tion; for 4-2-8-6-2-common difference of the couplets, 8—2— 6 difference of the consequent of one couplet and the antecedent of the next; also, 4, 2, 16, 8, are in discontinued geometrical pro 4 16 portion; for =—=2= common ratio of the couplets, and 2 2 8 ratio of the consequent of one couplet and the antecedent of the next. * ARITHMETICAL PROPORTION. THEOREM 1. IF any four quantities a, b, c, d, (2, 4, 6, 8) be in arithmetical proportion, the sum of the two means is equal to the sum of the two extremes.+ And if any three quantities, a, b, c, (2,4, 6,) be in arithmetical proportion, the double of the mean is equal to the sum of the ex tremes. THEOREM 2. In any continued Arithmetical Proportion (1, 3, 5, 7, 9, 11) the sum of the two extremes, and that of every other two terms, equally distant from them, are equal. Thus, 1+11=3+9=5+7.‡ When the number of terms is odd, as in the proportion 3. 8. 13. 18. 23, then, the sum of the two extremes being double to the mean or middle term, the sum of any other two terms, equally remote from the extremes, must likewise be double to the mean. THEOREM 3. In any continued Arithmetical Proportion, (a, a+b, a+2b, a+3b, at 4b, &c. 4, 4+2, 4+4, 4+6, 4+8, &c.) the last or greatest term is equal to the first or least more the common difference of the terms drawn into the number of all the terms after the first, or into the whole number of the terms, less one.§ THEOREM Although, in the comparison of quantities according to their differences, the term proportion is ufed: yet the word progreffion, is frequently fubftituted in its room, and is indeed more proper; the former form being, in the common acceptation of it, fynonymous with ratio, which is only used in the other kind of comparison. For fince b-a (4-2)=d-c(8-6) therefore b+c(4+6)=a+d(2+8.) Since, by the nature of progreffionals, the fecond term exceeds the first by just fo much as its corresponding term, the last but one, wants of the lait, it is evident that when thefe correfponding terms are added, the excefs of the one will make good the defect of the other, and so their fum be exactly the fame with that of the two extremes, and in the fame manner it will appear that the fum of any two other correfponding terms must be equal to that of the two extremes. For fince each term, after the first, exceeds that preceding it by the common difference, it is plain that the last must exceed the first by fo many times the comnon difference as there are terms after the first; and therefore must be equal to the first, and the common difference repeated that number of times. THEOREM 4. The sum of any rank, or series of quantities in continued Arithmetical Proportion (1. 3. 5. 7.9 11) is equal to the sum of the two extremes multiplied into half the number of terms.* ARITHMETICAL PROGRESSION. ANY rank of numbers, more than two, increasing by a common excess, or decreasing by a common difference, is said to be in Arithmetical Progression. If the succeeding terms of a progression exceed each other, it is called an ascending series or progression; if the contrary, a descend. ing series. 8. 2. 4. 8. 16. So 1. 8. 6. 4. 32. 16. 8. 4. The numbers which progression 2. 2. 10, &c. is an ascending arithmetical series. 32, &c. is an ascending geometrical scries. 0, &c. is a descending arithmetical series. 1, &c. is a descending geometrical series. form the series, are called the terms of the Note. The first and last terms of a progression are called the extremes, and the other terms the means. Any three of the five following things being given, the other twe may be easily found. 1. The first term. 2. The last term 3. The number of terms. 4. The common difference. 5. The sum of all the terms. PROBLEM For, because (by the fecond Theorem) the sum of the two extremes, and that of every other two terms, equally remote from them, are equal, the whole feries, confifting of half fo many fuch equal fums as there are terms, will therefore be equal to the fum of the two extremes, repeated half as many times as there are terins. The fame thing also holds, when the number of terms is odd, as in the feries 4, 8, 12, 16, 20; for then, the mean, or middle term, being equal to half the fum of any two terms, equally diftant from it on contrary fides, it is obvious that the value of the whole feries is the fame as if every term thereof were equal to the mean, and therefore is equal to the mean (or half the fum of the two extremes) multiplied by the whole number of terms; or to the fum of the extremes multiplied by half the number of terms. The fum of any number of terms (x) of the arithmetical feries of odd numbers 1, 3, 5, 7, 9, &c. is equal to the square (x2) of that number. For, 0+1 or the fum of 1 term = 12 or 1 16+9 or the fum of 5 terms === 52 or 25, &c. Whence, it is plain, that, let x be any number whatever, the fum of x term will be x. EXAMPLE. The first term, the ratio, and number of terms given, to find the fum of the feries. A gentleman travelled 29 days, the first day he went but 1 mile, and increased every day's travel 2 miles; How far did he travel? 