difference, and to the product add the first term, the sum will be the last term. 2. Multiply the sum of the two extremes by the number of terms, and half the product will be the sum of all the terms. EXAMPLE. 1. A merchant bought 50 yards of linen, at 2 cents for the first yard, 4 for the second, 6 for the third, &c. increasing two cents every yard; what was the price of the last yard, how much the whole amount, and what the average price per yard? 50 number of terms 2+100=102 sum of the two extremes Multiply by 50 number of terms 2)5100 50) 25,50 sum of all the terms Ans. 51 cents 100 cents the last yard 25,50 do. the whole amount 51 do. the average price per yd. 2. Bought 20 yards of calico at 3 cents for the first yard, 6 for the second, 9 for the third, &c.; what did the whole cost? Ans. 6 dolls. 30 cents. 3. If 100 apples were laid two yards distant from each other, in a right line, and a basket placed two yards distant from the first apple, what distance must a person travel to gather them singly into the basket? Ans. 11 miles, 3 furlongs, 180 yards. 4. A agreed to serve B 10 years, at the rate of 20 dollars for the first year, 30 for the second, 40 for the third, &c.; what had he the last year, how much for the whole time, and what per annum? Ans. 110 dolls. for the last year, 650 dolls. the whole amount, and 65 dolls. per annum. 5. A sold to B 1000 acres of land, at 10 cents for the first acre, 20 for the second, 30 for the third, &c.; what was the price of the last acre, and what did the whole come to? 100 dolls. the last acre, 50050 do. whole cost. Ans. { Case 2. When the two extremes, and number of terms are given, to find the common difference. RULE. Divide the difference of the extremes, by the number of terms, less one; the quotient will be the common difference. EXAMPLE. 1. A is to receive from B a certain sum to be paid in 11 several payments in arithmetical progression; the first payment to be 20 dollars, and the last to be 100 dollars; what is the common difference, what was each payment, and how much the whole debt? Operation 100 last term 20 first term No. of terms 11-1-10) 80 the difference 8 common difference 20+100×5660 whole debt 20 first payment 20+8 28 second do. 28+8 36 third do. &c. 2. There are 21 persons whose ages are equally distant from each other, in arithmetical progression; the youngest is 20 years old, and the eldest 60; what is the common difference of their ages, and the age of each man? Ans. 2 common difference 3. A man is to travel from Pittsburgh to a certain place in 12 days, and to go but three miles the first day, increasing each day's journey in arithmetical progression, making the last day's travelling 58 miles; what is the daily increase, and what the whole distance? Ans. { 5 miles daily increase 366 miles whole distance. SECTION 6. GEOMETRICAL PROGRESSION. ANY series of numbers increasing or decreasing by one continual multiplier, or divisor, called the ratio, is termed geometrical progression; as 2, 4, 8, 16, 32, &c. increase by the multiplier 2; and 32, 16, 8, 4, 2, decrease, continually, by the divisor 2. In geometrical progression there are five things to be carefully observed. To find the last term, and sum of all the series in geometrical progression, work by the following RULE. 1. Raise the ratio in the given sum, to that power whose index shall always be one less than the number of terms given; multiply the number so found by the first term, and the product will be the last term, or greater extreme. 2. Multiply the last term by the ratio, from that product subtract the first term, and divide the remainder by the ratio, less one; the amount will be the sum of the series, or of all the terms. EXAMPLE. 1. Suppose 20 yards of broadcloth was sold at 4 mills for the first yard, 12 for the second, 36 for the third, &c. what did the cloth come to, and what was gained by the sale, supposing the prime cost to have been $15 per yard? Note. In this question observe, the first term is 4, the ratio 3, and the number of terms 20, consequently the ratio 3 must be raised to 20-1-19th power. Thus, First cost of the cloth 6973568,80,0=sum of the series, or number of 300,00 mills for which the cloth was sold 6973268,80=gain. N 2. A father gave his daughter who was married on the first day of January, one dollar towards her portion, promising to double it on the first day of every month for one year; what was the amount of her whole portion? Ans. 4095 dollars. 3. A merchant sold 15 yards of satin; the first yard for 1s. the second for 2s. the third for 4s. &c. in geometrical progression; what was the price of the 15 yards? 4. A goldsmith sold 1 ounce, 4 for the second, 1 Ans. 16381. 7s. of gold at 1 cent for the first or the third, &c.; wna did it come to, and what did he gain, supposing he gave 2uollars per ounce ? Ans. He sold it for 55924 dollars 5 cents, and gained 55684 dollars 5 cents. 5. What sum would purchase a horse with 4 shoes and 8 nails in each shoe, at one mill for the first nail, 2 mills for the second, 4 for the third, &c. doubling in geometrical progression to the last? Ans. 4294967 dollars 29 cents 5 mills. 6. What sum would purchase the same horse, with the same number of shoes, and nails, at 1 mill for the first nail, 3 for the second, 9 for the third, &c., in a triple ratio of geometrical progression to the last? Ans. 926510094425 dollars 92 cents. 7. What sum would purchase the same horse, with the same number of shoes, and nails, at 1 mill for the first nail, 4 for the second, 16 for the 3d, &c., in a quadruple ratio of geometrical progression to the 'ast? Ans. 6148914691236′′ 17 dollars 20 cents 5 mills. 8. Sold 30 yards of silk v et, at 2 pins for the first yard, 6 for the second, 18 for third, &c., and these disposed of at 1000 for a farthing; what did the velvet amount to, and what was gained by the sale, supposing the prime cost to have been 1007. per yard? |