OF RATICS, PROPORTIONS, AND PROGRESSIONS.

NUMBERS are compared to each other in two different ways: the one comparison considers the difference of the two numbers, and is named Arithmetical Relation; and the difference sometimes the Arithmetical Ratio: the other considers their quotient, which is called Geometrical Relation; and the quotient is the Geometrical Ratio. So, of these two numbers 6 and 3, the difference, or arithmetical ratio, is 6-3 or 3, but the geometrical ratio is or 2.

There must be two numbers to form a comparison: the number which is compared, being placed first, is called the Antecedent; and that to which it is compared, the Consequent. So, in the two numbers above, 6 is the antecedent, and 3 the consequent.

If two or more couplets of numbers have equal ratios, or equal differences, the equality is named Proportion, and the terms of the ratios Proportionals. So, the two couplets, 4, 2 and 8, 6, are arithmetical proportionals, because 4 - 2 = 8 62; and the two couplets 4, 2 and 6, 3, are geometrical proportionals, because ៖ 2, the same ratio.

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To denote numbers as being geometrically proportional; a colon is set between the terms of each couplet, to denote their ratio; and a double colon, or else a mark of equality, between the couplets or ratios. So, the four proportionals, 4, 2, 6, 3 are set thus, 4:2:: 6:3, which means, that 4 is to 2 as 6 is to 3; or thus, 4:2 6: 3, or thus, both which mean, that the ratio of 4 to 2, is equal to the ratio of 6 to 3.

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Proportion is distinguished into Continued and Disconti nued. When the difference or ratio of the consequent of one couplet, and the antecedent of the next couplet, is not the same as the common difference or ratio of the couplets, the proportion is discontinued. So, 4, 2, 5, 6 are in discontinued arithmetical proportion, because 4-28-62, whereas 8-26; and 4, 2, 6, 3 are in discontinued geometrical proportion, because 2, but 3, which is not

the same.

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But when the difference or ratio of every two succeeding terms is the same quantity, the proportion is said to be Continued, and the numbers themselves make a series of Continued Proportionals,

Proportionals, or a progression. So 2, 4, 6, 8 form an arithmetical progression, because 4-26-48-6 = 2, all the same common difference; and 2, 4, 8, 16 a geometrical progression, because 162, all the same ratio.

When the following terms of a progression increase, or exceed each other, it is called an Ascending Progression, or Series; but when the terms decrease, it is a descending one. So, 0, 1, 2, 3, 4, &c. is an ascending arithmetical progression, -but 9, 7, 5, 3, 1, &c. is a descending arithmetical progression. Also 1, 2, 4, 8, 16, &c. is an ascending geometrical progression, and 16, 8, 4, 2, 1, &c. is a descending geometrical progression.

ARITHMETICAL PROPORTION and PROGRESSION.

IN Arithmetical Progression, the numbers or terms have all the same cominon difference. Also, the first and last terms of a Progression, are called the Extremes; and the other terms, lying between them, the Means. The mose useful part of arithmetical proportions, is contained in the following theorems :

THEOREM 1. When four quantities are in arithmetical proportion, the sum of the two extremes is equal to the sum of the two means. Thus, of the four 2, 4, 6, 8, here 2 + 8 = 4 + 6 = 10.

THEOREM 2. In any continued arithmetical progression, the sum of the two extremes is equal to the sum of any two/ means that are equally distant from them, or equal to double the middle term when there is an uneven number of terms

Thus, in the terms 1, 3, 5, it is 1 - 5 =3+3=6., And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4 +12= 6 + 10 = 8 + 8 = 16.

THEOREM 3. The difference between the extreme terms of an arithmetical progression, is equal to the common difference of the series multiplied by one less than the number of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, the common difference is 2, and one less than the number of terms 9; then the difference of the extremes is 20218, and 2 × 9 = 18 also.

Consequently,

Consequently the greatest term is equal to the least term added to the product of the common difference multiplied by 1 less than the number of terms.

THEOREM 4. The sum of all the terms, of any arithme tical progression, is equal to the sum of the two extremes multiplied by the number of terms, and divided by 2; or the sum of the two extremes multiplied by the number of the terms, gives double the sum of all the terms in the series.

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This is made evident by setting the terms of the series in an inverted order, under the same series in a direct order, and adding the corresponding terms together in that order. Thus, in the series 1, 3, 5, 7, 9, 11, 13, 15; ditto inverted 15, 13, 11, 9, 7, 5, 3, the sums are 16+16 +16 +16 + 16 + 16 + 16 + 16, which must be double the sum of the single series, and is equal to the sum of the extremes repeated as often as are the number of the terms.

From these theorems may readily be found any one of these five parts; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given; as in the following problems:

PROBLEM I.

Given the Extremes, and the Number of Terms; to find the Sum of all the Terms.

ADD the extremes together, multiply the sum by the num ber of terms, and divide by 2.

EXAMPLES.

1. The extremes being 3 and 19, and the number of terms 9; required the sum of the terms?

2. It is required to find the number of all the strokes a common clock strikes in one whole revolution of the index, or in 12 hours?

Ans. 78.

Ex.

Ex. 3. How many strokes do the clocks of Venice strike in the compass of the day, which go continually on from I to 24 o'clock ? Ans. 300.

4. What debt can be discharged in a year, by weekly payments in arithmetical progression, the first payment being 1s, and the last or 52d payment 5/ 35? Ans. 135/ 4s.

PROBLEM II.

Given the Extremes, and the Number of Terms; to find the Common Difference.

SUBTRACT the less extreme from the greater, and divide the remainder by 1 less than the number of terms, for the common difference.

EXAMPLES.

1. The extremes being 3 and 19, and the number of terms 9; required the common difference?

2. If the extremes be 10 and 70, and the number of terms 21; what is the common difference, and the sum of the series? Ans. the com. diff. is 3, and the sum is 840. 3. A certain debt can be discharged in one year, by weekly payments in arithmetical progression, the first payment being 1s, and the last 5/ 3s; what is the common difference of the terms? Ans. 2.

PROBLEM III.

Given one of the Extremes, the Common Difference, and the Number of Terms: to find the other Extreme, and the Sum of the Series.

MULTIPLY the common difference by 1 less than the num ber of terms, and the product will be the difference of the extremes: Therefore add the product to the less extreme, to give the greater; or subtract it from the greater, to give the less extreme.

VOL. I.

EXAMPLES.