RULE. Multiply half the sum of the extremes by the number of terms. EXAMPLES. 1. The first term of an arithmetical progression is 2, the last term is 50, and the number of terms is 17. What is the sum of all the terms? In this example, half the sum of the extremes is of (2+50)=26. This, multiplied by the number of terms, gives 26 x 17= 442, for the sum required. 2. The first term of an arithmetical progression is 13, the last term is 1003, the number of terms is 100. What is the sum of the progression? Ans. 50800. 3. A person travels 25 days, going 11 miles the first day, 135 the last day; the miles which he traveled in the successive days, form an arithmetical progression. How far did he go in the 25 days? Ans. 1825 miles. 4. Bought 7 books, the prices of which are in arithmetical progression. The price of the first was 8 shillings, and the price of the last was 28 shillings. all come to? What did they Ans. £6 6s. 5. What is the sum of 1000 terms of an arithmetical progression, whose first term is 7 and last term 1113? Ans. 560000, 6. The first term of an arithmetical progression is, and the last term 365+, and the number of terms 799. What is the sum of all the terms? Ans. 146217. GEOMETRICAL PROGRESSION 138. A SERIES of numbers which succeed each other regularly, by a constant multiplier, is called a geometrical progression. This constant factor, by which the successive terms are multiplied, is called the ratio. When the ratio is greater than a unit, the series is called an ascending geometrical progression. When the ratio is less than a unit, the series is called a descending geometrical progression. Thus, 1, 3, 9, 27, 81, &c., is an ascending geometrical progression, whose ratio is 3. And, 1,,, &c., is a descending geometrical progression, whose ratio is 4. In geometrical progression, as in arithmetical progres sion, there are five things to be considered. 1 The first term. 2. The last term. 3. The common ratio. 4. The number of terms. 5. The sum of all the terms. These quantities are so related to each other, that any three being given, the remaining two can be found. The solution of some of these cases requires a knowledge of higher principles of mathematics than can be detailed by arithmetic alone. We will give a demonstration of the rules of one or two of the most important cases. When are numbers in geometrical progression? What is the constant factor, by which the successive terms are multiplied, called? When this ratio exceeds a unit, the progression is called what? When this ratio is less than a unit, how is the progression called? Give an example of an ascending geometrical progression? Give an example of a descending geometrical progression. How many quantities are to be considered in geometrical progression? Mention these quantities. How mary of these must be known to enable us to find the others? CASE I. By the definition of a geometrical progression, it follows that the second term is equal to the first term, multiplied by the ratio; the third term is equal to the first term, multiplied by the second power of the ratio; the fourth term is equal to the first term, multiplied by the third power of the ratio; and so on, for the succeeding terms. Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this RULE. Multiply the first term by the power of the ratio, whose exponent is one less than the number of terms. EXAMPLES. 1. The first term of a geometrical progression is 1, the ratio is 2, and the number of terms is 7. What is the last term? In this example, the power of the ratio, whose exponent is one less than the number of terms, is 26-64, which, multiplied by the first term, 1, still remains 64, for the last term. 2. The first term of a geometrical progression is 5, the ratio is 4, and the number of terms 9. What is the last term? 27 Ans. 327680. 3. A person traveling, goes 5 miles the first day, 10 miles the second day, 20 miles the third day, and so on, increasing in geometrical progression. If he continue to travel in this way for 7 days, how far will he go the last day? Ans. 320 miles. CASE II. Let it be required to find the sum of all the terms of the geometrical progression 2, 6, 18, 54, 162, 486. If we multiply each term by 3, which is the ratio, we shall obtain this second progression, 6, 18, 54, 162, 486, 1458, the sum of whose terms is obviously 3 times as great as the sum of the terms of the first progression. Consequently, the difference between the sums of the terms of these two progressions is (3-1)=2 times the sum of all the terms of the first progression. If we omit the first term of the first progression, it will agree with the second progression, after omitting its last term. Hence, the difference between the sums of the terms of these two progressions may be found by subtracting 2, the first term of the first progression, from 1458, the last term of the second progression; but 1458 was obtained by multiplying 486, the last term of the first progression, by 3, the ratio. Hence, we finally obtain this condition : That the sum of all the terms of a geometrical progression, repeated as many times as there are units in the ratio, less one, is equal to the last term multiplied by the ratio, and diminished by the first term. Therefore, when we have given the first term of a geo metrical progression, the last term, and the ratio, to find the sum of all the terms, we have this RULE. Subtract the first term from the product of the last term into the ratio; divide the remainder by the ratio, less one. EXAMPLES. 1. The first term of a geometrical progression is 4, the ast term is 78732, and the ratio is 3. What is the sum of all the terms? In this example, the first term subtracted from the product of the last term into the ratio, is 236192, which, divided by the ratio, less one, gives 118096, for the sum of all the terms. 2. The first term of a geometrical progression is 5, the last term is 327680, and the ratio is 4. What is the sum of all the terms? Ans. 436905. 3. A person sowed a peck of wheat, and used the whole crop for seed the following year; the produce of this second year again for seed the third year, and so on. If in the last year, his crop is 1048576 pecks, how many pecks did he raise in all, allowing the increase to have been in a fourfold ratio? Ans. 1398101 pecks. 139. When the ratio of a geometrical progression is less than a unit, the first term will be the largest, and the last term the least; the progression will, in this case, bo descending; but if we consider the series of terms in a reverse order, that is, calling the last term the first, and the first the last, the progression may then be considered as ascending. If a decreasing geometrical progression. be continued to |