Then
Correct Average = Old Average + Error / n. Difference divided by count — one
division, no sums.
50/20=2.5 → correct average =
52.5. Check: wrong sum 1000, correct 1050, /20=52.5 ✓.
| Symbol | Meaning |
|---|---|
A |
Average |
S |
Sum of observations |
n |
Count |
x, y |
New values / speeds |
Sum from average: S = ?
Avg of 4 is 20, 5th joins and avg becomes 22 → new person?
Average rises after someone leaves → leaver was?
Replacement: 30 leaves, avg 20→22 for 5 → newcomer?
Weighted: 20@60 and 30@70 → combined?
Go 60 km/h, return 40 km/h (equal distance) → avg speed?
Average of AP 2,4,6,8,10 is?
Wrong value 30 instead of 80 for n=20, correction to average?
Avg ⇒ Sum/n ; Sum ⇒ A×n
a) New joins ⇒ value = Old ± Change×NewCount (+ if avg↑)
(eg 4@20→5@22: new=20+2×5=30)
b) Leaves ⇒ avg↑→leaver below avg ; avg↓→leaver above avg
c) Replacement ⇒ new = removed ± Change×n
(eg 30 leaves, 20→22, 5: new=30+2×5=40)
Weighted ⇒ (n₁A₁+n₂A₂)/(n₁+n₂) (not simple mean)
Avg speed ⇒ total D / total T ; equal D ⇒ 2xy/(x+y) (harmonic)
AP avg ⇒ (first+last)/2 = middle (if odd count)
Error ⇒ correct = wrong −wrongVal+correctVal ; shortcut +Error/n
(eg 30→80 for 20: +50/20=+2.5)
Primary source: none pinned yet — drop your preferred video/book resource into RESOURCES.md and it will be linked here. Ask me anything that's unclear.
Questions? Ask your agent — you can follow up on any concept, quiz answer, or get extra practice problems tuned to this module.