CASE XIII. To reduce a fraction, whose denominator is either a simple or a compound surd, to another that shall have a rational denominator. RULE. b Va' 1. When any simple fraction is of the form multiply each of its terms by a, and the resulting fraction 2. If it be a compound surd, find such a multiplier, by the last rule, as will make the denominator rational; and multiply both the numerator and denominator by it, and the result will be the fraction required. Here to a fraction, whose denominator 5-2 3 3 √5+√2_3√5+3√2__3√✓/5+3√√2_ √5−√2^√5+√2 √5+ √2 =√5+√2 the answer required. 1 6. Reduce to a fraction that shall have a to a fraction, the denominator of a+ √b OF ARITHMETICAL PROPORTION AND PROGRESSION. ARITHMETICAL PROPORTION, is the relation which two quantities, of the same kind, have to two others, when the difference of the first pair is equal to that of the second. Hence, three quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the second and third. Thus, 2, 4, 6, and a, a+b, a+2b, are quantities in arithmetical proportion. And four quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the third and fourth. Thus, 3, 7, 12, 16, and a, a+b, c, c+b, are quantities in arithmetical proportion. ARITHMETICAL PROGRESSION is when a series of quantities increase or decrease by the same common diffe rence. Thus, 1, 3, 5, 7, 9, &c. and a, a+d, a+2d, a+3d, &c. are increasing series in arithmetical progression, the common differences of which are 2 and d. And 15, 12, 9, 6, &c. and a, a-d, a—2d, a—3d, &c. are decreasing series in arithmetical progression, the common differences of which are 3 and d. The most useful properties of arithmetical proportion and progression are contained in the following theorems : 1. If four quantities are in arithmetical proportion, the sum of the two extremes will be equal to the sum of the two means. Thus, if, the proportionals be 2, 5, 7, 10, or a, b, c, d ; then will 2+10=5+7, and a+d=b+c. 2. And if three quantities be in arithmetical propor tion, the sum of the two extremes will be double the mean. Thus, if the proportionals be 3, 6, 9, or a, b, c, then will 3+9=2×6=12, and a+c=2b. 3. Hence an arithmetical mean between any two quantities is equal to half the sum of those quantities. Thus, an arithmetical mean between 2 and 4 is 2+4 5+6 3 ; and between 5 and 6 it is = =51. 2 2 a+b And an arithmetical mean between a and b is 4. In any continued arithmetical progression, the sum of the two extremes is equal to the sum of any two terms that are equally distant from them, or to double the middle term, when the number of terms is odd. Thus, if the series be 2, 4, 6, 8, 10, then will 2+10 =4+8=2×6=12. And, if the series be a, a+d, a+2d, a+3d, a+4d, then will a+(a+4d)=(a+d)+(a+3d)=2×(a+2d). 5. The last term of any increasing arithmetical series is equal to the first term plus the product of the common difference by the number of terms less one; and if the series be decreasing, it will be equal to the first term minus that product. Thus, the nth term of the series a, a+d, a+2d, a+3d, a+4d, &c. is a+ (n-1)d. -3d, And the nth term of the series a, a―d, a—2d, aa-4d, &c. is a-(n-1)d. 6. The sum of any series of quantities in arithmetical progression is equal to the sum of the two extremes multiplied by half the number of terms. 6 Thus, the sum of 2, 4, 6, 8, 10, 12, is = (2+12) X 14X3=42. And if the series be a+(a+d)+(a+2d)+(a+3d) H +(a+4d) &c. . . . +, and its sum be denoted by S, we shall have S=(a+1)×, where I is the last term, and n 2 Or, the sum of any increasing arithmetical series may be found, without considering the last term, by adding the product of the common difference by the number of terms less one to twice the first term, and then multiplying the result by half the number of terus. And, if the series be decreasing, its sum will be found by subtracting the above product from twice the first term, and then multiplying the result by half the number of terms. as before. Thus, if the series be a+(a+d)+(a+2d)+(a+3d) +(a + 4d), &c. continued to n terms, we shall have S= {2a+(n−1)d } × 32 X n And if the series be a+(a-d)+(a−2d)+(a−3d)+ (a-4d), &c. to n terms, we shall have (y) The sum of any number of terms (n) of the series of natural numbers 1, 2, 3, 4, 5, 6, 7, &c. is = n(n+1) 2 Thus, 1+2+3+4+5, &c. continued to 100 terms, is 50X1015050. 100 × 101 2 Also the sum of any number of terms (n) of the series of odd numbers 1, 3, 5, 7, 9, 11, &c. is n2. Thus, 1+3+5+7+9, &c. continued to 50 terms, is 502 2500 = And if any three of the quantities, a, d, n, S, be given, the fourth may be found from the equation S= = {2a= |