SECTION IX. MISCELLANEOUS RULES. 1. Arithmetical Progression. 275. When numbers increase by a common excess, or decrease by a common difference, they are said to be in Arithmetical Progression. When the numbers increase, as 2, 4, 6, 8, &c., they form an ascending series, and when they decrease, as 8, 6, 4, 2, &c., they form a descending series. The numbers which form the series are called its terms. The first and last term are called the extremes, and the others the means. 276. If I buy 5 lemons, giving for the first, 3 cents, for the second, 5, for the third, 7, and so on with a common difference of 2 cents; what do I give for the last lemon? Here the common difference, 2, is evidently added to the price of the first lemon, in order to find the price of the last, as many times, less 1 (3+2 +2+2+211 Ans.), as the whole number of lemons. Hence, I. The first term, the number of terms, and the common difference given to find the last term. RULE. Multiply the number of terms less 1, by the common difference, and to the product add the first term. 2. If I buy 60 yards of cloth, and give for the first yard 5 cents, for the next 8 cents, for the next, 11, and so on, increasing by the common difference, 3 cents, to the last, what do I give for the last yard? 59X3=177, and 177+5 182 cts. Ans. 3. If the first term of a series be 8, the number of terms 21, and the common difference 5, what is the last term? 20X5+8=108 Ans. 4. If the first term be 4, the difference 12, and the number of terms 18, what is the last term? Ans. 208. 277. If I buy 5 lemons, whose prices are in arithmetical progression, the first costing 3 cents, and the last 11 cents, what is the common difference in the prices? Here 11-3-8, and 5-1-4; 8 then is the amount of 4 equal differences, and 4)8(-2, the common difference. Hence, II. The first term, the last term, and the number of terms given to find the common difference. RULE. Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference. 2. If the first term of a series be 8, the last 108, and the number of terms 21, what is the common difference? 108-8-21-1-5 Ans. 3. A man has 12 sons whose ages are in arithmetical progression; the youngest is 2 years old, and the oldest 35; what is the common difference in their ages? Ans. 3 yrs. 278. If I give 3 cents for the first lemon, and 11 cents for the last, and the common difference in the prices be 2 cents, how many did I buy? The difference of the extremes divided by the number of terms, less 1, gives the common difference (277); consequently the difference of the extremes divided by the common difference, must give the number of terms, less 1 (11-3-8, and 8-2-4, and 4+1) 5 Ans. Hence, III. The first term, the last term, and the common difference given to find the number of terms. RULE.-Divide the difference of the extremes by the common difference, and the quotient, increased by 1, will be the answer. 2. If the first term of a series be 8, the last 108, and the common difference 5, what is the number of terms? 108—8—5—20, and 20+1= 21 Ans. 3. A man on a journey travelled the first day 5 miles, the last day 35 miles, and increased his travel each day by 3 miles; how many days did he travel? Ans. 11. 279. If I buy 5 lemons, whose prices are in arithmetical progression, giving for the first 3 cents, and for the last 11 cents, what do I give for The whole? The mean, or average price of the lemons will obviously be half way between 3 and 11 cents the difference between 3 and 11 added to 3 is (11-3-2-) 7, and 7, the mean price, multiplied by 5, the number of lemons, equals (75) 35 cents, the answer. Therefore, IV. The first and last term, and the number of terms given to find the sum of the series. RULE.-Multiply half the sum of the extremes by the number of terms, and the product will be the sum of the series. 2. How many times does a common clock strike in 12 hours? 1+12+2x12-78 Ans. 3. Thirteen persons gave presents to a poor man in arithmetical progression; the first gave 2 cents, the last 26 cents; what did they all give? Ans. $1,82. 2. Geometrical Progression. 280. A Geometrical Progression is a series of terms which increase by a constant multiplier, or decrease by a constant divisor, as 2, 4, 8, 16, 32, &c., increasing by the constant multiplier, 2, or 27, 9, 3, 1, 1, &c., decreasing by the constant divisor, 3. The multiplier or divisor, by which the series is produced, is called the ratio. 281. A person bought 6 brooms, giving 3 cents for the first, 6 cents for the second, 12 for the third, and so on, doubling the price to the sixth; what was the price of the sixth? or, in other words, if the first term of a series be 3, the number of terms 6, and the ratio 2, what is the last term? The first term is 3, the second, 3×2-6, the third, 6×2 (3×2×2) 12, the fourth, 12×2=(3×2×2×2) 24, the fifth, 24×2=(3×2×2×2×2 )48, and the sixth, 48×2=(3×2×2×2×2×2=) 96. Then 96 cents is the cost of the sixth broom. By examining the above, it will be seen, that the ratio is, in the production of each term of the series, as many times a factor, less one, as the number of terms, and that the first term is always employed once as a factor, or, in other words, any term of a geometrical series is the product of the ratío, raised to a power whose index is one less than the number of the term, multiplied by the first term. NOTE. If the second power of a number, as 22, be multiplied by the third power, 23, the product is 25. Thus, 22-2 ×2-4, and 23-2×2×2-8, and 8×4-32-2×2×2×2×2; and, generally, the power produced by multiplying one power by another is denoted by the sum of the indices of the given powers. Hence, in finding the higher powers of numbers, we may abridge the operation, by employing as factors several of the lower powers, whose indices added together will make the index of the required power. To find the seventh power of 2, we may multiply the third and fourth powers together, thus: 27 28X248x16-128. Ans. I. The first term and ratio given to find any other term. RULE. Find the power of the ratio, whose index is one less than the number of the required term, and multiply this power by the first term, the product will be the answer, if the series is increasing; but if it is decreasing, divide the first term by the power. 