PROBLEM III. Given the first term, the last term, and the common differ ence, to find the number of terms. RULE.—Divide the difference of the extremes by the common difference, and the quotient increased by 1, is the number of terms required. EXAMPLES. 1. The extremes are 3 and 19, and the common difference 2 ; what is the number of terms ? 19 Ans. 9 2. Suppose a man travel the first day 7 miles, the last 51 miles, and increase his journey each day by 4 miles ; how many days will he travel, and how far ? Ans. 12 days, and 343 miles. GEOMETRICAL PROGRESSION. Any series of numbers, the terms of which gradually increase or decrease by a constant multiplication or division, are said to be in Geometrical Progression. Thus, 4, 8, 16, 32, 64, &c. and 81, 27, 9, 3, 1, &c. are series in geometrical progression, the one increasing by a constant multiplication by 2, and the other decreasing by a constant division by 3. The number by which the series is constantly increased or diminished, is called the ratio. PROBLEM I. Given the first term, the last term, and the ratio, to find the sum of the series. RULE.-Multiply the last term by the ratio, and from the product subtract the first term, and the remainder divided by the ratio less 1, will give the sum of the series. EXAMPLES. 1. The extremes of a geometrical progression are 1 and 65536, and the ratio 4 ; what is the sum of the series? 65536 4 262144 4x65536-1 =87381 Ans. 4-1 4-1=3)262143 87381 Ans. 2. A man travelled 6 days; the first day he went 4 miles, and doubling his travel each day, his last day's ride was 128 miles; how far did he go in the whole ? Ans. 252 miles. 3. The extremes of a geometrical series are 1024 and 59049, and the ratio is 1}; what is the sum of the series ! Ans. 175099. PROBLEM II. Given the first term and the ratio, to find any other term assigned.* CASE 1.-When the first term of the series, and the ratio, are equal.t Rule.-1. Write down a few of the leading terms of the series, and place their indices over them, beginning the indices with a unit or 1, 2. Add together such indices as, in their sum, shall make up the entire index to the term required. a * As the last term in a long series of numbers is very tedious to be found by continual multiplication, it will be necessary for more readily finding it out, to have a series of numbers in arithmetical proportion, called indices, whose common difference is 1. | When the first term of the series, and the ratio, are equal, the indices must begin with a unit, and in this case, the product of any two terms is equal to that term, signified by the sum of their indices. Thus : S 1, 2, 3, 4, 5, &c. Indices arithmetical series. 2, 4, 8, 16, 32, &c. Geometrical series. Now, 2+3=5, the index of the fifth term, and 4x8= 32=the fifth term. 3. Multiply together the terms of the geometrical series belonging to those indices, and the product will be the term required. EXAMPLES. 1. The first term of a geometrical series is 2, and the ratio 2; required the 13th term. 1, 2, 3, 4, 5, 6, 7, indices. Then 6+7=index to the 13th term. And 64x123=8192 the answer. 2. A young man agreed with a farmer to work for him 11 years, with no other reward than the produce of one grain of wheat for the first year, allowing the increase to be tenfold, and that produce to be sowed the second year, and so on from year to year, until the end of the time; what is the sum of the whole produce, allowing 7680 grains to make a pivt, and what does it amount to at one dollar and fifty cents per bushel ? Aus. 2260561 +bush., and $339084,19cts. + Note.--In such questions, you first find the last term by one of the cases in Problem 2, and then the sum of the whole series by Problem 1. 3. A rich miser thought 20 guineas apiece too much for 12 fine horses, but readily agreed to give 4 cents for the first, 16 cents for the second, 64 cents for the third horse, and so on, in fourfold proportion, to the last ;what did they come to at that rate, and how much did they cost per head, one with another? Ans. The 12 horses came to $223696,20cts., and the average price was $18641,35cts. per head. Case 2.-When the first term of the series, and the ra tio, are different : that is, when the first term is either greater or less than the ratio.* RULE.-J. Write down a few of the leading terms of the series, as before, and begin their indices with a cipher; thus: 0, 1, 2, 3, &c. * When the first term of the series, and the ratio are different, the indices must begin with a cipher, and the sim of the indices made choice of, must be one less than the number of terms given in the question; because 1 in the indices stands over the second term, and 2 in the indices over the third term, &c.; and, in this { 2. Add together the most convenient indices to make an index, less by 1, than the number expressing the place of the term sought. 3. Multiply the terms of the geometrical series together, belonging to those indices, and make the product a dividend. 4. Raise the first term to a power whose index is one less than the number of terms multiplied, and make the result a divisor. 5. Divide the said dividend by the said divisor, and the quotient is the term required. NOTE.-If the first term of any series be unity, or 1, the term required is found by multiplying the terms of the geometrical series together which belong to those indices, without needing any division. EXAMPLES. 1. Required the 12th term of a geometrical series, whose first term is 3, and ratio 2. 0, 1, 2, 3, 4, 5, 6, indices. The number of terms multiplied is 2, and 2-1=1 is the power to which the term 3 is to be raised ; but the first power of 3 is 3=divisor ; therefore 18432:3=6144, the 12th term. 2. A goldsmith sold 1tb of gold, at 2cts. for the first ounce, 8cts. for the second, 32cts. for the third, &c., in quadruple proportion geometrical; what did the whole l come to ? Ans. $111843,10cts. 3. A man bought a horse, and by agreement was to give a farthing for the first nail, two for the second, four for the third, &c. There were four shoes, and eight nails in each shoe ;-What did the horse come to at that rate ! Ans. £4473924 5s. 3d. 3qrs. case, the product of any two terms, divided by the first term, is equal to that term beyond the first, signified by the sum of their indices. Thus : { i; So, 1, 2, 3, 4, &c., indices. 1, 3, 9, 27, 81, &c., geometrical series., Here, 4+3=7, the index of the 8th term. 81 X27=2187, the 8th term, or 7th beyond the 1st. 4. Suppose a certain body, put in motion, should move the length of one barleycoșu the first second of time, one inch the second, three inches the third second of time, and so continue to increase its motion in triple proportion geometrical ; how many yards would the said body move in the space of half a minute ? Ans. 953199685623yds. 1ft. lin. Ibar. s' which is no less than five hundred and forty-one millions of miles. ALLIGATION. Alligation teaches to mix several simples of different qualities, so that the composition may be of a middle quality; and is commonly distinguished into two principal cases, called Alligation Medial and Alligation Alternate. ALLIGATION MEDIAL. ALLIGATION MEDIAL is the method of finding the rate of the compound, from having the rates and quantities of the several simples given. Rule.--Multiply each quantity by its rate; then divide the sum of the products by the sum of the quantities, or the whole composition, and the quotient will be the rate of the compound required. EXAMPLES. 1. Suppose 20 bushels of wheat at 10s. per bushel, 36 bushels of rye at 6s. per bushel, and 40 bushels of barley at 4s. per bushel, were mixed together; what would a bushel of this mixture be worth ? 20x10=200 96 )576(6s. Answer. 576 2. A composition being made of 5 pounds of tea at 7s. per pound, 9 pounds at 8s. 6d. per pound, and 144 pounds at 6s. 104d. per pound ; what is a pound of it worth ? Ans. 7s. 4 d. + 3. A goldsmith mixes 8 pounds 54 ounces of gold of 14 carats fine, with 12 pounds 84 ounces of 18; what is the fineness of this mixture ? Ans. 16-24 carats. |