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5. Required to find four arithmetical means between 7 and 37.
Ans. 13, 19, 25, 31. 6. The first term of an arithmetical series is 3, the number of terms 60, and the sum of the terms 3720; required the common difference, and the last term.
Ans. d=2,1=121. 7. What will be the sum of the series if 9 arithmetical means be inserted between 9 and 109 ?
8. If three arithmetical means be inserted between 1 and 1, what will be the common difference?
9. What debt can be discharged in a year by paying 1 cent the first day, 3 cents the second, 5 cents the third, and so on, increasing the payment each day by 2 cents ? Ans. 1332 dollars 25 cents.
10. A footman travels the first day 20 miles, 23 the second, 26 the third, and so on, increasing the distance each day 3 miles. How many days must he travel at this rate to go 438 miles ? Ans. 12. 11. Find the sum of n terms of the progression of 1, 2, 3, 4,
Ans. S n 12. Find the sum of n terms of the progression 1, 3, 5, 7,.....
Ans. S = n'. 13. The sum of the terms of an arithmetical series is 950, the common difference is 3, and the number of terms 25. What is the first term ?
Ans. 2. 14. A man bought a certain number of acres of land, paying for the first $}; for the second, $ž ; and so on.
When he came to settle he had to pay $3775. How many acres did he purchase ?
Ans. 150 acres. 15. The 14th, 131th, and last terms of an arithmetical
progres sion are 66, 666, and 6666, respectively. Required the number of terms.
THE TEN CASES.
354. Given any three of the quantities, a, I, n, d, S, to find the other two.
This problem will present ten cases, each giving rise to two formulas, making in all twenty different formulas, or four values for each letter. The results in each case may be obtained directly from the two fundamental equations, or those of any particular case may be derived from some preceding case, as is most convenient. The whole will be left as an exercise for the student.
PROBLEMS IN ARITHMETICAL PROGRESSION
10 WHICH THE FORMULAS DO NOT IMMEDIATELY APPLY.
355. When in the conditions of a problem no three of the five parts, a, l, n, d, S, are directly given, the general formulas will not directly apply. It is usually necessary in such instances to represent the several terms of the series by means of two or more unknown quantities; and for this purpose there are two methods of notation. 1st. Let x denote the first term and
y the common difference; thus,
2, (x+y), (x+2y), (+3y).... This method of notation, however, is seldom the most expedient.
2d. When the number of terms is odd, denote the middle term by x, and the common difference by y; then we shall have, for three terms,
(x-y), x, (x+y); for five terms,
(x-2y), (x-y), a, (x+y), (+2y). And when the number of terms is even, represent the two middle terms by x—y and x+y respectively, 2y being the common difference; thus,
(*—34), (2-y), (•+y), (x+3y). The advantage of the second method is, that the sum of all the terms, or the sum and difference of two ternus equidistant from the extremes, will each contain but a single unknown quantity.
'1. There are three numbers in arithmetical progression; the sum of these numbers is 18, and the sum of their squares is 158. What are the numbers ?
Ans. 1, 6, 11. 2. There are five numbers in arithmetical progression; their sum is 65, and the sum of their squares 1005. What are the numbers ?
Ans. 5, 9, 13, 17, 21. 3. It is required to find four numbers in arithmetical progression, such that their common difference shall be 4, and their continued product 176985.
Ans. 15, 19, 23, 27.
4. There are four numbers in arithmetical progression ; the sum of the extremes is 8, and the product of the means 15. What are the numbers ?
Ans. 1, 3, 5, 7. 5. A person starts from a certain place and goes 1 mile the first day, 2 the second, 3 the third, and so on; in six days after, another sets out from the same place in pursuit, and travels uniformly 15 miles a day. How many days after the second starts before they are together?
Ans. 3 days, and 14 days. NOTE.-Reconcile these two values.
6. A man has borrowed $60. What sum shall he pay daily to cancel the debt in 60 days; interest being allowed on the sum borrowed for the whole time, and on each payment from the time it is made to the end of the 60 days, at the rate of 10 per cent. for 12 months of 30 days each ?
Ans. $1 is. 7. There are four numbers in arithmetical progression ; the sum of the squares of the extremes is 65, and the sum of the squares of the means is 61. Required the numbers. Ans. 4, 5, 6, 7.
8. The sum of four numbers in arithmetical progression is 24, and their continued product is 945; what are the numbers ?
Ans. 3, 5, 7, 9. 9. A certain number consists of three digits, which are in arithmetical progression ; if the number be divided by the sum of its digits the quotient will be 26, and if 198 be added to the number its digits will be inverted. What is the number? Ans. 234.
10. From two towns which were 102 miles apart, two persons, A and B, set out to meet each other; A went 3 miles the first day, 5 the next, 7 the next, and so on; B went 4 miles the first day, 6 the next, 8 the next, and so on. In how many days did they meet ?
Ans. 6. 11. A quantity of corn is to be divided among persons, calculated to last a certain time if each of them receive a peck every week; during the distribution it is found that one person dies at the end of every week, and then the corn lasts twice as long as was expected. Required to find the quantity of corn.
Ans. 231 pecks.
356. A Geometrical Progression is a series of quantities, each of which is equal to the preceding one multiplied by a constant factor.
357. The constant factor is called the ratio ; and if the first term is positive, the progression will be an increasing, or a decreasing series, according as the ratio is greater or less, than unity. Thus,
2, 6, 18, 54, 162, is an increasing geometrical series, in which the ratio is 3; and
81, 27, 9, 3, 1, 3, .... is a decreasing geometrical series, in which the ratio is f.
358. When a geometrical progression is supposed to terminate, the first and last terms are called extremes, and all the terms between the first and last are called geometrical means.
359. To find the last term of a geometrical progression.
Let a denote the first term, r the ratio, 2 the last term, and n the number of terms. Then the series will be represented thus :
a, ar, ar, ars, ....l. Now we perceive that in any term the exponent of r is equal to the number of preceding terms. Hence, we shall have 1= ar-1.
(4) 360. To find the sum of the terms in a geometrical progression. Denote the sum of the series by S; then
S=a+ar+ara tars.. tarha, (1) and
ar+ar+ar... tarni tar". (2) Hence by subtraction, remembering that ar=rl,
rS-S=rla. Thus we obtain two expressions for S, as follows: