CASE V. Where a, z, and d are given, to find n (the first and last term, and the common difference, given to find the number of terms). Exercises for the Slate or Black-board. 1. The first term of a progression by differences is 4, the last term 119, and the common difference 5. What is the number of terms? Ans. 24. Suggestive Questions.-What does the last term consist of besides the first term? If the first term, then, be subtracted from the last, how shall the number of terms less 1 be found? 2. The first term in a progression by differences is 3, the last term 363, and the common difference 3. What is the number of terms? Ans. 49. 3. A note was paid by annual instalments, whose common difference was $30. The first payment was $130, and the last $400. How many instalments were there? Ans. 10. 4. In a triangular field of maize, the number of hills in the successive rows forms a progression by differences. In the first row there is but one hill, in the last 81, and the common difference in the number of hills in a row is 2. What is the number of rows? Ans. 41. [The number of cases might be very much enlarged; but this is hardly necessary, as each pupil can readily arrange them for himself. The following equations, with the cases already given, comprehend all that are most common and useful. The exercises given above will answer for the new cases. But it would be more profitable for the student to furnish examples himself. The demonstration of the cases given below, by suggestive questions, would form one of the best exercises that a class could engage in.] 1. A series of numbers, which succeed each other regularly by a constant factor, which is called a ratio, is called a Progression by Ratios. Thus : 2 4 8 16 32 64 64 32 16 8 4 2 form two progressions by ratios, the ratio of the first being 2, and that of the second. 2. When the ratio is more than a unit, the series is called an ascending progression; when it is less, the series is called a descending progression. 3. As in the progression by differences, so in the progression by ratios, five things are to be considered, any three of which being known, the remaining two can be found, namely: Where the first term, the ratio, and the number of terms are given, to find the last term. Exemplification for the Black-board. 1. The first term of a progression by ratios is 2, the ratio is 3, and the number of terms is 6. What is the last term? 2 6 18 54 162 486 Suggestive Questions.-How often is the ratio a factor in the 2d term? By what power of the ratio, then, is the first term multiplied to form the 2d term? How often is the ratio a factor in the 3d term? By what power of the ratio, then, is the first term multiplied to form the 3d? By what power to form the 4th? By what to form the 5th? the 6th? Does every term, then, consist of the first term multiplied by a power of the ratio one less than its number? Find, then, the last term of the above progression, without finding the intermediate ones. Exercises for the Slate or Black-board. 1. The first term of a progression by ratios is 4, the ratio is 2, and the number of terms 7. What is the last term? [It is to be found without finding the intermediate terms.] Ans. 256. 2. A farmer sold 7 cows to a neighbor, for which he was to receive $5 for the first, $10 for the second, $20 for the third, and so on, increasing in progression by ratios. What would be the price of the last cow at this rate? Ans. $320. 3. A man once offered to sell a horse on condition that he should have a sufficiency of wheat to pay for the last nail in the animal's shoes, reckoning the first nail at 1 gill, the second at 2 gills, the third at 4, and so on, increasing in progression by ratios. It was found that the horse had 8 nails in each of his four shoes. What would he get for the horse at that rate, reckoning wheat at $125 per bushel? Ans. $10,485,760, COMPOUND INTEREST BY PROGRESSION. If the manner in which compound interest is computed, p. 314, be carefully examined, it will be evident that the successive amounts which are considered as new principals, form the terms of a series of progression by ratios, whose first term is the original principal, and the ratio the amount of $1, for one year, at the given rate per cent. The number of terms is equal to one more than the number of years, a circumstance which will be readily understood by a glance at the example referred to, the number of principals in the computation being 4, while the number of years is only three. Thus, to find the amount of a given principal at compound interest, for a given number of years, at a given rate per cent., is merely to find the last term of a progression by ratios, when the first term, the ratio, and the number of terms are given. Exercises for the Slate or Black-board. 1. Find the compound interest of $100 for 4 years at six per cent. Ans. $26'247. 2. Find the amount at compound interest of $100 for 17 years at six per cent. Ans. $269 277. 3. Find the compound interest of $600 for 20 years at 5 per cent. Ans. $1591'978. 4. Find the amount of $100 for 3 years at 6 per cent. per annum, when the interest is to be added at the end of every six months. Suggestive Questions.-If a new principal be formed at the end of every year, of how many terms would the series consist, reckoning the original principal of $100 as one? At what per cent., then, should the interest be reckoned? But, if the new principal be formed at the end of every half year, of how many terms would the series then consist? At what per cent., then, should the interest be reckoned for the half year? But, if the new principal were formed by adding in the interest every 3 months, how many terms would there then be in the 4th exercise above? How much per cent. must be added in for the 3 months, when the interest is 6 per cent. per annum? 5. Find the amount of $3705 at compound interest, in 3 years and 3 months, at 12 per cent. per annum, the interest being added every months. Ans. $5440 918. 