Work with geometric progressions

Recognise and generate geometric progressions

A geometric progression (GP) in this lecture consists of nonzero terms with a nonzero constant consecutive ratio:

$$\frac{u_{n+1}}{u_n}=r.$$

The common ratio can be positive or negative. With first term $a$, multiply by $r$ to obtain each next term. The Teacher’s Models callout connects this to a quantity changing repeatedly by a fixed percentage or ratio, whereas an AP models a fixed added amount.

The source table gives $5,25,125,625$ with $a=5,r=5$; $27,9,3,1,1/3,1/9,1/27$ with $a=27,r=1/3$; and $2,-6,18,-54,162$ with $a=2,r=-3$. The following question $2,4,6,10,14$ is not a GP: $4/2=2$ but $6/4\ne2$.

There are $n-1$ multiplications after the first term, so

$$u_n=ar^{n-1},\qquad n\in\mathbb Z^+.$$

Example 9 retains both complete table rows: $-1,2,-4,8,-16,\ldots$ has $a=-1,r=-2$ and $u_n=(-1)(-2)^{n-1}=\tfrac12(-2)^n$; $1,1/2,1/4,1/8,1/16,\ldots$ has $a=1,r=1/2$ and $u_n=1(1/2)^{n-1}=(1/2)^{n-1}$. Self-Practice 3 uses $a=10,r=2/3$ to find $u_6\approx1.32$ and $u_{13}\approx0.0771$. Keep the ratio exact until rounding the results.

Solve geometric term and threshold problems

Example 10 uses $a=2,r=8$, so $u_n=2(8)^{n-1}$ and $u_7=524288$. To find the first term exceeding one million, retain an exact inequality until the final integer choice:

$$\begin{aligned} 2(8)^{n-1}&>1000000,\\ 8^{n-1}&>500000,\\ (n-1)\ln8&>\ln500000,\\ n&>1+\frac{\ln500000}{\ln8}\approx7.31. \end{aligned}$$

The least integer is $8$. The source also displays $n-1>6.31$ and $n>7.31$ as rounded cutoffs; these do not replace the exact logarithmic inequality.

For its calculator method, enter Y1=2*8^(X−1) and inspect the table. The source explicitly requires showing the supporting values:

Highlight the Y1 value to read its actual integer. The Teacher warns against copying the rounded screen value 4.19E6 instead of 4194304. Since the terms increase, checking the two adjacent terms 7 and 8 establishes the first crossing. The screenshot’s X=0 row is a calculator sample; this sequence’s domain still starts at n=1.

Example 11 reverses the task: from $u_3=36$ and $u_6=243/2$, write

$$ar^2=36\quad\text{(1)},\qquad ar^5=\frac{243}{2}\quad\text{(2)}.$$

The illustrated source callout explains why division is useful: (2) divided by (1) eliminates the unknown first term $a$. It gives $r^3=27/8$, hence $r=3/2$. Substitute into either equation to recover $a$; using (1), $a=36/(3/2)^2=16$. The divisor is nonzero because the given third term is 36.

Compare AP and GP tests and constant sequences

A GP proof must show that $u_{n+1}/u_n$ is independent of $n$ for every valid neighbouring pair, with nonzero terms. Equivalently use $u_n/u_{n-1}=r$ where defined. The source asks why $u_2/u_1=u_3/u_2$ alone is insufficient: as in the AP argument, it tests only two pairs.

For three consecutive GP terms $x,y,z$,

$$\frac yx=\frac zy=r\quad\Longrightarrow\quad y^2=xz.$$

When all three terms are nonnegative, $y=\sqrt{xz}$ is the geometric mean of $x$ and $z$. Keep that sign condition: $y^2=xz$ alone does not choose the positive square root. The lecture’s nonzero GP convention also excludes zero denominators.

A constant sequence has all its terms equal. The source examples are $1,1,1,1,1,1,1,\ldots$ and the finite list $0,0,0,0,0,0,0$. Its unnumbered question about $2,2,2,\ldots,2$ has both classifications: AP with $a=2,d=0$, and GP with $a=2,r=1$. A zero constant sequence is an AP with difference zero, but it is excluded from GP under this lecture’s nonzero-term definition.