added to t d, which is equal to a b the first pile, to make cd the second, and roadded to r f, which is equal to the second pile, to make o f the third. In the same manner all the others are formed. As we do not add to make the first pile, we can ad 2 only 8 times, which is once less than the number of terms to make the last, or the 9th pile. Now if we suppose c t, or, &c. to be removed on to the column g l, as all the small squares are supposed equal, they would just fill that column, as is now represented, which represents 16 blocks, the number found by multiplying 2 the common difference, by 8 one less than the number of terms; and if a g and b h be parallel, g h is equal to the first term a b, and l h will represent the 9th pile or last term of the series. Froin this it appears that the last term of an increasing series, will always be made up of the common difference taken as many times as there are terms wanting one and the first term, for we may suppose a b to be any other number than 5, and g l will always be made up of the common difference, whether it be 2 or any other number. 2. If all the blocks represented by the small squares above the line m n, be removed and placed upon the dotted squares, there would be just enough to fill them, and in h b would represent all the blocks contained in the 9 piles, that is, each pile would average the same number as is contained in the middle one me; and if we multiply m e or n h by h b, the product would express the number of blocks contained in all the piles ; but if we add n h and i b together, the number would be twice n h, and multiplying twice n h by h b the number of piles, would give twice the whole number of blocks, or twice the sum of the series. It should be observed, that the number of blocks contained in n h and i b represent twice the number contained in the middle term 7?? 6. 3. Let the number of squares in a y equal the number in l g, then a y represents the number of
blocks added after the first term is subtracted. If the number contained in a y be divided by the number of times the additions were made, the quotient would be the number added, that is, the common difference; and if the number be divided by 2, the guotient would be 8, the number of times the common difference was added, which is once less than the number of terms. •eeGessARITHMETICAL PROPORTION. When two numbers are compared together to ascertain how much larger one is than the other, or how many times one contains the other, the first is called the antecedent, the second the consequent; both taken together, a couplet. If the difference between the antecedent and consequent of two or more couplets, be the same num: ber, this equality of difference is termed arithmetical proportion, and each number, considered in connection with the others, is called an arithmetical proportional. The proportion of numbers is expressed by placing a colon between the antecedent and consequent of each couplet, and a double colon between the couplets. Thus, 2 : 4 :: 6:8, are four numbers in arithmetical proportion; and are read, as 2 is to 4, so is 6 to 8. Proportion is distinguished into continued and discontinued. When the difference between the consequent of one couplet, and the antecedent of the next, is the same number as the common difference of the couplets, the proportion is continued; but when the difference between the consequent of one couplet and the antecedent of the next, is not the same number as the common difference of the couplets, the proportion is discontinued. Thus, 2 : 4 :: 6:8, is continued proportion; but, 2:4::10: 12, is discontinued proportion. PROMLEM I, To find a mean proportional between two given eactremes. RULE, Take half the sum of the two extremes, the quotient will be an arithmetical mean proportional between the two given extremes. 1. What is the mean proportional between 7 and 17 ? Ans. 12. 2. A certain debt can be discharged at 3 payments in arithmetical proportion. If the first payment be $11, and the third $31,17, what is the second 7 Ans. $21,085. PROBLEM II, To find any number of arithmetical means between two given terms. RULE, Subtract the less extreme from the greater, and divide the difference by one more than the number of means required ;-this quotient added to the first term, if the proportion is increasing, or subtracted, if it be decreasing, will give the first mean ; added to, or subtracted from the first mean, will give the second. Continue the same process till all the means are found. 3. Required to find four arithmetical means between 1 and 96. Ans. 20, 39, 58, 77. 4. If a man work 6 days for wages in arithmetical proportion, and receive 29 cents for his first day's - w work and 79 for the last, what does he receive for the intervening days 2 Ans. 39, 49, 59, 69 cents. -soeGEOMETRICAL PROGRESSION. When a series of numbers is produced by continually multiplying or dividing by the same number, it is called geometrical progression. The number by which a series in geometrical progression is increased or decreased, is called the ratio. Thus, 1, 2, 4, 8, 16, 32, is an increasing series in geometrical progression, having 2 for its ratio ; and 81, 27, 9, 3, 1, is a decreasing series, having 3 for its ratio. All the other terms have the same names as the corresponding terms in arithmetical progression. PROBLEM Is Raise the ratio to a power denoted by one less than the number of terms, and multiply that power by the first term, if the series be increasing, but if decreasing, divide by the same power. Illustration.—When the first term of an increasing series is one, the second term will always be the same number as the ratio; consequently, multiplying by the ratio to find the last term, is only raising the ratio to a power denoted by one less than the number of terms. But when the first term is any number more than one, the second term will contain the ratio as many times as the first term contains an unit; so that the last term will contain such a power of the ratio, as is denoted by one less than the number of terms, as many times as the first term contains an unit. Thus, in the series 1, 2, 4, 8; 8, the last term, is simply the third power of the ratio. But in the series, 2, 4, 8, 16, the last term contains twice the third power of the ratio. - When the progression is decreasing, it may be considered as increasing from the last term; consequently dividing the first term by that power of the ratio denoted by one less than the number of terms, will give the last term. * . 1. Let the first term of a geometrical series be 5, the ratio 3, number of terms 10; what is the last term 2 - Ans. 98.415. 2. Given the first term of a decreasing series, 1000, number of terms 5, the ratio 4, to find the last term. Last term, 3}}. 3. The first term of a series is ro, the ratio 1's, the number of terms 4 ; what is the last term 2 - - Ans. ####. 4. A man agrees to sell 10 bushels of wheat at 1 mill for the first bushel, and 5 times as much for each succeeding bushel as what he receives for the preceding one; what will he receive for the last 6. If a balloon descend from a certain height in 6 minutes, and descend r3+ of a mile the first minute, three times that distance the second, 9 times as far the third minute as the first, and increase its velocity during the remainder of its descent by the same ra tio; how many feet will it descend the last minute 2 Ans. 102644;. PROBLEM II, To find the sum of all the terms. Illustration.—Let it be required to find the sum of the four terms in the series 1, 2, 4, 8. If we add all the terms together, we shall obtain 15 for the sum. The second term contains the ratio once, the third |