The first term, the last term, and the number of terms, being given, to find the common difference;— 1. A man had 6 sons, whose several ages differed alike; the youngest was 3 years old, and the oldest 28; what was the common difference of their ages? The difference between the youngest son and the eldest evidently shows the increase of the 3 years by all the subsequent additions, till we come to 28 years; and, as the number of these additions are, of course, 1 less than the number of sons, (5), it follows, that, if we divide the whole difference (28-3=), 25, by the number of additions, (5), we shall have the difference between each one separately, that is, the common difference. Thus, 28-3=25; then, 2555 years, the common difference. A.5 yrs. Hence, 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. 2. If the extremes be 3 and 23, and the number of terms 11, what is the common difference? A. 2. 3. A man is to travel from Boston to a certain place in 6 days, and to go only 5 miles the first day, increasing the distance travelled each day by an equal excess, so that the last day's journey may be 45 miles; what is the daily increase, that is, the common difference? A. 8 miles. 4. If the amount of $1 for 20 years, at simple interest, be $2,20, what is the rate per cent. ? In this example we see the amount of the first year is $1,06, and the last year $2,20; consequently, the extremes are 106 and 220, and the number of terms 20. A. $,06: 6 per cent. 5. A man bought 60 yards of cloth, giving 5 cents for the first yard, 7 for the second, 9 for the third, and so on to the last; what did the last cost? Since, in this example, we have the common difference given, it will be easy to find the price of the last yard; for, as there are as many additions as there are yards, less 1, that is, 59 additions of 2 cents, to be made to the first yard, it follows, that the last yard will cost 2 X 59118 cents more than the first, and the whole cost of the last, reckoning the cost of the first yard, will be 118 +5 = $1,23. A. $1,23. Hence, when the common difference, the first term, and the number of terms, are given, to find the last term ; Multiply the common difference by the number of terms, less 1, and add the first term to the product. 6. If the first term be 3, the common difference 2, and the number of terms 11, what is the last term? A. 23. 7. A man, in travelling from Boston to a certain pace in 6 days, travelled the first day 5 miles, the second 8 miles, travelling each successive day 3 miles farther than the former; what was the distance travelled the last day? A. 20. 8. What will $1, at 6 per cent., amount to, in 20 years, at simple interest? The common difference is the 6 per cent.; for the amount of $1, for 1 year, is $1,06, and $1,06+ $,08 $1,12 the second year, and so on. A. $2,20. 9. A man bought 10 yards of cloth, in arithmetical progression; for the 1st yard he gave 6 cents, and for the last yard he gave 24 cents; what was the amount of the whole ? In this example it is plain that the cost of the first and last yards will be the average price of the whole number of yards; thus, 6 cts. +24 cts. 30 15 cts., average price; then, 10 yds. X 15. A. $1,50. 150 cts, = $1,50, whole 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 3 and 273, and the number of terms 40, what is the sum of all the terms? A. 5520. 11. How many times does a regular clock strike in 12 hours? A. 78. 12. A butcher bought 100 oxen, and gave for the first ox $1, for the second $2, for the third $3, and so on to the last; how much did they come to at that rate? A. $5050. 13. What is the sum of the first 1000 numbers, beginning with their natural order, 1, 2, 3, &c.? A. 500500. 14. If a board, 18 feet long, be 2 feet wide at one end, and come to a point at the other, what are the square contents of the board? A. 18 feet. 15. If a piece of land, 60 rods in length, be 20 rods wide at one end, and at the other terminate in an angle or point, what number of square rods does it contain? A. 600. 16. A person, travelling into the country, went 3 miles the first day, and increased every day's travel 5 miles, till at last he went 58 miles in one day; how many days did he travel? We found, in example 1, the difference of the extremes, divided by the number of terms, less 1, gave the common difference; consequently, if, in this example, we divide (583) 55, the difference of the extremes, by the cominon difference, 5, the quotient, 11, will be the number of terms, less 1; then, 1+11 12, the number of terms. A. 12. Hence, when the extremes and common difference are given, to find the number of terms;— Divide the difference of the extremes by the common difference, and the quotient, increased by 1, will be the answer, 17. If the extremes be 3 and 45, and the common difference 6, what is the number of terms? A. 8. 