Geometric progressions and compound percentages

Two increases of 10% turn a hypothetical 100 units into 121, not 120. The second percentage uses 110 as its base. A geometric progression preserves a multiplier, not a difference. Use percentages, powers and sequences.

Repeating one multiplication

For a nonzero first term and nonzero constant ratio q:

$b_{n+1}=qb_n,\qquad b_n=b_1q^{n-1}.$

There are n−1 multiplications from the first to the nth term. The sequence 3,6,12,24 has q=2; 8,4,2,1 has q=1/2. Negative q alternates signs. A positive progression decreases when 0<q<1.

Doubling

bₙ=3·2ⁿ⁻¹

Approaching a sum

Sₙ=2(1−2⁻ⁿ)

Discrete terms 3·2ⁿ⁻¹ and partial sums 1+1/2+…; uniform scales and integer n. Lower dashed y=2 is an unattained finite-sum bound.

Reconstructing the sum

Let $S_n=b_1+b_1q+\cdots+b_1q^{n-1}$ . Subtract qSₙ: intermediate terms cancel, giving $(1-q)S_n=b_1(1-q^n)$ .

$S_n=b_1\frac{1-q^n}{1-q}\ (q\ne1);\qquad S_n=nb_1\ (q=1).$

The sum of 3,6,12,24 is 45. An infinite nonzero geometric series converges precisely when |q|<1: its remainder contains qⁿ, which approaches zero.

$S_\infty=\frac{b_1}{1-q},\qquad |q|<1.$

Thus 1+1/2+1/4+…=2, although every finite partial sum is below 2. At q=−1 terms oscillate; at q=1 they do not shrink. The limit formula does not apply. OpenStax

A constant rate is not a constant increment

For n equal periods, rate r per period, no deposits, withdrawals, fees or rounding:

$P_n=P_0(1+r)^n,\qquad P_n^{\mathrm{simple}}=P_0(1+nr).$

With $P_0=100$ , $r=0.1$ , n=2, results are 121 versus 120. Simple percentages always use the initial base. A decrease of 20% gives q=0.8, not −0.2. This is a learning model, not an income forecast: identify the base and period before the multiplier.