29X29-841 miles, Ans PROBLEM I. The first term, the last term, and the number of terms being given, to find the common difference. RULE.* Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference sought. EXAMPLES. 1st. The extremes are 3 and 39, and the number of terms is 19: What is the common difference? 36 Divide by the number of terms less 1-19-1-18)36(2 Ans. 2d. A man had 10 sons, whose several ages differed alike; the youngest was 3 years old, and the eldest 48: What was the common difference of their ages? 3d. A man is to travel from Boston to a certain place in 9 days, and to go but 5 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 37 miles: Required the daily increase? The first term, the last term, and the number of terms being given, to find the sum of all the terms. RULE.+-Multiply the sum of the extremes by the number of terms, and half the product will be the answer. EXAMPLES. as The difference of the first and last terms evidently fhews the increase of the firft term by all the subrequent additions, till it becomes equal to the laft, a.. the number of thofe additions was one lefs than the number of terms, and the increafe, by every addition, equal, it is plain that the total increase, divided by the number of additions, must give the difference of every one feparately; when.ce the rule is manifeft. Suppofe another series of the fame kind with the given one be placed under it in an inverie order; then will the fum of any two corresponding terms be the fame as that of the first and laft; confequently, any one of thofe fums, multiplied by the number of terms, must give the whole fum of the two feries. Let 1, 2, 3, 4, 5, 6, 7, 8, be the given feries. 72 1+2+3+4+5+6+7+8=—=36. 2 EXAMPLES. 1st. The extremes of an arithmetical series are 3 and 39, and the number of terms 19: Required the sum of the series? 2d. It is required to find how many strokes the hammer of a clock would strike in a week, or 168 hours, provided it increased 1 at each hour? 168+1x168 = 14196 Ans. 2 3d. Suppose a number of stones were laid a yard distant from each other for the space of a mile, and the first a yard from a basket: What length of ground will that man travel over, who gathers them up singly, returning with them one by one to the basket? 3520+2×1760 2 =3099360 yards = 1761 miles, Ans. N. B. In this question, there being 1760 yards in a mile, and the man returning with each stone to the basket, his travel will be doub. led; therefore the first term will be 2, and the last 1760×2, and the number of terms 1760. 4th. A man bought 25 yards of linen in Arithmetical Progression; for the 4th yard he gave 12 cents, and for the last yard 75 cents: What did the whole amount to, and what did it average per yard? 75-12 22-1 3 the common difference by which the first term is found to [be 3. 75+3×25 Then9D. 75c. and the average price is 39 cents per yard. 2 5th. Required the sum of the first 1000 numbers in their natural order? PROBLEM III. Given the extremes and the common difference, to find the number of terms. RULE.*-Divide the difference of the extremes by the common difference, and the quotient increased by 1 will be the number of terms required. EXAMPLES. 1st. The extremes are 3 and 39, and the common difference 2; What is the number of terms? 2d. A man going a journey, travelled the first day 7 miles, the last day 51 miles, and each day increased his journey by 4 miles: How many days did he travel, and how far? The extremes and common difference given, to find the sum of all the series. RULE.-Multiply the sum of the extremes by their difference increased by the common difference, and the product divided by twice the common difference will give the sum. EXAMPLES. 1st. If the extremes are 3 and 39, and the common difference 2 : What is the sum of the series? 39+3=42 sum of the extremes 39-3-36-difference of extremes. 36+2=38=difference of extremes increased by the common differ By the firft Problem, the difference of the extremes, divided by the number of terms lefs 1, gave the common difference; confequently, the fame divided by the common difference, muft give the number of terms lefs 1; hence, this quo tient, augmented by 1, must be the answer to the question. |