774. When the numbers of a series increase or decrease by becoming a certain number of times greater or less, the series is called a geometrical progression, because each term is to the following one in the same geometrical ratio. The number expressing how many times each term is greater than the preceding is called the exponent: thus, if the first term = 1 and the exponent =2, the geometrical progression becomes, Terms 1 2 3 4 5 6 7 8 9 &c Progression 1, 2, 4, 8, 16, 32, 64, 128, 256, &c. In which the numbers 1, 2, 3, &c. mark the place which each term holds in the progression. Generally, if the first term a and the exponent =b, we have the following geome trical progression Thus, when the progression proceeds to n terms, the last term is ab"-1. If the exponent b be greater than unity, the terms continue to increase; if the exponent b=1, the terms are all equal; and, lastly, if the exponent b be less than 1, or a fraction, the terms continually decrease. So if a=1 and b, we have the geometrical progression 1,8 &c., wherein we have for consideration, FIRST-The first term, which has been called a. FOURTH-The last term, which has been found =ab”—1. Hence, if any three of these be given, the last term may be found by multiplying the n − 1 power of b, or b-1, by the first term a. 775. If, therefore, in the geometrical progression 1, 2, 4, 8, &c. the fiftieth term be required, we have a=1, b=2, and n=50, consequently the fiftieth term is =249. Now 29=512, and 210=1024. Wherefore the square of 220=1048576, and the square of this number or 1099511627776=240; and multiplying this value of 240 by 29 or 512, we have 249-562949953421312. 776. One of the most usual questions which occur relative to geometrical progression is to find the sum of the terms, the mode of doing which we shall now explain. Let the following progression of ten terms be given : 1, 2, 4, 8, 16, 32, 64, 128, 256, 512. We will represent the sum by s, that is, s=1+ 2 + 4 + 8 + 16 +32 +64 + 128 +256 + 512. Double both sides and we have 2s = 2 + 4 + 8 + 16 + 32 +64 + 128 + 256 + 512 +1024. Subtracting from this the progression represented by s we have s=1024 −1 =1023; wherefore the sum required is 1023. .... 777. Suppose in the same progression the number of terms is undetermined and =n, so that the sum in question or s=1 + 2 + 22 + 23 + 24 . . . . 2-1. If we multiply by 2 we have 28=2+22+ 23 + 24 . . . . 2′′, and subtracting the preceding from the last equation we have s=9-1. Hence we see 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. 778. Suppose, generally, the first term =a, the exponent =b, the number of terms =n, and their sum =s, so that 8= abn-a and subtracting the first equation, the remainder is (b−1)s=ab”—a, whence is easily deduced the sum required, s=' Whence it follows that the sum of any geometrical progression may be found by multiplying the last term by the exponent of the progression, subtracting the first term from the product and dividing the remainder by the exponent minus unity. = 779. Let there be a geometrical progression of seven terms, whereof the first = 3 and the exponent = 2. Then a 3, b=2, and n=7. The first term will = 3 x 26, or 3 × 64 =192, and the progression will be 3, 6, 12, 24, 48, 96, 192. Multiplying the last term 192 by the exponent 2 we have 384; subtracting the first term the remainder is 381; and dividing this by b-1 or by 1, we have 381 for the sum of the whole progression. 780. When the exponent is less than 1, and the terms of the progression consequently diminish, the sum of such a decreasing progression, which would go on to infinity, may nevertheless be accurately expressed. Thus, let the first term = 1, the exponent, and the sums, so that subtracting the preceding progression the remainder is s=2 for the sum of the proposed infinite progression. In general, suppose the first term =a and the exponent of the progression = (a fraction less than 1), and consequently c is greater than b, the sum of the progression will be found thus: ab ab2 ab3 abs ,&c.; ab 8 = c subtracting, the remainder is (1)s=a. ac fraction by c we have s= The sum, therefore, of the progression is found by dividing the first term a by 1 minus the exponent, or by multiplying a by the denominator of the exponent, and dividing the product by the same denominator, diminished by the numerator of the exponent. 781. So are found the sums of progressions whose terms are alternately affected by the signs + and -. For example: ab ab2 ab3 abs c 8=α- + or s= Thus, if the first term a=3, and the exponent =; that is to say, b=2 and c=5, we shall have for the sum of the progression ++++, &c. =1: for by subtracting the exponent from 1 there remains; and by dividing and dividing the first term by that remainder the quotient is 1. 