1. The first term of a geometrical series is 5, the ratio 3; what is the tenth term? 2. The first term of a decreasing series is 1000, the ratio 4, and the number of terms 5; what is the least Ans. 333. 3934X3581x248= term? 19683, and 19683X598415 Ans. 282. A person bought 6 brooms, giving 3 cents for the first, and 96 cents for the last, and the prices form a geometrical series, the ratio of which was 3; what was the cost of all the brooms ? The price would be the sum of the following series: 3+6+12+24+ 48+96-189 cents, Ans. If the foregoing series be multiplied by the ratio 2, the product is 6+12+24+4896=192, whose sum is twice that of the first. Now, subtracting the first series from this, the remainder is 192— 3=189 the sum of the first series. Had the ratio been other than 2, the remainder would have been as many times the sum of the series as the ratio, less 1, and the remainder is always the difference between the first term and the product of the last term by the ratio. Hence, II. The first and last term and ratio given to find the sum of the series. RULE.-Multiply the last term by the ratio, and from the product subtract the first term, the remainder divided by the ratio, less 1, will give the sum of the series. 2. The first term of a geo- | second, $4 the third, and so on, metrical series is 4, the last each succeeding payment beterm 972, and the ratio 3;ing double the last; and what what is the sum of the series? will be the last payment? 3-1)972X3-4(-1456 Ans. NOTE. The marks drawn over the numbers show, that 4 must be taken from the product of 972, by 3, and the remainder divided by (3-1 =) 2. This mark is called a vincu lum. Ans. {$1095 the debt. $2048 last pay't. 5. A gentleman, being asked to dispose of a horse, said he would sell him on condition of having 1 cent for the first 3. The extremes of a geo-nail in his shoes, 2 cents for metrical progression are 1024 and 59049, and the ratio 1; what is the sum of the series? Ans. 175099. 4. What debt will be discharged in 12 months, by paying $1 the first month, $2 the the second, 4 cents for the third, and so on, doubling the price of every nail to 32, the number of nails in his four shoes; what was the price of the horse at that rate? Ans. $42949672.95. 283. If a pension of 100 dollars per annum be forborne 6 years, what is there due at the end of that time, allowing compound interest at 6 per cent. ? Whatever the time, it is obvious that the last year's pension will draw no interest; it is, therefore, only $100; the last but one will draw interest one year, amounting to $106; the last but two, interest (compound) for 2 years, amounting to $112,36; and so on, forming a geometrical progression, whose first term is 100, the ratio 1.06, and the sum of this series will be the amount due. To find the last term (281) say, 1.065×100— 133.82255776, the sixth term; and to find the sum of the series (282) say, 133.82255777X1.06-100-41.8519112256, which, divided by 1.06-10.06, gives $667.5318576 Ans. or sum due. 284. A sum of money payable every year, for a number of years, is called annuity. When the payment of an annuity is forborne, it is said to be in arrears. 1. What is the amount of an annuity of $40, to continue 5 years, allowing 5 per cent. compound interAns. $221.025. est ? 2. If a yearly rent of $50 be forborne 7 years, to what does it amount, at 4 per cent. compound interest? Ans. $394.91. 3. Duodecimals. 285. Of the various subdivisions of a foot, the following is one of the most common: 1 foot 1 inch TABLE. is 12 inches, or primes, () = of 1 foot. 1 second" 12 thirds, (''') 1 third 12 fourths, (!!) 1 of 1 of 1=17289 &c. forming a decreasing geometrical progression, whose first term is 1, and ratio 12. Hence they are called Duodecimals. 286. How many square feet in a floor, 10ft. 4in. long, and 7ft. 8in. wide? 10ft. 4/ 6 10 8 72 4 Here we wish to multiply 10ft. 4' by 7ft. 8; we therefore write them as at the left hand, and multiply 4 by 8=32; but 4' being of a foot, and 8', the product is (X) 32 of a foot, or 32", which reduced gives 2' 8'; putting down 8, we reserve the 2' to be 79ft. 21 8 Ans. added to the inches. Multiplying 10ft. by 8, the product is (223) 89, to which being added, we have 6ft. 10'. Next, multiplying 4' by 7-23=2ft. 4', writing the 4' in the place of inches, and reserving the 2ft., we say 7 times 10 are 70, and two added are 72, which we write under the 6ft., and the sum of these partial products is 79ft. 2/8/ Ans. NOTE. When feet are concerned, the product is of the same denomination as the term multiplying the feet; and when feet are not concerned, the name of the product will be denoted by the sum of the indices of the two factors, or strokes over them. Thus, 4X2"-8"". Therefore, |