Hence, 232-336=6-1476 inches nearly, the diameter of the ball required. .. 3. A cooper having a cask 40 inches long and 32 incl.es at the bung diameter, wishes to make another cask of the same shape, which shall contain just twice as much. What will be the dimensions of the new cask ? 5 4072=50-3968 inches, nearly, for length. ? 3252=40-3175 inches, nearly, for diameter. 4. What is the side of a cube, which will contain as much as a chest 8 feet 3 inches long, 3 feet wide, and 2 feet 7 inches deep ? Ans. 47.984 inches, nearly. 5. How many cubic quarter inches can be made out of a cubic inch? Ans. 64. 6. Required the dimensions of a rectangular box, which shall contain 20000 solid inches, the length, breadth, and depth being to each other, as 4, 3, and 2. SOLUTION. 2epee xx}x)=2500, whose cube root is 5v = 9.4103, nearly 79.4103x4=37.6412, length. Ans. 9.4103 x 3=28.2309, breadth. | 9:4103 x2=18.8206, depth. Or, as follows: If we were to augment the width of this box, so as to make it as wide as it is long, its volume would become of 20000=26666}. Again, if we augment the depth of this new box, so that it may be as deep as it is wide, and as it is long, its volume will become 2 times 26666 =53333}, which is the contents of a cubical box, whose side is equal to the length of the original box. Hence, 533331=37.641, nearly, for the length. The width is of this length, and the depth is this length. Note.—For a more complete treatise on the square and cube root, as well as the roots of all powers, see Higher Arithmetic. ARITHMETICAL PROGRESSION. 137. A SERIES of numbers, which succeed each other reguleriy, by a common difference, is said to be in arithmetical progression. When the terms are constantly increasips, the series is an arithmetical progression ascending. When the terms are constantly decreasing, the series is an arithmetical progression descending. Thus, 1, 3, 5, 7, 9, &c., is an ascending arithmetical progression; and 10, 8, 6, 4, 2, is a descending arithmetical progression. The terms of an arithmetical progression may be fractional. Thus, in the progressions, 1, 1, 13, 2, 27, 3, 31, 4, 47, &c.; $ $, 1, 13, 14, 2, 23, 24, 3, &c. The first has a common difference of }; the second has it common difference of }. In arithmetical progression, there are five things to be considered: 1. The first term. 2. The last term. & The comnon difference. These quantities are so related to each other, that any three of them being given, the remaining two can be found We will demonstrate one or two of the most important cases. When are numbers in arithmetical progression ? When is the progression ascending? When is it descending? Are the numbers 1, 3, 5, 7, 9, &c., in ascending or descending arithmetical progression ? Mention the five quantities to be considered in arithmetical progression. How many of these must be given in order to be able to find the others ? CASE I. By our definition of an ascending arithmetical progression, it follows that the second term is equal to the first, increased by the common difference; the third is equal to the first, increased by twice the common difference; the fourth is equal to the first, increased by three times the common difference; and so on, for the succeeding term. Hence, when we have given the first term, the common difference, and the number of terms, to find the last term, we have this RULE. To the first term add the product of the common difference into the number of terms, less one. EXAMPLES 1. What is the 100th term of an arithmetical progreso sion, whose first term is 2, and common difference 3 ? In this example, the number of terms, less one, is 99, which, multiplied by the common difference, 3, gives 297, which, added to the first term, 2, makes 299 for the 100th term. 2. What is the 50th term of the arithmetical progression, whose first term is 1, the common difference ? Ans. 251 3. A man buys 10 sheep, giving $1 for the first, $3 for the second, $5 for the third, and so on, increasing in arithmetical progression. What did the last sheep cost at that rate ? Ans. $19. 4. The first term of an arithmetical progression is the common difference, and the number of terms 26. What is the last term ? Ans. 37. CASE II. From the nature of an arithmetical progression, we see that the second term added to the next to the last term is equal to the first added to the last ; since the second term is as much greater than the first, as the next to the last is less than the last. After the same method of reasoning, we infer that the sum of any two terms equidistant from the extremes, is equal to the sum of the extremes. Hence, it follows that the terms will average just half the sum of the extremes. Therefore, when we have given the first term, the last term, and the number of terms, to find the sum of all the terms, we have this RULE. Multiply half the sum of the extremes by the number of terms. EXAMPLES. 1. The first term of an arithmetical progression is 2 che 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. 'What did they all come to ? Ans. £6 6s. 5. What is the sum of 1000 terms of an arithmetical pmgression, whose first term is 7 and last term 1113? Ans. 560000. 6. The first term of an arithmetical progression is , and the last term 3654, and the number of terms 799. What is the sum of all the terms ? Ans. 146217. |