Why Folding a Sheet of Paper in Half Again and Again Makes It Thick So Fast

In an AP, you add the same number each time. In a geometric progression (GP), you multiply by the same number each time. That number is the common ratio $r$:

$$r=\frac{a_2}{a_1}=\frac{a_3}{a_2}=\dots$$

A GP looks like $a,ar,a{r}^2,a{r}^3,\dots$ Because each step multiplies, **a GP with $r>1$ grows very fast.**

This chapter covers testing for a GP, the nth term, the sum of n terms, and the geometric mean.

Divide each term by the one before it; if every ratio is the same, the sequence is a GP, and multiplying by that ratio gives the next terms.

Worked example 1. $3,6,12,24,\dots$

$$\frac{6}{3}=\frac{12}{6}=\frac{24}{12}=2\text{next: }48,96,192$$

Worked example 2. $81,27,9,\dots$: $r=\frac{1}{3}$, so the next terms are $3,1,\frac{1}{3}$.

Worked example 3. $2,-6,18,\dots$: $r=-3$, so the next terms are $-54,162,-486$.

Worked example 4. $1,2,4,7,\dots$: $\frac{2}{1}=2$ but $\frac{7}{4}\ne 2$, so not a GP.

An everyday example. A message forwarded on a family chat group, where each person sends it to $3$ others, reaches $1,3,9,27,\dots$ new people at each step — a GP with $r=3$.

The substance. The common ratio can be a fraction or negative, but no term of a GP can be zero, since you cannot divide by it.

**Use $a_n=a{r}^{n-1}$; to find which term equals a value, set $a{r}^{n-1}$ equal to it and write both sides as powers of $r$.

Worked example 1.** The $8$th term of $3,6,12,\dots$

$$a_8=3\times 2^7=3\times 128=384$$

Worked example 2. The $6$th term of $81,27,9,\dots$

$$a_6=81\times {\left(\frac{1}{3}\right)}^5=\frac{81}{243}=\frac{1}{3}$$

Worked example 3. Which term of $2,6,18,\dots$ is $1458$?

$$2\times 3^{n-1}=14583^{n-1}=729=3^{6}n=7$$

Worked example 4. Which term of $\frac{1}{4},\frac{1}{2},1,\dots$ is $64$?

$$\frac{1}{4}\times 2^{n-1}=642^{n-1}=256=2^{8}n=9$$

An everyday example. **A ball dropped from $16$ m bounces back to $\frac{3}{4}$ of its height each time.** After the third bounce it rises to

$$16\times {\left(\frac{3}{4}\right)}^3=16\times \frac{27}{64}=6.75\text{ m}$$

The substance. **The power is $n-1$**, because the first term has not yet been multiplied by $r$.

**For $r\ne 1$, use $S_n=\frac{a(r^n-1)}{r-1}$, or the equivalent $\frac{a(1-r^n)}{1-r}$, which is easier when $r<1$.

Worked example 1.** Sum of the first $8$ terms of $3,6,12,\dots$

$$S_8=\frac{3(2^8-1)}{2-1}=3\times 255=765$$

Worked example 2. Sum of the first $6$ terms of $1,\frac{1}{2},\frac{1}{4},\dots$

$$S_6=\frac{1-{\left(\frac{1}{2}\right)}^6}{1-\frac{1}{2}}=2\left(1-\frac{1}{64}\right)=\frac{63}{32}$$

Worked example 3. Sum of the first $5$ terms of $2,-6,18,\dots$

$$S_5=\frac{2[(-3{)}^5-1]}{-3-1}=\frac{2(-244)}{-4}=122$$

Check: $2-6+18-54+162=122$.

An everyday example. **A shopkeeper gives away $1$ box of sweets on the first day of a festival and doubles it each day for $10$ days.** The total is $\frac{1(2^{10}-1)}{2-1}=1023$ boxes.

The boundary case. **When $r=1$ the formula divides by zero**; then every term equals $a$ and $S_n=na$.

**The geometric mean between two positive numbers $a$ and $b$ is $\sqrt{ab}$, so that $a$, the mean and $b$ form a GP; GPs model anything that grows or shrinks by a fixed percentage each period.

Worked example 1.** Geometric mean of $4$ and $25$: $\sqrt{100}=10$, giving the GP $4,10,25$.

Worked example 2. If $5,x,45$ are in GP: $x^2=225$, so $x=15$ or $x=-15$.

Worked example 3 — growth. Suppose a town of $20000$ people grows by $10\%$ each year.

$$\text{after }3\text{ years}=20000\times 1.1^3=20000\times 1.331=26620$$

Worked example 4 — depreciation. A machine bought for ₹50000 loses $20\%$ of its value each year.

$$\text{value after }3\text{ years}=50000\times 0.8^3=50000\times 0.512=₹25600$$

An everyday example. A phone bought for a family loses value year after year — each year's value is a fixed fraction of the previous one, a GP.

The substance. The geometric mean is never larger than the arithmetic mean for positive numbers: for $4$ and $25$, $10<14.5$.