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2. The least term is 5, the greatest 205, and the number of terms 11, being given, to find the sum of the terms. Thus 5 + 205 = 210 = sum of the extremes. Also 54 = half the number of terms. Wherefore 210 × 53 = 1155 = the sum required.
3. The extremes are 4 and 800, and the number of terms 40, to find the sum. Thus 4 + 800 x 20 = 16080, the sum required.
4. A man paid a debt which he owed at 20 payments in arithmetical progression; the first payment was 3l, and the last 18l. what was the debt Ans. 210l.
5. I bought 100 peaches, and paid for them in arithmetical progression, viz. for the first 4d. and for the last 6d. what sum did the whole amount to ? Ans. ll. 7s. 1d.
6. What must be given for 120 elm trees, the prices whereof are in arithmetical progression, that of the first being 5s. and that of the last 10l. Ans. 615l.
290. The least term, the greatest term, and the number of terms, being given, to find the common difference.
Rule. Subtract the least term from the greatest, and divide the remainder by 1 less than the number of terms; the quotient will be the common difference required.
7. In an arithmetical progression, the least term is 3, the greatest 17, and the number of terins 8 ; required the common difference 2 Thus 17–3 = 14 = the difference of the extremes. And 8–1 = 7 = the number lessened by 1.
8. The least term is 5, the greatest 205, and the number of terms 11, in an arithmetical progression; required the common difference 2 O5–5 2OO Thus * I = -ia-- 20 = the common difference required. 9. A man had 5 sons, whose ages were in arithmetical progression, the youngest was 3 years old, when the eldest was 13;
required the common difference of their ages 2
o- - o = 24 years, the com, difference required. 10. Bought 120 sheep, and gave a shilling for the first, and five pounds for the last; if the prices are in arithmetical progression, what is the common difference Ans, 9++#d. 11. Required the common difference of 40 terms in arithmetical progression, whereof the least is 4, and the greatest 800 Ans. 20; }.
12. A farmer bought 100 oxen, for the first he paid 3!, and for the last 48l. supposing the prices in arithmetical progression, what was the common difference Ans. 9s. 1+1+d.
291. The least term, the greatest term, and the common difference being given, to find the number of terms.
RULE. Subtract the less term from the greater, and divide the remainder by the common difference; increase the quotient by 1, and it will be the number of terms required.
13. The least term of an arithmetical progression is 4, the greatest 39, and the common difference 5; required the number of terms ? 39–4 35
Thus —- - 5 = 7. Then 7 -- 1 = 8, the number re
quired. 14. A grazier sold a certain number of oxen, the prices of which were in arithmetical progression; for the first he received 21, and for the last 50l. how many were there, supposing the common difference of the prices to be 4l.
48 + 1 = ++ 1 = 12 + 1 = 13 = the number required. 15. The ages of a family are in arithmetical progression, the youngest is 5 years old, the eldest 27, and the common difference 2; required the number of persons?
3 + 1 = 11 + 1 = 12 = the answer. 16. The two extremes of an arithmetical progression are 27
and 38, and the common difference 1; required the number of
terms ? Ans. 12.
17. A bill was paid by instalments in arithmetical progression, the least payment was 5l. the greatest 291. and the common difference 4l. ; how many payments were made 2 Ans. 7.
18. Bought a lot of books, and paid ls. 6d. for the first, 5s. 8d. for the last, and each (beginning at the first) cost 5d. more than the preceding; how many books were there Answer, 11.
292. Given the number of terms, the sum of the terms, and the common difference, to find the least term.
Rule I. Divide the sum of the terms by the number of terms.