782. Suppose the terms were alternately positive and negative, thus the first term is, and the exponent. Subtract this last from 1, and the remainder If we divide the first term by this fraction the quotient is gression; so that by taking only one term of the progression, But taking two terms, +10=, there would still be wanting to make the for the sum of the pronamely, the error is We shall conclude with another example in the infinite progression: and the exponent Then 1 minus the exponent and 9 10, the sum required. This series is thus expressed by a decimal fraction 9.9999999, &c. INFINITE DECIMAL FRACTIONS. 783. It has already been seen that decimal fractions are used in logarithmic calculations, in which vulgar fractions would be useless and cumbersome. In other calculations they are of such importance that we shall here dwell upon them, and show how to transform a vulgar fraction into a decimal, and the converse. 784. Generally let it be required to change the fraction into a decimal fraction. Now, as this fraction expresses no more than the quotient of a divided by b, instead of a let us write a 0000000, whose value does not at all differ from a, since it contains neither tenth nor hundredth parts. Let us divide this by b by the common rules of division, taking care to put in the proper place the point which separates the decimals and the integers, and the operation is performed. Let the fraction, for example, be equivalent to, the division in From this it appears that =0.5000000 or 0·5; which is sufficiently manifest, since this decimal fraction represents, which is the same as 785. Let the given fraction be, and we have 3)1'0000000 786. From this it is seen that the decimal fraction equivalent to cannot be discontinued, but that the number 3 is repeated ad infinitum. Indeed, it has been already seen, in the preceding article, that the fractions +18+100+ added together make} In the same way, the decimal fraction which expresses the value of is 0·6666666, evidently the double of 787. Suppose to be the proposed fraction, we have Thus we see =0.75, that is, 7+1 = 700 = 1 for 789. The fraction is changed into a decimal fraction by making 4)5-0000000 790. So will be found =0·2,3=0·4, }=0·6, }=0·8, }=1,}=1·2, &c. In the occurrence of the denominator 7, the decimal fractions become a little more complicated; thus we have 4=0.142857142857, &c., in which the six figures are continually repeated. By transforming this decimal fraction into a geometrical progression, we may see that it precisely expresses 142857, and the exponent the value, the first term of this progression 1000000 Toooooo Hence 142857 the sum = 1- - 1000000 = 142857 (multiplying both terms by 10000000)=4. 1000000 There is, how ever, a simpler mode of proving that this decimal fraction is exactly 4, by substituting for its value the letter s, as under: Now, dividing by 999999, we shall have 8= 1428374; hence s=4. 791. The same will be seen by trial upon other fractions whose denominator is 7, the decimal fraction being infinite and six figures continually repeated. The reason is, that in continuing the division, we must return to a remainder which has already been had; and in that division only 6 different numbers can form the remainder, namely, 1, 2, 3, 4, 5, 6; so that after the sixth division the same figures must return. 792. With the denominator 8 we have the following decimal fraction: =0·125, 0.25, 1=0.375,1=0·5, }=0·625, =0·75 7=0.875, &c. With 9 for the denominator, we have =0·11111, &c.; =0·22222, &c.; }=0.33333, &c. With 10 for the denominator, we have=0·1,=0·2, =0·3, which, indeed, is manifest from the nature of the thing; as also that must be 0.01, and 0.37; that 73=0.472, and that =0·0015. If the denominator be 11, then =0·0909090, &c. Suppose we desired to know the value of this decimal fraction, call it 8, then 8 = 0.0909090, &c. 10s = 0.9090909, &c. 100s = 9 0909090, &c. Subtracts, and we have 99 s=9; consequently s== so with, &c. 8= 793. There are many of these decimal fractions which are called recurring, sometimes with two, and at other times with more, figures. Their values may be found without difficulty. Thus in the case of a single figure constantly repeated, let it be represented by a, so that s=0‘aaaaaaa, we have 795. So that if a decimal fraction occurs, it is easy to find its value; for instance, of 0.296296, the value will be 3, which fraction, it may easily be proved, will give again the decimal fraction required. 796. We shall close this section with a curious example of changing into a decimal fraction the vulgar fraction the operation whereof is as follows: 1 1x2x3x4x5x6x7x8x9x 10' 2) 1·00000000000000 3)0.50000000000000 4)0.16666666666666 5)0.04166666666666 6)0.00833333333333 7)0-00138888888888 8)0.00019841269841 9)0.00002480158730 10)000000275573192 0.00000027557319. CALCULATION OF INTEREST. 