444. It is observed, that in the free descent of bodies, a body falls 16* feet in a second. that in two seconds of time it falls 64 feet, and that in three seconds it falls 144 feet; hence it is concluded, that the heights are to one another as the squares of the times; and that, reciprocally, the times are in the subduplicate ratio of the heights, or as the square roots of the heights. If, therefore, it be required to find how long a stone must take to fall from the height of 2304 feet; we have 16: 2304 = 1 to the square of the time sought. So that the square of the time sought is 144; and, consequently, the time required is 12 seconds. 445. It is required to find how far, or through what height, a stone will pass, by descending for the space of an hour; that is, 3600 seconds. We say, therefore, as the squares of the times, that is, 12: 36002; so is the given height 16 feet, to the height required. 1:12960000 = 16: ... 207360000 height required. 16 77760000 207360000 If we now reckon 19200 feet for a league, we shall find this height to be 10800; and consequently, nearly four times greater than the diameter of the earth. 446. It is the same with regard to the price of precious stones, which are not sold in the proportion of their weight; every body knows that their prices follow a much greater ratio. The rule for diamonds is, that the price is in the duplicate ratio of the weight, that is to say, the ratio of the prices is equal to the square of the ratio of the weights. The weight of diamonds is expressed in carats, and a carat is equivalent to 4 grains; if, therefore, a diamond of one carat is worth 10 livres, a diamond of 100 carats will be worth as many times 10 livres, as the square of 100 contains 1; so that we shall have, according to the rule of three, * 15 is used in the original, as expressing the descent in Paris feet. It is here altered to English feet. There is a diamond in Portugal, which weighs 1680 carats; its price will be found, therefore, by making 12 : 16802 10 liv. : .... or 447. The posts, or mode of travelling, in France furnish examples of compound ratios, as the price is according to the compound ratio of the number of horses, and the number of leagues or posts. For example, one horse costing 20 sous per post, it is required to find how much is to be paid for 28 horses and 4 posts. 1 We write first the ratio of horses, And, compounding the two ratios, we have : 28, : 9, 2 : 252, Or, 1 : 126 = 1 livre to 126 francs, or 42 crowns. Another Question. If I pay a ducat for eight horses, for 3 German miles, how much must I pay for thirty horses for four miles? The calculation is as follows: 1 : 5,1 ducat: the 4th term, which will be 5 ducats. 448. The same composition occurs, when workmen are to be paid, since those payments generally follow the ratio compounded of the number of workmen, and that of the days which they have been employed. If, for example, 25 sous per day be given to one mason, and it is required to find what must be paid to 24 masons who have worked for 50 days; we state this calculation; As, in such examples, five things are given, the rule, which serves to resolve them, is sometimes called, in books of arithmetic, The Rule of Five. CHAPTER X. Of Geometrical Progressions. 449. A SERIES of numbers, which are always becoming a certain number of times greater or less, is called a geometrical progression, because each term is constantly to the following one in the same geometrical ratio. And the number which expresses how many times cach term is greater than the preceding, is called the exponent. Thus, when the first term is 1 and the exponent metrical progression becomes, 2, the geo &c. Terms 1 2 3 4 5 6 7 8 9 the numbers 1, 2, 3, &c. always marking the place which each term holds in the progression. 