6. What will $1000 amount to in 15 years, at 8 per cent. per annum, the interest being compounded half-yearly? Ans. $3243'398. CASE II. Where the last term, the ratio, and the number of terms, are given, to find the first term. Exercises for the Slate or Black-board. 1. The last term of a progression by ratios is 256, the ratio 2, and the number of terms 7; what is the first term? Ans. 4. Suggestive Questions.-How many terms are there in the above progression? How many times, then, is the ratio a factor in the last term? What other number is a factor in the last term? How, then, can the first term be found? 2. A man bought 7 cows, for which he engaged to pay a certain sum for the first, double the sum for the second, and so on in regular progression to the 7th, the price of which was found to amount to $320. What was the price of the first? Ans. $5. 3. A horse was once offered for sale, on condition that the purchaser should give a small quantity of wheat for the first of the 32 nails in his shoes, doubling the wheat for the second, and so on in regular progression to the last. No one offering to take the horse on these conditions, the owner said he would sell it for the price of the last nail. On calculation, it was found that the horse would cost, on this last condition, $10,485,760, reckoning the wheat at $125 per bushel. How much wheat was to be given for the first nail, agreeably to the first offer? COMPOUND DISCOUNT. Definitions. Ans. 1 gill. 1. Compound Discount is an allowance made for the payment of money before it is due, on the supposition that the money draws compound interest. 2. When compound interest is reckoned, the present worth of a debt payable at some future time without interest, is such a sum, as being put out at compound interest, will, in the given time, at the given rate per cent., amount to the debt. Finding the present worth of a debt, then, resolves itself as follows: given the amount at compound interest, the amount of $1 at the given rate, and the number of terms, to find the principal, or, which is the same thing, given the last term of a progression by ratios, the ratio, and the number of terms, to find the first term. Observe that the number of terms is always one more than the number of years. Why? 1. What is the present worth of a debt of $126 247, due 4 years hence, at 6 per cent. compound interest? Ans. $100. Suggestive Questions.-How often is the amount of $1 a factor in the above amount? What other number is a factor in it? If the first be known, then, how shall the latter be found? 2. What is the present worth of $269 277, due 17 years hence, at 6 per cent. compound interest? Ans. $100. 3. What is the present worth of $1000, due 20 years hence, at 5 per cent. compound interest? Ans. $376 889. 4. What is the present worth of $1593'30, due 20 years hence, at 5 per cent. compound interest? CASE III. Ans. $600 50. Where the first term, the last term, and the ratio are given, to find the sum of all the terms. Exemplification for the Black-board. 1. What is the sum of a series of progression by ratios, whose first term is 2, the ratio 3, and the number of terms 6? Write the series at length, with its proper signs of addition, and then form a new series, by multiplying each term of the old series by the ratio, placing one over the other, so that each term shall be removed one step to the right of that by which it was produced. Thus : From new series 6+18+54+162+486+1458-3s (or s\r) Take old series 2+6+18-+-54162486 2 = 8 from 1458-2s (or sX(r−1)) Suggestive Questions.-How was the new series formed? Ans. By each term of the old series by the. What is the ratio in this problem? Is the sum of the new series, then, 3 times the amount of the sum of the old series? Had the ratio been 5, or 7 (or any other number), would the old series have been 5 or 7 (or any other number) times the amount of the new? How many times the sum of the old series is the difference between the sums of the old and new series? Does that depend on the number of the ratio? Had the ratio been 5, or 7, how many times the sum of the old series would then have been the difference between the two series? Would it always be sX(r-1)? What is the difference between the first term of the old series, and the last of the new, ascertained at a glance? Is this number the difference between the two series? Is it equal to twice the sum of the old series? What, then, is the sum of the old series? How, then, can you ascertain the sum of any series of progression by ratios, from the first term, the last term, and the ratio, without forming the intermediate terms? May not the following, then, be considered the first principle in Progression by Ratios? I. The sum of any series in a Progression by Ratios is equal to the last term, multiplied by the ratio, diminished by the first term, and divided by the ratio less one. 1. What is the sum of the series, 1, 3, 9, 27, &c., to 12 terms? Ans. 265720. 2. What is the sum of 10 terms of the series, whose first term is 2, and the ratio 2? Ans. 2046. 3. A man bought a horse, agreeing to give 1 dollar for the first nail in one of his shoes, 3 for the second, 9 for the third, and so on. The shoe contained 8 nails. What was the cost of the horse? Ans. $3280. 4. A man agreed to buy 30 bushels of wheat: the first bushel for 2 cents, the second 4 cents, the third 8 cents, doubling the price of each preceding bushel for that of the next. What would the 30 bushels cost, and what would be the average price per bushel ? Ans. to the last question, $815,827 88+. 5. What sum would purchase a horse with 4 shoes, and 8 nails in each shoe, at 1 mill for the first nail, 2 for the second, 4 for the third, &c., doubling to the last? Ans. $5,668,109 139. 6. A lady, who was married on new-year's day, received from her father one dollar towards her marriage portion, which he promised to double on the first day of every month during the year. What would her portion amount to? Ans. $4095. |