XIV.

Geometrical Progression.

A series of numbers is said to be in Geometrical Progression when the quotient of any term divided by the preceding term is the same throughout the series.

Thus $a$ , $b$ , $c$ , $d$ , etc., are in geometrical progression if $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$ , etc.

This quotient is called the common ratio, and is represented by $r$ .

State the common ratio of the following series: $r * 4 > r 1, & 3, & 9, & 27, & \dots 2, & 4, & 8, & 16, & \dots 16, & 8, & 4, & 2, & \dots \frac{2}{3}, & 1, & \frac{3}{2}, & \frac{9}{4}, & \dots 4, & - 2, & 1, & - \frac{1}{2}, & \dots$

If the first term of a geometrical progression is represented by $a$ , and the common ratio by $r$ , then

&\text{the \emph{second} term will be~ $a r$ ,} \ &\text{the \emph{third} term will be~ $a r^{2}$ ,} \ &\text{the \emph{fourth} term will be~ $a r^{3}$ ,}

and so on, the index of $r$ being always less by $1$ than the number of the term in the series.

Hence the $n$ th term will be $a r^{n - 1}$ .

If we denote the $n$ th term by $l$ , we have $l = a r^{n - 1} . Formula (1)$

If the first term and common ratio are given, or if any two terms are given, we can find the series.

Find the 5th term of a geometrical progression if the first is $3$ and the common ratio $2$ .

In formula (1), put $5$ for $n$ , $3$ for $a$ , and $2$ for $r$ .

Therefore the 5th term is $48$ .

Find the geometrical series if the 5th term is $48$ and the 7th term is $192$ .

The 5th and 7th terms are $a r^{4}$ and $a r^{6}$ , respectively.

lint Whence ar^4 = 48, (1) lintand ar^6 = 192. (2) lint Divide (2) by (1), r^2 = 4. r = ±2. lint From (1), a = 4816 = 3.

Therefore the series is $3$ , $\pm 6$ , $12$ , $\pm 24$ , $\dots$ .

176. Geometrical Mean.

If three numbers are in geometrical progression, the middle number is called the geometrical mean of the other two numbers. Hence, if $a$ , $G$ , $b$ are in geometrical progression, $G$ is the geometrical mean of $a$ and $b$ .

By the definition of a geometrical progression,

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