Ex. 3. How many strokes do the clocks of Venice strike in the compass of a 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 5l 3s? Ans. 135l 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 57 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 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. I. 16 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 5l 3s, and the debt is 135l 4s. PROBLEM IV. To find an Arithmetical Mean 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 Extremes. 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 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 com. dif. continually, the means are found 4, 6, 8, 10, 12. com. dif. 2 See more of Arithmetical progression in the Algebra. GEOMETRICAL PROPORTION AND GEOMETRICAL PROGRESSION Or Progression by equal Ratios. IN Geometrical Progression the numbers or terms have all the same multiplier or divisor. The most useful part of Proportion, is contained in the following theorems. THEOREM 1. When four quantities are in 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 × 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 12 2 = 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 them, 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 X 4 = 16. And in the series 2, 4, 8, 16, 32, 64, 128, it is 2 X 1284 X 648 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 term is 9; then the quotient of the extremes is 1024 ÷ 2 = 512, and 2o 512 also. THEOREM THEOREM 4. The sum of all the terms, of any geometrical progression, is found by adding the greatest term to the difference of the extremes divided by 1 less than the ratio. So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1024-2 (whose ratio is 2), is 1024 + 2-1 1024 +1022=2016. The foregoing, and several other properties of proportion, are demonstrated more at large in the Algebraic part of this work. A few examples may here be added of the theorems, just delivered, with some problems concerning mean proportionals. EXAMPLES. 1. The least of ten terms, in geometrical progression, being 1, and the ratio 2; what is the greatest term, and the sum of all the terms? Ans. The greatest term is 512, and the sum 1023. 2. What debt may be discharged in a year, or 12 months, by paying 11 the first month, 2l the second, 4l the third, and so on, each succeeding payment being double the last; and what will the last payment be? Ans. The debt 40951, and the last payment 20481. PROBLEM I. To find One Geometrical Mean Proportional between any two Numbers. MULTIPLY the two numbers together, and extract the square root of the product, which will give the mean proportionat sought. EXAMPLE. To find a geometrical mean between the two numbers 3 and 12. |