2. Divide the number 210 into three parts, so that the last shall exceed the first by 90, and the parts be in geometrical progression. Let x = the first part, and y = the ratio of the progres. sion, then the parts will be x, xy, and xy2, and we shall have, 2 By putting x + 90 in the place of xy2, in the first equation, From the equation, xy2 = x + 90, we find, from which we obtain, by clearing of fractions and reducing, 3. The sum of four numbers in geometrical progression, is 15, and the sum of their squares 85. Required the numbers. Let x = the second term, and y = the third term of the , and be the first and fourth, since = xy (95). Putting, for the sake of brevity, 15 = a, and 85 = b, we shall have the equations If in the first equation, we call the sum of the middle terms s, we shall have, in the first place, adding them together, and transposing 4ry, we have, y + x2 + y2 + = s2+(as)2 - 4xy; whence, s2+(as)2 - 4ху = b. In order to eliminate zy, take the two equations, x2+3x2y + 3xy2 + y3 = s3; transposing, x3 + y3 = s3 (3x2y + 3xy2) Clearing the second equation of fractions, we have, wherefore, (a - s)xy = s3 =s3-3xy(x + y) = s3-3xys. x3 + y3 = (a - s)xy; 3xys, Clearing this equation of fractions, and reducing, it becomes, Giving to a and b their numerical values, we have, Having found the value of s, we substitute it in the equation, $3 2s+a; from which we find We then have the two equations xy = 8, and x + y find the values of x and y. 6, to y in the second, we have, 4 1, and, y 16 8. 2 The numbers are therefore 1, 2, 4, and 8. If we had taken the negative value of y, we should have had, so that the result would have been the same, except that the pro 4. The sum of four numbers in geometrical progression is 30; and the last term divided by the sum of the mean terms What are the numbers? Ans. 2, 4, 8, and 16. 1 5. There are three numbers in geometrical progression, whose product is 64, and the sum of their cubes 584. What are the numbers? Ans. 2, 4, and 8. 6. The sum of three numbers in geometrical progression is 13; and the sum of the extremes being multiplied by the mean term, the product is 30. What are the numbers? Ans. 1, 3, and 9. 7. 120 dollars are divided between four persons, in such a way, that their shares may be in arithmetical progression; but if the second and third had received 12 dollars less each, and the fourth 24 dollars more, the shares would have been in geometrical progression. Required each share. Ans. Their shares were 3, 21, 39, and 57, respectively. 8. There are three numbers in geometrical progression, whose sum is 31, and the sum of the first and last is 26. What are the numbers? Ans. 1, 5, and 25. 9. The sum of the first and second of four numbers in geometrical progression is 15, and the sum of the third and fourth 60. What are the numbers? Ans. 5, 10, 20, 40. 10. The sum of four numbers in geometrical progression is equal to the common ratio +1, and the first term is What are the numbers? Ans. ΕΧΡΟΝΕΝTIAL EQUATIONS. (104.) An exponential equation is one in which the unknown quantity is an exponent; as a = b, 3" = 243. When the exponent is a whole number, such an equation presents no difficulty; for example, if we would find the value of x in the equation 3" = 243, we are only to raise 3 to such a power as will equal 243; and the exponent of this power is the value of x. We find 35 = 243, and therefore x = 5. If, however, we have 3" = 54, we can only obtain the value of x by approximation, and the approximate value will be a fraction. We now proceed to solve the equation 3 = 54. If we raise 3 to the third power, we have, 33 = 27: x is therefore greater than 3; and it is less than 4, for 34 = 81. We must therefore take for the value of x, 3+ a 1 fraction. Let that fraction be, in which x is greater than x 1 = 54, or 33 × 3 = 54, 1 27×37 = 54. Raising both members of this equation to the 'th power, it Here the value of x' is less than 2; for 22=4; and it is greater than one, since 21 = 2; x' therefore equals 1 + a fraction. 1 we have, x is less; wherefore, 1 x" = 1 + a fraction. Making 2" = 1 +, we have, 2; <; we may therefore put x" = 1 +, and then we have, We shall here find a'"'"' greater than 2, and less than 3; for <, and > . Putting x"""' = 2 + |