2. The sum of four numbers in geometrical progression is 15 or a, and the sum of their squares 85 or b. What are the numbers ? Taking the proper notation for an even number of terms, we have Assume +y=s, and xy = p; then by (301), x +y = $_2p, x +y = $_3sp. Substituting the values of (x+y) and (x++y'), in (1) and (2), 20% (3) =b--$+2p. + -8, Squaring (3), and then transposing 2xy, or 2p, * y" +*=(a—s)?—2p; (5) whence, from (4) and (5), (a--8)-2p=b-$*+2p; or, a'-2a8+23-4p=b. (6) Clearing (5) of fractions, and putting xy=p in second member, x +y8 = apps ; or, 58—3sp = ap-ps; whence, P = (V) a+2s Substituting this value of p in (6), and reducing, we have a'-2as = ab+268 ; or, as'+bs= (a'—6). 2 Restoring the numerical values of a and b, 158° +858 = 70 X 15, whence, Substituting the values of a and s in (9), and we obtain p=8; that is, x+y=6, ay=8; whence, 2, Therefore, the required numbers are 1, 2, 4, 8, Ans. a S = 6. y = 4. 3. There are three numbers in geometrical progression; their sum is 21, and the sum of their squares is 189. Find the nuinbers. Ans. 3, 6, 12. 4. Divide the number 210 into three parts, so that the last shall exceed the first by 90, and the parts be in geometrical progression. Ans. 30, 60, and 120. 5. The sum of four numbers in geometrical progression is 30; and the last term divided by the sum of the mean terms is 13. What are the numbers ? Ans. 2, 4, 8, and 16. 6. The sum of the first and third of four numbers in geometrical progression is 148, and the sum of the second and fourth is 888. What are the numbers ? Ans. 4, 24, 144, and 864. 7. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Ans. 2, 4, and 8. 8. There are four numbers in geometrical progression, the sec. ond of which is less than the fourth by 24; and the sum of the extremes is to the sum of the means as 7 to 3. What are the numbers ? Ans. 1, 3, 9, and 27. 9. There are three numbers in geometrical progression; the sum of the first and second is 20, and the difference of the second and third is 30. What are the numbers ? Ans. 5, 15, 45. 10. The continued product of three numbers in geometrical progression is 216, and the sum of the squares of the extremes is 328. What are the numbers ? Ans. 2, 6, 18. 11. The sum of three numbers in geometrical progression is 13, and the sum of the extremes being multiplied by the mean, the product is 30. What are the numbers ? Ans. 1, 3, and 9. 12. There are three numbers in geometrical progression ; their continued product is 64, and the sum of their cubes is 581. What are the numbers ? Ans. 2, 4, 8. 13. There are three numbers in geometrical progression; their continued product is 1, and the difference of the first and second is to the difference of the second and third as 5 to one. What are the numbers ? Ans. 1, 1, 5. 14. The sum of 120 dollars was divided between four persons in such a manner that the shares were in arithmetical progression ; if the second and third had each received 12 dollars less, and the fourth 24 dollars more, the shares would have been in geometrical progression. Find the shares. Ans. $3, $21, $39, and $57. 15. 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. 16. The sum of six numbers in geometrical progression is 189, and the sum of the second and fifth is 54. What are the numbers ? Ans. 3, 6, 12, 24, 48, and 96. 17. The sum of six numbers in geometrical progression is 189, and the sum of the two means is 36. What are the numbers ? Ans. 3, 6, 12, 24, 48, and 96. 18. A man borrowed p dollars ; what sum must he pay yearly in order to cancel the debt in n years, interest being allowed on the anpaid parts of the principal at r cents per annum on a dollar ? pr(1+r)" Ans. dollars. (1+r)*_1 IDENTICAL EQUATIONS. 366. An Identical Equation is one in which the two members are either the same algebraic expression, or the one member is merely another form for the other. In every case, either the one member may be reduced to the other directly, or the two members may be reduced to some expression different from either, from which both members may be supposed to originate. Thus, ax+b=ax+b, 1 xo x*(1+x)' are identical equations. In the first, the two members have exactly + X the same form. In the second, the second member may be reduced to the form of the first, by performing the multiplication indicated. 1 In the third, each member may be reduced to the fraction, 1+x 367. There are certain properties of identical equations, which are of great importance in the further treatment of series, and in the general theory of equations. In order to investigate these properties, let us first consider what any term containing the variable x, as ax", will become when x = 0, under the various conditions of the exponent. 1.–Suppose n to be positive; then if x = 0, we have ax" =a:0= 0. 2.-Suppose n to be negative; then if x=0, we have 3.-Suppose n to be nothing or zero; then if x = 0, we have ax" = a:0° = a:1= a. 368. We are now prepared to demonstrate the following propo sitions : I. An identical equation is satisfied for any value whatever of the unknown quantity. The truth of this proposition follows directly from the definition of an identical equation. It is implied in all algebraic transformations, that the value of a function is not changed by changing its form, whatever quantities the symbols represent. Hence, if the two members of an equation are the same in form, or reducible to the same expression, they must be equal, whatever value be substituted for the unknown quantity. To illustrate this principle, we will take the following identical equation, {(2—3)+1+(«—2)"}' = 2{(x—3)*+1+(x—2)"}, where the form is such that the identity of the two members is not apparent from inspection. If in this equation we make x equal to 1, 2, 3, 4, 5, etc. succes sively, we shall have, {4+1+1}' = 2{16+1+ 1}, {4+1+9} = 2{16+1+81}, etc., every result being a true equation. II. Conversely :-Every equation which is satisfied for any value whatever of the unknown quantity, is an identical equation. Suppose the given equation to be cleared of fractions, and each member arranged according to the ascending powers of the unknown quantity. Then the equation may be represented thus : A.xo+BX+Cx+.... = A'x®+B'xb' + C'*'+.... 1 (1) in which, by hypothesis, we have a< b < c...., and a' <b' < d'.... It is implied, also, that the coefficients, A, B, C, etc., and A', B', C", etc., are all finite quantities greater than zero, and independent of x; and the number of terms may be limited or unlimited. Divide both members of equation (1) by xa; we have A+Bxbma + Cromat A'ma'-a +B'xbl-a + C'xa-at..., (2) in which the exponents, 6–a, –a, etc., in the first member, are all positive, because a < b < c..... Now by hypothesis, the given equation is true for all values of X; hence every modification of it will be true for all values of x. 0; then in the first member of equation (2), every term after the first will reduce to zero, (367, 1), and we shall have A = A'xala+B+C - +.... (3) Now since a' <' < c...., we must have (a'—a) < (b'—a)< (d-a) <.... Hence, in equation (3), the first exponent, a-a', is the least of all. But we observe, J.st. The exponent, a'-a, can not be a positive quantity; for in Make x = |