II. Subtract 1 from the number of terms, and multiply the remainder by half the common difference.
III. From the quotient (found above) subtract this product, and the remainder will be the least term.
19. The sum of an arithmetical progression, consisting of 11 terms, is 154, and the common difference 2; required the least
term 2 154
II–– 14 the quotient of the sum, by the number of
— 2 And 11–1 x + = 10 = the number of terms minus 1,
multiplied by half the common difference. Then 14–10 = 4, the least term, as was required. 20. The sum of the terms 366, the number of terms 12, and the common difference 5, of an arithmetical progression being given, required the least term 2
21. The sum of the ages of 9 persons is 162, and the common difference 3 years; required the age of the youngest ?
22. The sum of 6 numbers in arithmetical progression is 108,
and the common difference 4; required the least term 2 Answer, 8. 23. Seven poor persons received among them 63 shillings, their shares were in arithmetical progression, the common difference being 2; required the least share 2 Ans. 3 shillings.
293. Given the least term, the number of terms, and the common difference, to find the greatest term. Rule. Multiply the number of terms by the common difference, to the product add the least term, and from this sum subtract the common difference; the remainder will be the greatest term.
24. In an arithmetical series of 10 terms, the least term is 8, and the common difference 3; required the greatest term Thus 10 x 3 + 8 – 3 = 30 + 8 – 3 = 38–3 = 35 = the greatest term.
25. The least term of an arithmetical series is 2; ; the common difference 44; and the number of terms 16 : required the greatest term 2 Thus 16 × 4 + 2} – 44 = 70 = the greatest term. 26. A man has 12 children, the youngest is three quarters of a year old, and each was born when the preceding was fifteen months old; required the age of the eldest ? . . . Thus 12 x 14 + 4 – 14 = 144, the answer. 27. The least term is 3, the common difference 2, and the number of terms 7 ; to find the greatest term 2 Ans. 15. 28. The age of the youngest of 9 persons is 6, and the common difference 3; required the age of the eldest? Ans. 30. 29. The least term is 24, the common difference 34, and the number of terms 20, being given to find the greatest term 2 Ans. 64.
294. PR omiscuous ExAMPLEs for PRACTICE. 1. How many times does the hammer of a clock strike in 12 hours ? Ans. 78. 2. The clocks at Venice go to 24 o'clock; how many times does the hammer of one of them strike the bell in that space of time 2 Ans. 300, 3. If 1000 stones be placed in a straight hine, the first a yard distant from a basket, and the rest in succession, each a yard distant from the preceding; what length of ground must a man go over, to pick up the stones one by one, and return with them singly to the basket? Ans. 560 miles, 1540 yds. 4. Bought 9 books, the prices in arithmetical progression, that of the least being 3 shillings, and that of the best 19; what sum did I pay for the whole; and what is the common difference of the prices Ans. Paid 4l. 19s. Com. diff. 2s. 5. A man travelled 2 miles the first day, and 53 miles the last, and increased every day's journey three miles more than the preceding; how many days, and what distance, did he travel? Ans. 18 days, and travelled 495 miles. 6. There are 64 squares on a chess board; now if I lay half a crown on the first square, three shillings on the second, and so on, increasing successively by sixpence, how much will there be on the last square, and on the whole board 2 Ans. On the last square ll. 14s. On the whole board 58l. 8s. 7. A debt of 15l. is to be discharged at 12 payments, each succeeding payment to exceed the former by 4 shillings; what will the first and last payment be Ans. The first payment 3 shillings. The last 21.7s. 8. A poet, who had agreed with a bookseller to receive 40l. for every thousand verses he should write, set to work on new year's day, and composed 10 verses; next day he composed 12, and so on, increasing every day by 2; now allowing 70 days for sundays and other holidays, what sum would be due to him at the year's end?
295. A series of numbers is said to be in Geometrical Progression, when the terms successively increase by the constant multiplication by some number called the ratio, or decrease by constantly dividing by the same number or ratio".
h The fundamental property of a Geometrical Progression is this; namely, “The product of the two extreme terms is equal to the product of any two in“ termediate terms, equally distant from the extremes;” from this property the rest are derived, as will be shewn when we resume the subject in the Alge