450. If we suppose, in general, the first term = a, and the exponent=b, we have the following geometrical progression ; .... n 1, 2, 3, 4, 5, 6, 7, 8 Prog. a, ab, ab2, ab3, ab1, abs, ab", ab.... ab"-1. So that, when this progression consists of n terms, the last term is =ab"-1. We must remark here, that if the exponent b be greater than unity, the terms increase continually; if the exponent b = 1, the terms are all equal; lastly, if the exponent b be less than 1, or a fraction, the terms continually decrease. Thus, when a 1 and = b, we have this geometrical progression ; = 1, 1, 1, 1, 1, 3, 4, T, &c. 451. Here therefore we have to consider ; ខ I. The first term, which we have called a. II. The exponent, which we call b. III. The number of terms, which we have expressed by n. So that, when the three first of these are given, the last term is found by multiplying the n-1 power of b, or bn-1, by the first term a. If, therefore, the 50th term of the geometrical progression 1, 2, 4, 8, &c. were required, we should have a = 1, b = 2, and n = 50; consequently the 50th term 2. Now 2 being =512; 21° will be 1024. Wherefore the square of 21o, or 2a°, =1048576, and the square of this number, or 1099511627776 = 2. Multiplying therefore this value of 24° by 2o, or by 512, we have 24 equal to 562949953421312. 452. One of the principal questions, which occurs on this subject, is to find the sum of all the terms of a geometrical progression; we shall therefore explain the method of doing this. Let there be given, first, the following progression, consisting of ten terms; 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, the sum of which we shall represent by s, so that s=1+2+4+8+ 16 +32 +64 + 128 +256 + 512; doubling both sides, we shall have 2s=2+4+8+ 16 +32 +64 + 128 +256 +512 +1024. Subtracting from this the progression represented by s, there remains s = 1024 — 1 = 1023; wherefore the sum required is 1023. If we multiply by 2, and subtracting from 21. We see, 453. Suppose now, in the same progression, that the number of terms is undetermined and n, so that the sum in question, or 8,1+2+22 + 23 +24.... 2n-1. we have 2s2+22+23+24.... 2", this equation the preceding one, we have s therefore, that the sum required is found, by multiplying the last term, 2-1, by the exponent 2, in order to have 2", and subtracting unity from that product. 454. This is made still more evident by the following examples, in which we substitute successively, for n, the numbers 1, 2, 3, 4, &c. 1=1; 1+2=3; 1+2+4=7; 1+2+4+8= 15; 1+2+4+8+16=31;1+2+4+8+16+32=63, &c. 455. On this subject the following question is generally proposed. A man offers to sell his horse by the nails in his shoes, which are in number 32; he demands 1 liard for the first nail, 2 for the second, 4 for the third, 8 for the fourth, and so on, demanding for each nail twice the price of the preceding. It is required to find what would be the price of the horse? This question is evidently reduced to finding the sum of all the terms of the geometrical progression, 1, 2, 4, 8, 16, &c. continued to the 32d term. Now this last term is 231; and, as we have already found 2201048576, and 210 1024, we shall have 220 × 210 = 23° equal to 1073741824; and multiplying again by 2, the last term 231 2147483648; doubling therefore this number, and subtracting unity from the product, the sum required becomes 4294967295 liards. These liards make 1073741823 sous, and dividing by 20, we have 53687091 livres, 3 sous, 9 deniers for the sum required. 456. Let the exponent now be = 3, and let it be required to find the sum of the geometrical progression 1, 3, 9, 27, 81, 243, 729, consisting of 7 terms. Suppose it = s, so that S=1+3+9+27+81 +243 + 729; we shall then have, multiplying by 3, 3s=3+9+27+81 +243 +729+2187; and subtracting the preceding series, we have So that the double of the sum is 2186, and consequently the sum required 1093. .... 3n. 457. In the same progression, let the number of terms = n, and the sum = s; so that s = 1 + 3 + 3a + 33 + 3a + 3n-1. If we multiply by 3, we have 3 s = 3 + 3a +33 + 3a + Subtracting from this the value of s, as all the terms of it, except the first, destroy all the terms of the value of 3 s, except the last, we .... So that the sum required is found by multiplying the last term by 3, subtracting 1 from the product, and dividing the remainder by 2. This will appear, also, from the following examples; 1=1; 1+3= 3 x 3-1 =4; 1+3+9= 3 X 9-1 = 13; 1 + 3 + 9 + 27 = 3 × 27—1—40; 1+3+9+27+81 = 458. Let us now suppose, generally, the first term = a, the exponent=b, the number of terms = n, and their sums, so that s=a=ab + a b3 + a b3 + aba + ..a bn-1. If we multiply by b, we have ... |