Let us take, as an example, the proportion

9 18 20 40,

in which the ratio is 2.

If we augment the antecedent and consequent by 15 and 30, which have the same ratio, we shall have

9+15 18+30 :: 20: 40;

that is,

24 :

:

48 20 40,

in which the ratio is still 2.

If we diminish the second antecedent and consequent by the same numbers, we have

A B

C D, which gives AXD=Bx C,

A(D+F+H)=B(C+E+G);

and by separating the factors,

A B C+E+G: D+F+H.

That is In any number of proportions having the same ratio, any antecedent will be to its consequent, as the sum of the antecedents to the sum of the consequents.

Let us take, for example,

Then,

2 46: 12 and 1: 2 :: 3 : 6, &c

that is,

246+3 : 12+6;

249: 18,

in which the ratio is still 2.

164. If we have four proportional quantities

A B C : D, we have

B D

=

A

and raising both members to any power, as n, we have

That is If four quantities are proportional, any like powers or roots will be proportional.

in which the terms are proportional, the ratio being 4.

165. Let there be two sets of proportions,

QUEST.-163. In any number of proportions having the same ratio, how will any one antecedent be to its consequent ?-164. In four proportional quantities, how are like powers or roots?

Multiply them together, member by member, we have

which gives AE : BF :: CG : DH.

That is: In two sets of proportional quantities, the products of the corresponding terms will be proportional. Thus, if we have the two proportions

166. We have thus far only required that the ratio of the first term to the second should be the same as that of the third to the fourth.

If we impose the farther condition, that the ratio of the second term to the third shall also be the same as that of the first to the second, or of the third to the fourth, we shall have a series of numbers, each one of which, divided by the preceding one, will give the same ratio. Hence, if any term be multiplied by this quotient, the product will be the succeeding term. A series of numbers so formed is called a geometrical progression. Hence,

A Geometrical Progression, or progression by quotients, is a series of terms, each of which is equal to the product of

QUEST.-165. In two sets of proportions, how are the products of the corresponding terms?

that which precedes it by a constant number, which number is called the ratio of the progression. Thus,

1:39 27: 81: 243, &c,

is a geometrical progression, which is written by merely. placing two dots between each two of the terms. Also,

64 32 16 8: 4:2:1

:

is a geometrical progression, in which the ratio is one-half. In the first progression each term is contained three times in the one that follows, and hence the ratio is 3. In the second, each term is contained one-half times in the one which follows, and hence the ratio is one-half.

The first is called an increasing progression, and the second a decreasing progression.

Let a, b, c, d, e, f, . . . be numbers in a progression by quotients; they are written thus:

and it is enunciated in the same manner as a progression by differences. It is necessary, however, to make the distinction, that one is a series of equal differences, and the other a series of equal quotients or ratios. It should be remarked that each term is at the same time an antecedent and a consequent, except the first, which is only an antecedent, and the last, which is only a consequent.

QUEST.-166. What is a geometrical progression? What is the ratio of the progression? If any term of a progression be multiplied by the ratio, what will the product be? If any term be divided by the ratio, what will the quotient be? How is a progression by quotients written? Which of the terms is only an antecedent? Which only a consequent How may each of the others be considered?

q being when the progression is increasing, and q<1 when it is decreasing. Then, since

b=aq, c=bq=aq2, d=cq=aq3, e=dq=aq1,
f=eq=aq5 . . .;

that is, the second term is equal to aq, the third to aq2, the fourth to aq3, the fifth to aq1, &c; and in general, any term n, that is, one which has n-1 terms before it, is expressed by aq-1.

Let be this term; we then have the formula

l=aq"-1,

by means of which we can obtain any term without being obliged to find all the terms which precede it. Hence, to find the last term of a progression, we have the following

RULE.

I. Raise the ratio to a power whose exponent is one less than the number of terms.

II. Multiply the power thus found by the first term: the product will be the required term.

QUEST.-167. By what letter do we denote the ratio of the progression? In an increasing progression is q greater or less than 1? In a decreasing progression is q greater or less than 1 ? If a is the first term and q the ratio, what is the second term equal to? What the fourth? What is the last term equal to finding the last term.

What the third?

Give the rule for