Hence, when the first term, the common difference, and the number of terms, are given, to find the last term,-Multiply the number of terms, less 1, by the common difference, and add the first term to the product for the last term. 2. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term? Ans. 301. 3. There are, in a certain triangular field, 41 rows of corn; the first row, in 1 corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2; what is the number of hills in the last row? Ans. 81 hills. 4. A man puts out $1, at 6 per cent. simple interest, which, in 1 year, amounts to $1'06, in 2 years to $1'12, and so on, in arithmetical progression, with a common difference of $'06; what would be the amount in 40 years? Ans. $3'40. Hence we see, that the yearly amounts of any sum, at simple interest, form an arithmetical series, of which the principal is the first term, the last amount is the last term, the yearly interest is the common difference, and the number of years is 1 less than the number of terms. 5. A man bought 100 yards of cloth in arithmetical pro gression; for the first yard he gave 4 cents, and for the last 301 cents; what was the common increase of the price on each succeeding yard? 201 This question is the reverse of example 1; therefore, 4297, and 297 ÷ 99 = 3, common difference. Hence, when the extremes and number of terms are given, to find the common difference,-Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference. 6. If the extremes be 5 and 605, and the number of terms 151, what is the common difference? Ans. 4 7. If a man puts out $1, at simple interest, for 40 years, and receives, at the end of the time, $3'40, what is the rate? If the extremes be 1 and 3'40, and the number of terms 41, what is the common difference? Ans. '06. alike; the what was Ans. 5 years. 8. A man had 8 sons, whose ages differed youngest was 10 years old, and the eldest 45; the common difference of their ages? 9. A man bought 100 yards of cloth in arithmetical series; he gave 4 cents for the first yard, and 301 cents for the last yard; what was the average price per yard, and what was the amount of the whole? Since the price of each succeeding yard increases by a com stant excess, it is plain, the average price is as much less than the price of the last yard, as it is greater than the price of the first yard; therefore, one half the sum of the first and last price is the average price. 15250 cts.= One half of 4 cts.+301 cts. = 152 cts. average price; and the price, 152 cts. X 100 $152'50, whole cost. Hence, when the extremes and the number of terms are given, to find the sum of all the terms,-Multiply the sum of the extremes by the number of terms, and the product will be the answer. 10. If the extremes be 5 and 605, and the number of terms 151, what is the sum of the series? Ans. 46055. 11. What is the sum of the first 100 numbers, in their natural order, that is, 1, 2, 3, 4, &c. ? Ans. 5050. 12. How many times does a common clock strike in 12 hours? Ans. 78. 13. A man rents a house for $50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent., simple interest, for the use of the money? The last year's rent will evidently be $50 without interest, the last but one will be the amount of $50 for 1 year, the last but two the amount of $50 for 2 years, and so on, in arithmetical series, to the first, which will be the amount of $50 for 19 years = $107. If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series? Ans. $1570. 14. What is the amount of an annual pension of $100, being in arrears, that is, remaining unpaid, for 40 years, allowing 5 per cent. simple interest? Ans. $7900. 15. There are, in a certain triangular field, 41 rows of eorn; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills; how many hills of corn in the field? Ans. 1681 hills. 16. If a triangular piece of land, 30 rods in length, be 20 rods wide at one end, and come to a point at the other, what number of square rods does it contain? Ans. 300. 17. A debt is to be discharged at 11 several payments, in arithmetical series, the first to be $ 5, and the last $75; what is the whole debt? the common difference be tween the several payments? Ans. whole debt, $440; common difference, $7. 18. What is the sum of the series 1, 3, 5, 7, 9, &c., to 1001 ? Ans. 251001. Note. By the reverse of the rule under ex. 5, the difference of the extremes 1000, divided by the common difference 2, gives a quotient, which, increased by 1, is the number of terms 501. 19. What is the sum of the arithmetical series 2, 21, 3, 84, 4, 41, &c., to the 50th term inclusive? Ans. 712. 20. What is the sum of the decreasing series 30, 294, 291, 29, 283, &c., down to 0? Note. 30+1 = 91, number of terms. Ans. 1365. QUESTIONS. 