18. A man, being asked how many children he had, replied, that the youngest was 4 years old, and the eldest 32, the increase of the family having been I in every 4 years; how many had he? A. 8. GEOMETRICAL PROGRESSION. ▼ LXXXIX. Any rank or series of numbers, increasing by a constant multiplier, or decreasing by a constant divisor, is called Geometrical Progression. Thus, 3, 9, 27, 81, &c., is an increasing geometrical series; And 81, 27, 9, 3, &c., is a decreasing geometrical series. There are five terms in Geometrical Progression; and, like Arithmetical Progression, any three of them being given, the other two may be found, viz. 1. The first term. 2. The last term. 3. The number of terms. 4. The sum of all the terms. 5. The ratio; that is, the multiplier or divisor, by which we form the series, 1. A man purchased a flock of sheep, consisting of 9; and, by agreement, was to pay what the last sheep came to, at the rate of $4 for the first sheep, $12 for the second, $36 for the third, and so on, trebling the price to the last; what did the flock cost him? We may perform this example by multiplication; thus, 4×3×3×3×3 X3X3X3X3=$26244, Ans. But this process, you must be sensible, would be, in many cases, a very tedious one; let us see if we cannot abridge it, thereby making it easier. In the above process we discover that 4 is multiplied by 3 eight times, one time less than the number of terms; consequently, the 8th power of the ratio 3, expressed thus, 38, multiplied by the first term, 4, will produce the last term. But, instead of raising 3 to the 8th power in this manner, we need only raise it to the 4th power, then multiply this 4th power into itself; for, in this way, we do, in fact, use the 3 eight times, raising the 3 to the same power as before; thus, 381; then, 81 X 816561; this, multiplied by 4, the first term, gives $26244, the same result as before. A. $26244. Hence, when the first term, ratio, and number of terms, are given, to find the last term; I. Write down some of the leading powers of the ratio, with the numbers 1, 2, 3, &e. sver them, being their several indices. II. Add together the most convenient indices to make an index less by 1 than the number of terms sought. III. Multiply together the powers, or numbers standing under those indices; and their product, multiplied by the first term, will be the term sought. 2. If the first term of a geometrical series be 4, and the ratio 3, what is the 11th term? Note. The pupil will notice that the series 3, 9, 27, 81, 243, powers. does not commence with the first term, but with the ratio. 1, 2, 3, 4, 5, indices.) The indices 5+3+2=10, and the powers under each, 243 27 X9= 59049; which, multiplied by the first term, 4, makes 236196, the 11th term required. A. 236196. 3. The first term of a series, having 10 terms, is 4, and the ratio 3; what is the last term? A. 78732. 4. A sum of money is to be divided among 10 persons; the first to have $10, the second $30, and so on, in threefold proportion; what will the last have? A. $196830. 5. A boy purchased 18 oranges, on condition that he should pay only the price of the last, reckoning 1 cent for the first, 4 cents for the second, 16 cents for the third, and in that proportion for the whole; how much did he pay for them? A. $171798691,84. 6. What is the last term of a series having 18 terms, the first of which is 3, and the ratio 3? A. 387420489. 7. A butcher meets a drover, who has 24 oxen. The butcher inquires the price of them, and is answered, $60 per head; he immediately offers the drover $50 per head, and would take all. The drover says he will not take that; but, if he will give him what the last ox would come to, at 2 cents for the first, 4 cents for the second, and so on, doubling the price to the last, he might have the whole. What will the oxen amount to at that rate? A. $167772,16. 