contain all the relations of the five quantities a, l, 9, n, and S; any three of which being given, the other two may be found. It would however be difficult to find n, without the aid of logarithms, which will be explained hereafter. Indeed logarithms will greatly facilitate the calculations in most cases of geometrical progression. Therefore we shall give but few examples, until we have explained them. If we substitute a qn~ in place of l, in the expression of S, it becomes When q is greater than unity, the quantity or will become greater as n is made greater, and S may be made to exceed any quantity we please, by giving na suitable value; that is, by taking a sufficient number of terms. But if q is a fraction less than unity, the greater the quantity n, the smaller will be the quantity.q". Suppose q = -, m being a number greater than unity, then Substituting - in place of q” in the expression of S, and it becomes Changing the signs of the numerator and denominator, and multiplying both by m, It is evident that the larger n is or the more terms we take in the progression, the smaller will be the quantity , and а от consequently the nearer the value of S will approach from which it differs only by the quantity a quantity quantity (m—1) maBut it can never, strictly speaking, be equal to it, for the _will always have some value, however large n may be ; yet no quantity can be assumed, but this expression may be rendered smaller than it. The quantity_am_ is therefore the limit which the sum sm - 1 of a decreasing progression can never surpass, but to which the value continually approximates, as we take more terms in the series. In the progression 1, 3, 4, i, ib, &c. a=1!= Hence S= 1 1 X 2nIn this example the more terms we take, the nearer the sum of the series will approach to 2, but it can never be strictly equal to it. Now if we consider the number of terms infinite, noir 1 will be so small that it may be omitted without any sensible error, and the sum of the series may be said to be equal to 2. By taking more and more terms we approach 2 thus, the quantity 1 Examples. What is the sum of the series 1, 4, ), a'r, &c. continued to an infinite number of terms ? 13. 3- 1 2. What is the sum of the series, 5, , , , &c. continued to an infinite number of terms ? 3. What is the sum of the following series continued to infinity ? 35, 7, 1, 1, &c. 4. What is the sum of the following series continued to in finity? 208, 26, 34, , &c. 5. What is the sum of the following series continued to infinity ? 38, 45, 41, , &c. 5, 15, 45, &c. ? 35, 175, 875, &c. ? When three numbers are in geometrical progression, the middle term is called a mean proportional between the other two. Let three numbers, a, b, c, be in geometrical progression, so that We have and b = (ac). 8 Find a mean proportional between 4 and 9. Ans. 6. * = 36 x = 6. 9. Find a mean proportional between 7 and 10. 10 Find a mean proportional between 2 and 3 XLVIII. Logarithms. We have seen, Art. XXXVI, with what facility multiplication, division, the raising of powers, and the extraction of roots may be performed on literal quantities consisting of the same letter, by operating on the exponents. We propose now to apply the same principle, though in a way a little different, to numbers. Multiplication, we observed, is performed by adding the exponents, and division by subtracting the exponent of the divisor from that of the dividend. Thus a x all is a 46 = a'. And is aans = a'. In the same manner 23 x 2 = 2048 = 28, 27 and = 27-5 = 22. Let us make a table consisting of two columns, the first containing the different powers of 2, and the second the exponents of those powers. 14 15 16 arco 12 Powers. | Expon. || Powers. | Expon. | Powers. | Expon. 16,384 32,763 4 1 2 512 9 65,536 1024 131,072 262,144 18 524,288 19 6 11 8192 | 13 | 1,048,576 | 20 Suppose now it is required to multiply 256 by 64. We find by the table that 256 is the 8th power of 2, that is 28, and that 64 is 28. Now 28 x 24 = 2846 = 24. Returning to the table again and looking for 14 in the column of exponents, against it we find 16384 for the 14th power of 2. Therefore the product of 256 by 64 is 16384. This we may easily prove. 16384 Multiply 256 by 128. Finding these numbers in the table in the column of powers, and looking in the other column for the exponents, we find that 256 is the Sth power of 2, and 128 the 7th power. Adding the exponents 8 and 7, we have 15 for the exponent of the product. Now looking for 15 in the column of exponents, we find against it in the column of powers, 32768 for the 15th power of 2, which is the product of 256 by 128. Let the learner prove this by multiplying 256 by 128. Divide 8192 by 32. |