Proportionals, or a progression. So 2, 4, 6, 8 form an arithmetical progression, because 4 – 2 = 6 4 = 8-6= 2, all the same cominon difference; and 2, 4, 8, 16 a geometrical progression, because === 2, 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 : TiteoREM 1. When four quantities are in arithmetical proportion, the sum of the two extremes is equal to the gum 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 froin 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. THE REM 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 20 -2 = 18, 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 arithmetical 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. 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, HI, 13, ditto inverted 15, 13, 11, 9, 7, 5, 3, the sums are 16 + 16 fe 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: 153 1; PROBLEM I. Given the Extremes, and the Number of Termis ; to find the Sum of all the Terms. Add the extremes together, multiply the sum by the number of terms, and divide by 2. EXAMPLES. 1. The extremes being 3 and 19, and the number of terins 9; required the sum of the terms ? 19 3 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. 3. How many strokes do the clocks of Venice strike in the compass of the day, which go continually on from 1 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 5/38? Ans. 1351 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 tèrms, for the common difference. EXAMPLES. 1. The extremes being 3 and 19, and the number of terms 9; required the common difference? 19 19 3 Or; 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 is, and the last 51 3s; what is the common difference of the terms ? Ans. 2. Given one of the Extremes, the Common Difference, and the Number of Terms : to find the other Extreme, and the Sum of the Series. Multiplý the common difference by 1 less than the number of termis, 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 EXAMPLES. 1. Given the least terin 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series: 2 16 19 the greatest termi 22 sun 2 ) 198 99 the sum of the series. 2. If the greatest term be 70, the common difference 3, and the number of terms 21, what is the least term, and the sum of the series? Ans. The least term is 10, and the sum is 840. 3. A debt can be discharged in a year, by paying 1 shilling the first week, 3 shillings the second, and so on, always 2 shillings more every week; what is the debt, and what will the last payment be? Ans. The last payment will be 51 3s, and the debt is 1351 4s. 1 PROBLEM IV. To find an Arithmetical Mean Proportional between Two Given Terms. Add the two given extremes or terms together, and take half their sum for the arithmetical mean required. EXAMPLE. To find an arithmetical mean between the two numbers 4 and 14. Here 2) 18 Ans. 9 the mean required. PROBLEM PROBLEM V. To find Two Arithmetical Means between Two Given Extreines. SUBTRACT the less extreme from the greater, and divide the difference by 3, so will the quotient be the common difference; which being continually added to the less extreme, or taken from the greater, gives the means. EXAMPLE To find two arithmetical means between 2 and s. 2 3) 6 Then 2 + 2 = 4 the one meane and 4 + 2 = 6 the other mean. com. dif. 2 PROBLEM VI. To find any Number of Arithmetical Means between Two Given, Terms or Extremes. SUBTRACT the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required. EXAMPLE. To find five arithmetical means between 2 and 14. 2 6) 12 Then by adding this com. dif. continually, the means are found 4, 6, 8, 10, 12. cam. dif. 2 See more of Arithmetical progression in the Algebra. I 2 GFOMETRICAL |