8. A man was to travel to a certain place in 4 days, and to travel at whatever rate he pleased; the first day he went 2 miles, the second 6 miles, and so on to the last, in a threefold ratio; how far did he travel the last day, and how far in all? In this example, we may find the last term as before, or find it by adding each day's travel together, commencing with the first, and proceeding to the last, thus: 2+ 6 + 18 +54=80 miles, the whole distance travelled, and the last day's journey is 54 miles. But this mode of operation, in a long series, you must be sensible, would be very troublesome. Let us examine the nature of the series, and try to invent some shorter method of arriving at the same result. By examining the series 2, 6, 18, 54, we perceive that the last term, (54,) less 2, (the first term,) 52, is 2 times as large as the sum of the remaining terms; for 2+ 6 + 18 = 26; that is, 54 — 2 = 52 ÷ 2 = 26; hence, if we produce another term, that is, multiply 54, the last term, by the ratio 3, making 162, we shall find the same true of this also; for 162-2, (the first term,) 160280, which we at first found to be the sum of the four remaining terms, thus: 2+6+18+54=80. In both of these operations it is curious to observe, that our divisor, (2,) each time, is 1 less than the ratio, (3.) Hence, when the extremes and ratio are given, to find the sum of the series, we have the following easy RULE. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio, less 1; the quotient will be the sum of the series required. 9. If the extremes be 5 and 6400, and the ratio 6, what is the whole amount of the series? 10. A sum of money is to be divided among 10 persons in such a manner, that the first may have $10, the second $30, and so on, in threefold proportion; what will the last have, and what will the whole have? The pupil will recollect how he found the last term of the series by a foregoing rule; and, in all cases in which he is required to find the sum of the series, when the last term is not given, he must first find it by that rule, and then work for the sum of the series, by the present rule. A. The last, $196830; and the whole, $295240. 11. A hosier sold 14 pair of stockings, the first at 4 cents, the second at 12 cents, and so on in geometrical progression; what did the last pair bring him, and what did the whole bring him? A. Last, $63772,92; whole, $95659,36. 12. A man bought a horse, and, by agreement, was to give a cent for the first nail, three for the second, &c.; there were four shoes, and in each shoe eight nails; what did the horse come to at that rate? A. $9265100944259,20 13. At the marriage of a lady, one of the guests made her a present of a half-eagle. saying, that he would double it on the first day of each succeeding month throughout the year, which he said would amount to something like $100; now, how much did his estima differ from the true amount? A. $20375. 14 If our pious ancestors, who landed at Plymouth, A. D, 1620, being 101 in number, had increased so as to double their number in every 20 years, how great would have been their population at the end of the year 1840? A. 206747. ANNUITIES AT SIMPLE INTEREST. TXC. An annuity is a sum of money, payable every year, for a certain number of years, or forever. When the annuity is not paid at the time it becomes due, it is said to be in arrears. The sum of all the annuities, such as rents, pensions, &c., remaining unpaid, with the interest on each, for the time it has been due, is called the amount of the annuity. Hence, to find the amount of an annuity;— Calculate the interest on each annuity, for the time it has remained unpaid, and find its amount; then the sum of all these several amounts will be the amount required. 1. If the annual rent of a house, which is $200, remain unpaid, (that is, in arrears,) 8 years, what is the amount? In this example, the rent of the last (8th) year being paid when due, of course, there is no interest to be calculated on that year's rent. The amount of $200 for 7 years= $284 5 The amount of $200 6 .... 3....... = $272 $260 = $248 The amount of $200 $236 The amount of $200 .... 2 ....... $224 The amount of $200 2. If a man, having an annual pension of $60, receive no part of it till the expiration of 8 years, what is the amount then due? A. $580,80. 3. What would an annual salary of $600 amount to, which remains unpaid (or in arrears) for 2 years? (1236) For 3 years? (1908) For 4 years? (2616) For 7 years? (4956) For 8 years? (5808) For 10 years? (7620) Ans. $24144. 4. What is the present worth of an annuity of $600, to continue 4 years? The present worth, (T LXVII.,) is such a sum as, if put at interest, would amount to the given annuity; hence, $600 $1,06 $566,037, present worth, 1st year. 2d $600 $1,18=$508,474, $600 $1,24= $483,870, ........ 3d 4th Ans., $2094,095, present worth required. Hence, to find the present worth of an annuity;— Find the present worth of each year by itself, discounting from the time it becomes due, and the sum of all these present worths will be the answer. 5. What sum of ready money is equivalent to an annuity of $200, to con tinue 3 years, at 4 per cent.? A. $556,063. 6. What is the present worth of an annual salary of $800, to continue 2 years? (1469001) 3 years? (2146967) 5 years? (3407512) A. $7023,48. |