1. What is an arithmetical progression? 2. Wher is the series called ascending? 3. when descending? 4. What are the numbers, forming the progression, called? 5. What are the first and last terms called? 6. What are the other terms called? 7. When the first term, common difference, and number of terms, are given, how do you find the last term? 8. How may arithmetical progression be applied to simple interest ? 9. When the extremes and number of terms are given, how do you find the common difference? how do you find the sum of all the terms? 10. GEOMETRICAL PROGRESSION. 113. Any series of numbers, continually increasing by a constant multiplier, or decreasing by a constant divisor, is called a Geometrical Progression. Thus, 1, 2, 4, 8, 16, &c. an increasing geometrical series, and 8, 4, 2, 1,,, &c is a decreasing geometrical series. As in arithmetical, so also in geometrical progression, there are five things, any three of which being given, the other two may be found :— 1st. The first terni. 2d. The last term. 3d. The number of terms. 4th. The ratio. 5th. The sum of all the terms. -The ratio is the multiplier or divisor, by which the series formed. 1. A man bought a piece of silk, measuring 17 yards, and, by agreement, was to give what the last yard would come to, reckoning 3 cents for the first yard, 6 cents for the second, and so on, doubling the price to the last; what did the piece of silk cost him? 3 X 2 X 2 X 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 X2 X2 X2 X2 196608 cents, $1966'08, Answer. = = In examining the process by which the last term (196608) has been obtained, we see, that it is a product, of which the ratio (2) is sixteen times a factor, that is, one time less than the number of terms. The last term, then, is the sixteenth power of the ratio, (2,) multiplied by the first term (3.) Now, to raise 2 to the 16th power, we need not produce all the intermediate powers; for 24 2 X 2 X 2 X 2 = 16, is a product of which the ratio 2 is 4 times a factor; now, if 16 be multiplied by 16, the product, 256, evidently contains the same factor (2) 4 times + 4 times, = 8 times; and 256 × 256 = 65536, a product of which the ratio (2) is 8 times 8 times, 16 times, factor; it is, therefore, the 16th power of 2, and, multiplied by 3, the first term, gives 196608, the last term, as before. Hence, When the first term, ratio, and number of terms, are given, to find the last term,— I. Write down a few leading powers of the ratio with their indices over them. II. Add together the most convenient indices, to make an index less by one than the number of the term sought. III. Multiply together the powers belonging to those indices, and their product, multiplied by the first term, will be the term sought. 2. If the first term be 5, and the ratio 3, what is the 8th term? Powers of the ratio, with . their indices over them. 1 2 3+4= 7 3, 9, 27, × 81 = 2187 × 5 first term, = 10935, Answer. 3. A man plants 4 kernels of corn, which, at harvest, produce 32 kernels; these lie plants the second year; now, supposing the annual increase to continue 8 fold, what would be the produce of the 16th year, allowing 1000 kernels to a pint? Ans. 2199023255'552 bushels. 4. Suppose a man had put out one cent at compound interest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years? 217 131072. Ans. $131072. 5. A man bought 4 yards of cloth, giving 2 cents for the first yard, 6 cents for the second, and so on, in 3 fold rawhat did the whole cost him? 2+6+18+54 80 cents. tio ; Ans. 80 cents. In a long series, the process of adding in this manner would be tedious. Let us try, therefore, to devise some shorter method of coming to the same result. If all the terms, excepting the last, viz. 2+6+18, be multiplied by the ratio, 3, the product will be the series 6+18+54; subtracting the former series from the latter, we have, for the remainder, 542, that is, the last term, less the first term, which is evidently as many times the first series (2+6+18) as is expressed by the ratio, less 1: hence, if we divide the difference of the extremes (54 2) by the ratio, less 1, (31,) the quotient will be the sum of all the terms, excepting the last, and, adding the last term, we shall have the whole amount. Thus, 54 - 2 52, and 3-1= 2; then, 52 ÷÷ 2 = 26, and 54 added, makes 80, Answer, as before. Hence, when the extremes and ratio are given, to fend the m of the series,-Divide the difference of the extremes by the ratio, less 1, and the quotient, increased by the graxter term, will be the answer. 6. If the extremes be 4 and 131072, and the ratio 8, what is the whole amount of the series? 131072 -4 8-1 + 131072 = 149796 Answer. |