797. Interest, or the value of the use of money, is usually expressed per cent., or after the rate per hundred on the principal lent. Thus, if we put out 500 pounds sterling at 5 per cent., it signifies that for every hundred pounds the lender is to receive five pounds per annum during the continuance of the loan. The solution of this question, which is one merely of simple interest, is so obvious, that it is unnecessary further to detain the reader upon it; and we therefore pass on to compound interest, or interest upon interest, which arises from the principal and interest taken together, as it becomes due at the end of each stated time of payment. 798. In the resolution of this question, we are to consider that 100%. at the end of a year becomes 1051. Let a = principal. Its amount at the end of the year is found by saying, if 100 gives 105, what will a give; and we answer which may be also expressed xa, or a + a. 105a 21a 799. Thus, by adding its twentieth part to the original principal, we have the principal at the end of the first year; adding to this last its twentieth, we know the amount of the given principal in two years, and so on. Hence the annual increases to the principal may be easily computed. Suppose, for instance, the principal of 1000l. Expressing the values in decimal fractions, it will be worth The method above exhibited would, however, in calculations for a number of years, become very laborious, and it may be abridged in the following manner. 212 800. Let the present principal a; now, since a principal of 20% will amount to 217. at the end of a year, the principal a will amount to xa at the end of that time. At the end of the following year the same principal will amount to (21) x a. This principal of two years will, the year after, amount to (2)3 x a, which will therefore be the principal of three years; increasing in this manner, at the end of four years the principal becomes (2)+ x a. After a century it will amount to (3)100 x a, and in general (3)" x a is the amount of the principal after n years; a formula serving to determine the amount of principal after any number of years. n 801. The interest of 5 per cent., which has been taken in the above calculation, determined the fraction. Had the interest been reckoned at 6 per cent. the principal a would at the end of a year be (188) × a; at the end of two years to (188)2 × a; and at the end of n years to (188) X a. Again, if the interest be at 4 per cent. the principal a will, after n years, be (18) " × a. Now all these formulæ are easily resolved by logarithms; for if, according to the first supposition, the question be (2})" × a, this will be L.(?!)" + L.a, and as ()" is a power, we have L.(2) "=nL. : so that the logarithm of the principal required is =nx L.+L.a, and the logarithm of the fraction = L.21 - L. 20. 802. We shall now consider what the principal of 1000l. will amount to at compound interest of 5 per cent. at the end of 100 years. Here n=100. Hence the logarithm of the principal required will be =100L.+L.1000, calculated as under : : 5.1189300 Logarithm of the principal required; from the characteristic whereof the principal must be a number of six figures, and by the tables it will appear to be 131,5011. In the case of a principal of 34521. at 6 per cent. for sixty-four years, we have a=3452 and n=64. Principal at the end of the first year therefore == Hence the logarithm of the principal sought=64 × L.§}+ L.3452, which will be found to amount to 143,7637. 803. When the number of years is very great, errors of considerable magnitude may arise from the logarithms not being sufficiently extended in the decimal places; but as our object here is only to show the principle on which these calculations are founded, we do not think it necessary further to pursue that subject. 804. There is another case which now requires our consideration; it is that of not only adding the interest annually to the principal, but increasing it every year by a new sum =b. The original principal a would then increase in the following manner: — After 1 year, a+b After 2 years, (}})2a + }}b+b After 3 years, (}})3a+ (}})2+ 21b+b After 4 years, (?})+a + (3})3b + (}} ) b + ?fb+b This principal evidently consists of two parts, whereof the first =()"a, and the other, taken inversely, forms the series b + }}b + (}} ) b + ({})3b+ (31) b. This last series is evidently a geometrical progression, whose exponent. Its sum, therefore, will be found by first multiplying the last term (2)"-16 by the exponent 2, which gives (?)"b. Subtract the first term b, and we have the remainder (3)"b-b; and lastly, dividing by the ex |