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XXXII.

EQUIDIFFERENT SERIES. A series of numbers composed of any number of terms, which uniformly increase or decrease by the same number, is called an EQUIDIFFERENT SERIES. This series has, very commonly, but without any propriety, been called Arithmetical Progression.

When the numbers increase, they form an ascending series; but when they decrease, a descending series. Thus, the natural numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, form an ascending series, because they continually increase by 1; but 9, 8, 7, 6, &c. form a descending series, because they continually decrease by 1.

The numbers, which form the series, are called the terms of the series. The first and last terms in the series are called the extremes; and the other terms, the means.

The number, by which the terms of the series are continually increased or diminished, is called the common difference. Therefore, when the first term and common difference

are given, the series may be continued to any length. For instance, let 1 be the first term in an equidifferent series, and 3 the common difference, and we shall have the following increasing series; 1, 4, 7, 10, 13, &c., in which each succeeding term is found by adding the common difference to the preceding term.

THEOREM I. When four numbers form an equidifferent series, the sum of the two extremes is equal to the sum of the two means. Thus, 1, 3, 5, 7, is an equidifferent series, and 1+7=3+5. Also in the series 11, 8, 5, 2, 11+2=8+5.

THEOREM II. ' In any equidifferent series, the sum of the two extremes is equal to the sum of any two means, that are equally distant from the extremes; and equal to double the middle term, when there is an uneven number of terms. Take, for example, the equidifferent series, 2, 4, 6, 8, 10, 12, 14; 2+14=4 +12; and 2.4-14 6.+10; also 2 +14=8+8.

Since, from the nature of an equidifferent series, the second term is just as much greater or less than the first, as the last but one is less or greater than the last, it is evident, that when these two means are added together, the excess of the one will make good the deficiency of the other, and their sum will be the same with that of the two extremes. In the same manner it appears, that the sum of any other two means equally distant from the extremes, must be equal to the sum of the extremes.

THEOREM III. The difference between the extreme terms of an equidifferent series is equal to the common difference multiplied by the number of terms less 1. Thus, of the six terms, 2, 5, 8, 11, 14, 17, the common difference is 3, and the number of terms less 1, is 5; then the difference of the extremes is 17-2, and the common difference multiplied by the number of terms less 1, is 3 X 5; and 17-2=3X 5.

The difference between the first and last terms, is the increase or diminution of the first by all the additions or subtractions, till it becomes equal to the last term: and, as the number of these equal additions or subtractions is one less than the number of terms, it is evident that this common difference being multiplied by the number of terms less 1, must give the difference of the extremes.

THEOREM IV. The sum of all the terms of any equidifferent series is equal to the sum of the two extremes inultiplied by the number of terms and divided by 2; or, which amounts to the same, the sum of all the terms is equal to the sum of the extremes multiplied by half the the number of terms. For example, the sum of the following series, 2, 4, 6, 8, 10, 12, 14, 16, is 2+16 X:45:72. This is made evident by writing under the given series the same series inverted, and adding the corresponding terms together as follows.

The given series, 2, 4, 6, 8, 10, 12, 14, 10 Same series inverted, 16, 14, 12, 10, S, 6, 4, 2

Sums of the series, 18, 18, 18, 18, 18, 18, 18, 18 This series of equal terms, (18), is evidently equal tu twice the sum of the given series; but the sum of these

equal terms is 18 X8=144; and since this sum is twice as great as that of the given series, the sum of the given series must be 72.

Any three of the five following things being given, tho other two may be readily sound.

The first term.
The last term.
The number of terms.
The common diference.
The sum of all the terms.

PROBLEM. I. The extremes and number of terms being given, to find the sum of all the terms.

RULE. Multiply the sum of the extremes by the number of the terms, and half the product will be the sum of all the terms. See Theorem 4th.

1. The first term in an equidifferent series, is 3, the last term 19, and the number of terms is 9. What is the sum of the whole series ?

2. How many strokes does a common clock strike in 12 hours ?

3. A hundred cents were placed in a right line, a yard apart, and the first a yard from a basket. What distance did the boy travel, who, starting from the basket, picked them up singly, and returned with them one by one to the basket ?

4. If a number of dollars were laid in a straight line for the space of a mile, a yard distant from each other. and the first a yard from a chest, what distance would the man travel, who, starting from the chest, should pick them up singly, returning with them one by one to the chest ?

PROBLEM II. The extremes and number of terms given, to find the common difference.

RULE. Subtract the less extreme from the greater, and divide the remainder by the number of terms less 1,

and the quotient will be the common difference.

It has been shown under Theorem 3d. that the Hiffer

ence of the extremes is found by multiplying the cominon difference by the number of the terms less 1, consequently, the common difference is found by dividing the difference of the extremes by the number of the terms less 1.

5. A man had 10 sons, whose ages differed alike; the voungest was 2 years old, and the eldest 29. What was the difference of their ages ?

6. The extremes in an equidifferent series are 3 and 87, and the number of terms 43. Required the common difference.

7. A man is to travel from Boston to a certain place in 9 days, and to go but 5 miles the first day, and to increase his journey every day alike, so that the last day's journey may be 37 miles. Required the daily increase, and also the number of miles travelled.

PROBLEM III. The extremes and common difference given, to find the number of terms.

RULE. Divide the difference of the extremes by the common difference, and add 1 to the quotient; the sum will be the number of terms.

The difference of the extremes divided by the number of the terms less 1, gives the common difference; consequently, the same divided by the common difference inust give the number of terms less 1: hence, this quotient augmented by 1, must give the number of terms.

8. The extremes in an equidifferent series are 3 and 39, and the common difference is 2: what is the number of terms ?

9. A man going a journey, travelled 7 miles the first day, and increased his journey every day by 4 miles, and the last day's journey was 51 miles. How many days did he travel, and how far ?

10. A man commenced a journey with great animation, and travelled 55 miles the first day; but on the second day he began to be weary, and travelled only 51 miles, and thus continued to lose 4 miles a day, till his last day's journey was only 15 miles. How many days did he travol ?

PROBLEM IV. To find an equidifferent mean between two given terms.

RULE. Add the two given terms together, and half their sum will be the equidifferent mean required.

11. Find an equidifferent mean between 3 and 15. 12. What is the equidifferent mean between 7 and 53? 13. Find an equidifferent mean between 5 and 18.

PROBLEM V. To find two equidifferent means between the given extremes.

Divide the difference of the extremes by 3, and the quotient will be the common difference, which, being continually added to the less extreme, or subtracted, from che greater, gives the two required means.

14. Find two equidifferent means between 4 and 13. 15. Find two equidifferent means between 5 and 22. 16. Find two equidifferent means between 4 and 53.

RULE.

PROBLEM VI. To find any numbers of equidifferent means between the given extremes.

RULE. Divide the difference of the extremes by the required number of means plus 1, and the quotient will be the common difference, which being continually added to the less extreme, or subtracted from the greater, will give the mean terms required.

17. Find five equidifferent means between 4 and 23 18. Find six equidifferent means between 6 and 55. 19. Find 3 equidifferent means between 34 and 142. 20. Find one equidifferent mean between 56 and 100

XXXIII.

CONTINUAL PROPORTIONALS. The numbers of a series in which the successive terms increase by a common multiplier, or decrease by a com mon divisor, are CONTINUAL PROPORTIONALS.

'This series of numbers bas been commonly called a Geometrical Progression; but, perceiving no appropriate

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