340. If four quantities be in proportion, like powers or roots of those quantities will be proportionals. 4. Find a fourth proportional to ab, cd and ax. Ans. x=4. 5. Find the number to which, if 2 and 5 be successively added, the resulting sums will be the ratio of 5:11. Ans. †. 6. Find a fourth proportional to, and . 7. Find a mean proportional to 2 and 8. 8. Divide $1000 between two persons so that their shares shall be in the ratio of 7 to 9. By conditions, By Theorem I, whence, and, SOLUTION. x= · number of dollars in first person's share; 1000-x= number of dollars in second person's share. x: 1000-x::7:9 9x=7000-7x x=437.50 1000-x-562.50 Hence, the first person's share is $437.50, and the second person's share, $562.50. 9. Divide the number 56 into two parts, such that one shall be to the other as 3 to 4. Ans. 24; 32. If A's age 10. A and B are at present of the same age. be increased by 36 years, and B's by 52 years, their ages will be as 3 to 4. What is the present age of each? 11. A, B and C make a joint stock. A puts in $60 less than B, and $68 more than C; and the sum of the shares of A and B is to the sum of the shares of B and C as 5 to 4. What did each put in? Ans. A, $140; B, $200; C, $72. 12. Find two numbers whose sum is to their difference as 3 to 2, and whose difference is to their product as 1 to 5. 13. A person in a railway-car observes that another train running on a parallel line in the opposite direction occupies 2 seconds in passing his train. But if the two trains had been proceeding in the same direction, one would have taken 30 seconds to pass the other. Compare the speed of the two That is, the speed of the two trains is as 8 to 7. 14. There is a rectangular field which contains 360 square rods, and whose length is to its breadth as 8 to 5. What are the length and breadth ? Ans. Length, 24 rods; breadth, 15 rods. of which is 10 yards areas are as 25 to 9. 15. In a court there are two square grass-plots, a side of one longer than a side of the other, and their What are the lengths of their sides? Ans. 15 yards; 25 yards. SECTION XLIII. ARITHMETICAL PROGRESSION. 341. A Progression is a succession or series of quantities increasing or decreasing according to some fixed law. 342. The terms of a progression, or series, are the quantities of which the series is formed. 343. An Arithmetical Progression is a series formed by adding a constant quantity. The constant quantity added is called the Common Difference, and the progression is ascending when the common difference is positive, and descending when it is negative. Thus, a, a+d, a+2d, a+3d,... is an ascending arithmetical progression in which the common difference is +d, and a, a-d, a-2d, a-3d,... is a descending arithmetical progression in which the common difference is - d. 344. In an Arithmetical Progression having a limited number of terms there are five elements to be considered: 1. The first term, a; 2. The last term, l; 3. The number of terms, n; 4. The common difference, d; 5. The sum of the terms, S. and any three of these being known, the other two can be found. The first and last terms are called Extremes, and all the other terms are called Arithmetical Means. Theorem I. 345. The last term of an arithmetical progression is equal to the first term plus the product of the common difference by the number of terms less one. The coefficient of d in the last term is one less than the number of terms; hence, the nth term equals 346. The sum of all the terms of an arithmetical progression is equal to half the sum of the two extremes multiplied by the number of terms. Let an arithmetical progression be α, a+d, a+2d, a+3d, Then, since the sum of a series is the same, whether written in the direct or in the inverse order, S=a+(a+d)+(a+2d)+(a+3d)+ ... +1 S=1+(l−d) + (1 − 2d) + (1 − 3d) + + a Here, a+l is taken as many times as there are terms, or n times. Hence, 28=n(a+1); whence, S= Theorem III. (a + D) n (2) 347. Any number of arithmetical means may be inserted between two given terms of an arithmetical progression. The number of terms in a series must consist of the two extremes and all the intermediate terms, or of two more terms than the number of means. Hence, if m be made to denote the number of means, or the whole number of terms. m+2=n, Substituting the value of n in formula (1), we have Hence, the required means may be obtained by the continued addition of the value of d. Theorem IV. 348. An arithmetical mean between any two quantities is one Adding a to each member, and reducing, we have (5) (6) But a+d is the second term of a series whose first term is a and |