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, ditto inverted 15, 13, 3, 5, 7, 9, 9, 7, 11, 13, 15; 5, 3, 1; 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 number 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 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 is, and the last or 52d payment 5/ 38 ? Ans. 135/ 48. 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 scries? 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 18, and the last 5/ 38; 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 number 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. 1 EXAMPLES. EXAMPLES. 1. Given the least term 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series. 2 8 16 3 19 the greatest term 3 the least 22 sum 9 number of terms. 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 38, and the debt is 135/ 48. 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 4 2) 18 Ans. 9 the mean required PROBLEM PROBLEM V. To find two Arithmetical Means between Two Given Extremes, SUBTR ACT 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 8. 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 terms required. EXAMPLE. To find five arithmetical means between 2 and 14. 2 6) 12 Then by adding this cem. dif. continually, com, dif. 2 See more of Arithmetical progression in the Algebra. GEOMETRICAL GEOMETRICAL PROPORTION AND PROGRESSION. IN Geometrical Progression the numbers or terms have all the same multiplier or divisor. The most useful part of Geometrical Proportion, is contained in the following theorems. THEOREM 1. When four quantities are in geometrical proportion, the product of the two extremes is equal to the product of the two means. Thus, in the four 2, 4, 3, 6, it is 2 x 6 3 X 4 = 12. And hence, if the product of the two means be divided by one of the extremes, the quotient will give the other extreme. So, of the above numbers, the product of the means 122 = 6 the one extreme, and 126 2 the other extreme; and this is the foundation and reason of the practice in the Rule of Three. THEOREM 2. In any continued geometrical progression, the product of the two extremes is equal to the product of any two means that are equally distant from then, or equal to the square of the middle term when there is an uneven number of terms. Thus, in the terms 2, 4, 8, it is 2 X 8 = 4 4 16. And in the series 2, 4, 8, 16, 32, 64, 128, it is 2 X 128 = 4 × 64 = 8 X 32 = 16 X 16 = 256. THEOREM 3. The quotient of the extreme terms of a geometrical progression, is equal to the common ratio of the series raised to the power denoted by 1 less than the number of the terms. Consequently the greatest term is equal to the least term multiplied by the said quotient. So, of the ten terms, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, the common ratio is 2, and one less than the number of terms is 9; then the quotient of the extremes is 1024$512, and 2° 512 also. THEOREM |