The Average That's Never Bigger

A chain-referral scheme starts with 4 people, and each of them refers 3 new people in the next round: 4, 12, 36, 108, ... a geometric progression with first term a=4 and common ratio r=3. Finding any single round's count is easy using the nth-term formula already familiar from earlier study, but adding up ALL the people across several rounds needs a new trick. Let Sn = a + ar + ar^2 + ... + ar^(n-1). Multiplying every term by r gives rSn = ar + ar^2 + ... + ar^n. Subtracting the first equation from the second, almost every term cancels: rSn - Sn = ar^n - a, so Sn(r-1) = a(r^n - 1), giving the sum formula Sn = a(r^n - 1)/(r - 1), valid whenever r is not 1. Checking the referral chain: S5 = 4(3^5 - 1)/(3-1) = 4(243-1)/2 = 4(242)/2 = 484 people total after 5 rounds.

The same formula works whether a GP is growing or shrinking. Consider the GP 8, 4, 2, 1, ..., with a=8 and r=1/2 (each term is half the one before). Using Sn = a(1 - r^n)/(1 - r), the form more convenient when r is less than 1, the sum of the first 6 terms is S6 = 8(1 - (1/2)^6)/(1 - 1/2) = 8(1 - 1/64)/(1/2) = 8 times (63/64) times 2 = 63/4 = 15.75. Notice the formula needs r to NOT equal 1: if r were exactly 1, every term would just equal a, dividing by (r-1) would mean dividing by zero, and the honest answer is simply Sn = na instead, no formula required.

Just as the arithmetic mean of two numbers a and b is (a+b)/2, sitting exactly halfway between them by ADDITION, the geometric mean is square-root(ab), sitting halfway between them by MULTIPLICATION. The geometric mean of 5 and 45 is square-root(5 times 45) = square-root(225) = 15. Checking this makes sense: 5, 15, 45 form a genuine GP, since 15/5 = 3 and 45/15 = 3, the same common ratio both times. This is exactly why it is called a MEAN: inserting it between the two original numbers produces a perfectly balanced geometric progression, the multiplicative equivalent of an arithmetic mean sitting evenly between two numbers by addition.

It is possible to insert not just one, but several geometric means between two numbers, so that the whole resulting list forms one single GP. To insert 2 numbers between 4 and 108, treat 4 as the first term and 108 as the FOURTH term of a 4-term GP (since 2 inserted numbers plus the original 2 endpoints makes 4 terms total): 108 = 4 times r^3, so r^3 = 27, giving r = 3. The two inserted means are then 4 times 3 = 12 and 12 times 3 = 36. Checking the full sequence 4, 12, 36, 108: every consecutive ratio is exactly 3, confirming a genuine GP. In general, to insert n geometric means between a and b, the common ratio is r = (b/a) raised to the power 1/(n+1), since b becomes the (n+2)th term of the combined sequence.

Compute both means for 9 and 16: the arithmetic mean is (9+16)/2 = 12.5, while the geometric mean is square-root(9 times 16) = square-root(144) = 12. The arithmetic mean came out bigger. This is not a coincidence: for ANY two positive numbers a and b, calling the arithmetic mean A and the geometric mean G, A minus G equals (square-root(a) minus square-root(b)) squared, all divided by 2. Since anything squared can never be negative, this difference A minus G can never be negative either, meaning A is always greater than or equal to G, with equality happening only in the one special case where a and b are already equal (making square-root(a) minus square-root(b) exactly zero).

If only the arithmetic mean and geometric mean of two numbers are known, the numbers themselves can still be recovered. Suppose two positive numbers have AM=17 and GM=15. Then a+b = 2(17) = 34, and ab = 15^2 = 225. Using the identity (a-b)^2 = (a+b)^2 - 4ab gives (a-b)^2 = 34^2 - 4(225) = 1156 - 900 = 256, so a-b = 16 (taking the positive root). Solving a+b=34 and a-b=16 together: adding the two equations gives 2a=50, so a=25, and then b=9. Checking: (25+9)/2 = 17 matches the given AM, and square-root(25 times 9) = square-root(225) = 15 matches the given GM exactly.

Suppose two positive numbers must multiply to exactly 100: what is the smallest their sum could possibly be? By the AM-GM inequality, (a+b)/2 is at least square-root(ab) = square-root(100) = 10, so a+b is at least 20, and this minimum is achieved exactly when a equals b. Since ab=100 and a=b, this forces a^2=100, so a=b=10, giving the minimum sum of 20. This kind of reasoning, using AM-GM to pin down a best-possible value without checking every case, is genuinely useful far beyond this one example. This inequality has deep roots: evidence shows the Babylonians knew of arithmetic and geometric sequences some 4000 years ago, the Indian mathematician Aryabhata gave formulas involving such sequences around 499 CE, the Fibonacci sequence was named after the Italian mathematician Leonardo Fibonacci (1170-1250), and the term 'infinite series' was first used by James Gregory in 1671.