SOLUTION 2D.-Since la+ (n-1)d, we have 76a+(25-1)x3, or 76=a+72; whence a=76—72, or 4. NOTE.-It may also be solved by deriving the formula and subtracting the value of the terms in the formula. See Art. 843. Rule. To find the first term, diminish the last term by the common difference multiplied by the number of terms less one. 2. The common difference is .05, number of terms 100, and one extreme is 5; what is the other? Ans. .05. 3. The amount of a certain sum for 25 years, at an annual interest of $121, is $5621; what is the principal? Ans. $250. CASE III. 845. Given, the first term, the last term, and the number of terms, to find the common difference. 1. Required the common difference, the first term being 4, the last term 76, and number of terms 25. SOLUTION.-By Case I. we have 76=4+(24 times the common difference); hence the common OPERATION. 76=4+24x d 76-4 hence, d= = = 3 24 SOLUTION 2D.—Since l = a+(n−1)d, we have 76=4+(25—1)d, or 76=4+24d; whence 24d-76-4-72, or d=72÷÷24, or 3. NOTE. It may also be solved by deriving the formula, and substituting the values of the given terms. See Art. 843. Rule. To find the common difference, divide the difference of the extremes by the number of terms less one. 2. $1600 in 60 years amounts to $8320; required the annual interest. Ans. $112. 3. A begins business with $4000; at the end of 15 years he has $9400; required his average annual income. Ans. $360. CASE IV. 846. Given, the first term, the last term, and the common difference, to find the number of terms. 1. The first term is 7, last term 39, and common difference 3; required the number of terms. SOLUTION 2D.-Since la+(n−1)d, 39=7+(n−1)×3}; hence (n−1)×3}=39—7, or 32, and n 32÷3+1, or 11. NOTE.-Require the pupils to derive the formula and solve the problems by substituting the values of the terms in the formula. See Art. 843. Rule. To find the number of terms, divide the difference between the extremes by the common difference, and add 1. 2. How many days will it take a student to walk 51 miles a day, if he goes 3 miles the first day, 6 miles the second day, etc. Ans. 20 days. 3. How many pigs must a man buy, giving $2.25 for the first, $2.37 for the second, etc., that the last may cost $4.75? CASE V. Ans. 21. 847. To insert a given number of arithmetical means between two given numbers. 1. Insert 3 arithmetical means between the numbers 4 and 12. OPERATION. 12-4 d = 2 4 4 .. 4, 6, 8, 10, 12. SOLUTION. Since there are 3 means, there are 3+2, or 5 terms in the whole series; hence by 12-4 Case III., the common difference equals or 2; hence the means are 6, 8, and 10. Rule. Take the given numbers as the extremes, and the number of means plus 2 as the number of terms; find the common difference by Case III., add this to the smaller number for the 1st mean, and so complete the series. 2. Insert 6 arithmetical means between 3 and 24. Ans. 6, 9, 12, 15, 18, 21. 3. If 2 means be found between the successive terms of the series, 1, 7, 13, 19, what will the new series be? Ans. 1, 3, 5, 7, 9, 11, 13, etc. 4. Form an arithmetical series by writing 3 means between the successive terms of the series 3, 15, 27. Ans. 3, 6, 9, 12, 15, 18, etc. 5. A man bought teas at prices increasing in arithmetical progression, the cheapest costing 25 cents, and the dearest $1.10 a pound; what were the prices of the four intermediate kinds? Ans. 42, 59, 76, and 939. CASE VI. 848. Given, the first term, the last term, and the number of terms, to find the sum of the series. 1. The first term is 3, the last term 19, and the number of terms 5; required the sum of the series. Rule. To find the sum of an arithmetical series, multiply half the sum of the extremes by the number of terms. a+l NOTE. This is expressed in the following formula: S xn. The problem may be solved by substituting the values of the terms in this formula. 2. How many strokes does an ordinary clock strike in 24 hours? Ans. 156. 3. The last term of a series is 18.75, the common difference .25, and the number of terms 18; required the sum of the series. Ans. 299.25. 4. The clocks in Venice strike from 1 to 24; strokes does such a clock strike in a day? how many Ans. 300. last pay 5. I discharge a mortgage in 15 payments; my ment was $850, and each payment was $50 greater than the preceding; what was the mortgage? Ans. $7500. 6. A stone falling from an altitude will descend 16 feet in 1 second, 3 times as far the next second, 5 times as far the next second, etc., how far will it fall in half a minute? Ans. 2 mi. 237 rd. 41 ft. 7. 150 apples are placed in a row 21 yards apart, the first being 3 yards from a basket; how far will a boy travel, starting from the basket, to gather them singly into the basket? Ans. 32 mi. 455 yd. 8. I wish to set out 75 fruit trees 4 yards apart around a circular field which will exactly contain them in its circumference; how far shall I have walked when the last one is planted, if I plant the first one at the starting point, and always go on the circumference, returning to the starting point every time? Ans. 12 mi. 196 rd. 2 yd. 9. Suppose, in the last example, I had returned to the starting point every time, but had taken the shortest distance on the circumference of the circle; how far would I have walked ? Ans. 6 mi. 125 rd. 1 ft. 6 in. CASE VII. 849. Given, the sum and any two of these threethe first term, the last term, or the number of terms— to find the one not given. 1. The sum of an arithmetical series is 63, the first term 3, and the last term 18; what is the number of terms? a+i; 2, we have 2.S=(a+l)×n; dividing by a+l, we have n= substituting the values of a and 7, we have n=63×2÷(3+18), which equals 126-21, or 6. Rule. To find the number of terms, divide twice the sum of the terms by the sum of the extremes. NOTE. The other cases are solved in a similar manner. Let the pupils derive and state the rules. 2. How long will it take to pay a debt of $3500, the payments being made yearly in a decreasing series, if the first and last are respectively $575 and $125? Ans. 10 years. 3. If I travel 660 miles in 15 days, going 65 miles the last day, increasing regularly each day, how far did I go the first day? Ans. 23 miles. 4. The sum of the terms is 4935, the first term 197, and number of terms 21; what is the last term? Ans. 273. 5. I owe a debt of $7200; I wish to cancel it in 16 payments, increasing regularly at each payment, the first being $300; required the last payment. Ans. $600. 850. Since there are five quantities in Arithmetical Series, any three of which being given the other two may be found, there are twenty distinct cases. 851. The rules for the eight simple cases are expressed GEOMETRICAL PROGRESSION. 852. A Geometrical Progression is a series of numbers which vary by a common multiplier; as, 2, 6, 18, 54, etc. 853. The Rate or Ratio is the common multiplier; thus, in the above series, the rate is 3. 854. In an Ascending series, the rate is greater than a unit; in a Descending series, the rate is less than a unit. 855. The Quantities considered are five, any three of which being given, the others may be found. QUANTITIES CONSIDERED. 1. The first term, a. 3. The number of terms, Symbols. n. r. |
