or a2 b2 a c, hence a ca2: b2; also, a2: ac:: b2: c2; : therefore, ac :: 62: c2. 2d. a ca: b2; but c: d::a: b; therefore, 3d. But de ad: a b3 :: b3 : c3 :: c3: d3. a3: 63; therefore, ae: a b1 : : b1 : c1 : : c1 : d1 :: d1 : e1. : :: The above may be easily illustrated by numbers. PROBLEMS FOR PROPORTION. 1. Divide 50 into two such parts that the greater, increased by 3, shall be to the less, diminished by 3, as 3 to 2. Let x the greater number, and 50-x the less. Multiplying extremes, 2x+6=150-3x-9. 2. What number is that to which if 3, 8, 12, and 20, be severally added, their sums shall be proportional? Let Then, x= the number. x+3x+8:: x+12: x+20. Multiplying extremes, x2+23x+60=x2+20x+96. Transposing, Dividing, 12+3 23x-20x=96-60. x=12. Ans. VERIFICATION. 12+8 :: 12+12 : 12+20=15 : 20 :: 24: 32. 3. If Mars, when in opposition to the sun, is 49,000,000 miles from the earth, and the quantity of matter in the earth is 11 times greater than that in Mars, at what distance from the earth, in a direction towards Mars, will a body remain at rest? See Art 213. Let x the distance from the earth. Then 49,000,000—x = the distance from Mars. And let a 49,000,000. Then, a2 (a-x): 1 : 11. Multiplying extremes, 11x2-a2-2ax+x2. And, by supplying the value of a, we have a 10° 4. There are two numbers which are to each other as 5 to 3; and, if 4 be added to the greater and 8 to the less, they will then be to each other as 6 to 5. What are the numbers? Ans. 20 and 12. 5. Divide the number 60 into two such parts that their product shall be to the difference of their squares as 2 to 3. Ans. 40 and 20. 6. I have two square house-lots, which, together, contain 208 square rods; and the area of the greater is to the area of the less as 9 to 4. How many more square rods are there in the greater than in the less? Ans. 80 square rods. 7. The product of two numbers is 12, and the difference of their cubes is to the cube of their difference as 13 to 4. What are the numbers? Ans. 2 and 6. 8. Divide the number 100 into two such parts that 6 times their product shall be to the sum of their squares as 24 to 17. What are those parts? Ans. 80 and 20. 9. There are two numbers, whose product is 35, and the difference of their squares is to the square of their difference as 6 to 1. What are the numbers? Ans. 7 and 5. 10. There are two numbers in the triplicate ratio of 4 to 1, whose mean proportional is 32. What are the numbers ? Ans. 256 and 4. 11. Divide 20 into two such numbers, that the quotient of the greater divided by the less shall be to the quotient of the less divided by the greater as 9 to 4. What are those numbers? Ans. 12 and 8. 12. Divide 26 into three such parts, that the first shall have the same ratio to the second that the second has to the third, and that the first term shall be the third term. SECTION XX. Ans. 2, 6, and 18. ARITHMETICAL PROGRESSION. ART. 251. An Arithmetical Progression is a series of numbers or quantities, increasing or decreasing by a constant difference. It is sometimes called Progression by Difference. 252. The constant difference is called the Common Difference, or ratio of the progression. Ratio here used is an Arithmetical rate. Thus, let there be the two following series. First series, (1) (2) (3) (4) (5) (6) (7) (8) Second series, 30, 26, 22, 18, 14, 10, 6, 2=128. 253. The numbers which form the series are called the terms of the progression. 254. The first is called an ascending series of progression, where the first term is 1, the common difference 3, the number of terms 8, the last term 22, and the sum of the series 92. 255. The second is called a descending series of progression, where the first term is 30, the common difference -4, the number of terms 8, the last term 2, and the sum of the series 128. 256. The first and last terms of the progression are called extremes, and the other terms are the means. 257. The number of common differences in any number of terms is one less than the number of terms. Hence, if there be 8 terms, the number of common differences will be 7, and the sum of the differences will be equal to the difference of the extremes. We therefore infer, that if the difference of the extremes be added to the first term, the sum will be the last term; also, if the difference of the extremes be taken from the last term, the remainder will be the first term. 258. Also, if the sum of the common differences be divided by the number of common differences, the quotient will be the common difference. To illustrate this, we will examine the following series: (1) (2) (3) (4) (5) (6) (7) 2, 5, 8, 11, 14, 17, 20. Here the first term is 2, the last term 20, the number of terms 7, and the common difference 3. Now, if we had only the first term, number of terms, and common difference, to find the last term, we should have only to add the difference of the extremes to the first term. The common difference is 3; and, as there are 7 terms, the number of common differences is 6. The difference of the extremes will, therefore, be 6×3=18, and the last term will be 2+18=20. Hence, having the first term, common difference, and number of terms given, to find the last term, we have the following RULE. Multiply the number of terms, less one, by the common difference, and to the product add the first term. Again, if we invert the terms, we have (1) (2) (3) (4) (5) (6) (7) 20, 17, 14, 11, 8, 5, 2. Here we have 20 for the first term, -3 for the common difference, and 7 for the number of terms, to find the last term. 6x-3=-18; 20-18-2 the last term. The pupil will perceive that 18 is a negative term; and to add a negative term to a positive is to write their difference. Again, we have given the extremes 2 and 20, and number of terms 7, to find the common difference. Here the number of common differences is 6; for we have before shown that the number of common differences is always one less than the number of terms; therefore, 18÷6=3, the common difference. 259. The principles of an arithmetical progression may be well illustrated by literal terms. Let a be the first term of an ascending series, and d the common difference; then the second term will be a+d, and the the third term a+2d, and the series will be If it be required to form a descending series, when the first term is a and the common difference ―d, it will be thus: 260. It is evident that the last term in both series is equal to the first term with the common difference repeated as many times, wanting me, as there are terms in the series. Hence, if n represent the number of terms, the following will be the formula to find L, the last term. L=a+(n−1)d. EXAMPLES. 1. If the first term be 7, the common difference 4, and the number of terms 20, required the last term. La+(n-1)d=7+(20-1)4-83. Ans. |