7 3. Find the 8th term of the series, 1, &c. Ans. 12. 3' 12' 4. Find the 30th term of the series -27,-20, -13, &c. 2 14. If a falling body descends 16 feet the first second, three times this distance the next, five times the next, and so on, how far will it fall the 30th second, and how far altogether in half a minute? Ans. 948, and 14475 ft. 15. Two hundred stones being placed on the ground in a straight line, at the distance of 2 feet from each other; how far will a person travel who shall bring them separately to a basket, which is placed 20 yards from the first stone, if he starts from the Ans. 19 miles, 4 fur., 640 ft. spot where the basket stands ? 16. Insert 3 arithmetical means between 2 and 14. To solve this problem generally, let it be required to insert m arithmetical means between a and b. Since the required terms, and those which are given, form an arithmetical series, if we insert m terms between a and b, we shall have a series consisting of (m+2) terms. Then to find d, the common difference, substitute m+2 instead of n in Formula fore, the common difference (d) will be equal to the difference of the extremes (b—a) divided by the number of terms plus one. In the particular example we find d=3; hence, the terms are 5, 8, and 11. 17. Insert 4AR. means between 3 and 18. Ans. 6, 9, 12, 15. 18. Insért 9AR. means between 1 and -1. Ans.,, &c., to -. 19. How many terms of the series 19, 17, 15, &c., amount to 91? Ans. 13, or 7. Explain this result. 20. How many terms of the series .034, .0344, .0348, &c., amount to 2.748? Ans. 60. 21. The sum of the first two terms of an arithmetical progression is 4, and the fifth term is 9; find the series. Ans. 1, 3, 5, 7, 9, &c. 22. The first two terms of an arithmetical progression being together 18, and the next three terms =12, how many terms must be taken to make 28? Ans. 4, or 7. 23. The nth term of an arithmetical series is (3n−1), find the first term, the common difference, and the sum of n terms. 24. In the series 1, 3, 5, &c., the sum of 2r terms: the sum of r terms::x:1; determine the value of x. Ans. 4. 25. Find the ratio of the latter half of 2n terms of any AR. series, to the sum of 3n terms of the same series. Ans. 3. 26. The sum of n arithmetical means between 1 and 19: sum of the first n-2 of them :: 5:3; required n. Ans. 8. 27. A traveler sets out for a certain place, and travels 1 mile the first day, 2 the second, and so on. In 5 days afterward another sets out, and travels 12 miles a day. How long and how for must he travel to overtake the first? Ans. 3 days or 10 days, and travel 36 miles, or 120 miles. Explain these results. GEOMETRICAL PROGRESSION. ART. 295. A Geometrical Progression is a series of terms each of which is derived from the preceding, by multiplying it by a constant quantity termed the ratio. Thus, 1, 2, 4, 8, 16, &c., is an increasing geometrical progression, whose common ratio is 2. 54, 18, 6, 2, &c., is a decreasing geometrical progression, whose common ratio is 1. In general, a, ar, ar2, ar3, &c., is a geometrical progression, whose common ratio is r, and which is an increasing series when r is greater than 1; but a decreasing series when r is less than 1. It is evident that in any given geometrical series, the common ratio will be found by dividing any term by the term next preceding. REMARK.— An Arithmetical Progression may be defined to be a series in which the difference between any two consecutive terms is the same; and a Geometrical Progression a series in which the ratio of any two consecutive terms is the same. Hence, a geometrical progression is a continued proportion. (Art. 266.) ART. 296. To find the last term of a geometrical progression. Let a denote the first term, r the common ratio, l the nth term, and S the sum of n terms; then the respective terms of the series That is, the exponent of r, in the second term is 1, in the third term 2, in the fourth term 3, and so on; hence, the nth term of the series will be larn-1; that is, any term of a geometrical series is equal to the product of the first term, by the ratio raised to a power, whose exponent is one less than the number of terms. Ex. Let it be required to find the 6th term of the geometrical progression whose first term is 7, and common ratio 2. 25=2×2×2×2×2=32; and 7×32=224, the 6th term. ART. 297. To find the sum of all the terms of a geometrical progression. If we multiply any geometrical series by the ratio, the product will be a new series, of which every term, except the last, will have a corresponding term in the first series. Thus, take the series, 1, 3, 9, 27, 81, and represent its sum by S; then S=1+3+9+27+81. Multiplying each term by the ratio 3, we have The terms of the two series are identical, except the first term of the first series, and the last term of the second series. If, then, we subtract the first equation from the second, all the other terms will disappear, and we shall have To generalize this method, let a, ar, ar2, ar3, &c., be any geometrical series, and S its sum; then, S=a+ar+ar2+ar3. . . . +arn-2+arn-1. Multiplying this equation by r, we have RULE, FOR FINDING THE SUM OF A GEOMETRICAL SERIES.Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less one. Ex. Find the sum of 6 terms of the progression 3, 12, 48. &c. The last term 3X45-3X1024=3072. lr_a_3072×4—3—4095. Ans r -1 4-1 ART. 298. If the ratio r is less than 1, the progression is decreasing, and the last term l, or arm-1, is less than a. In order that both terms of the fraction rl-a arna or may be positive, the signs of the terms may be changed, (Art. 124), and we have S= —rl 1-r a-arn Therefore, the sum of the series, when the progression is decreasing, is found by the same rule, as when it is increasing, except that the product of the last term by the ratio, is to be subtracted from the first term, and the ratio subtracted from unity. ART. 299. The formula S=" ,by separating the numera a-arn tor into two parts, may be placed under the form Now when is less than 1, it must be a proper fraction, which may be represented by ; then r"= ( 2 ) n Since Р is less than q, the higher the power to which the fraction is raised, the less will be the numerator compared with the denominator; that is, the less will be the value of the fraction; ph therefore, when n becomes very large, the value of org, will qn pr be very small; and when n becomes infinitely great, the value of , or rm, will be infinitely small; that is, 0. But, when the qn numerator of a fraction is zero (Art. 135) its value is 0. This reduces the value of S to α Hence, when the number of terms of a decreasing geometrical series is infinite, the last term is zero, and the sum is equal to the first term divided by one minus the ratio. Ex. Find the sum of the infinite series 1+1+1++, &c. That the sum of an infinite number of terms of a geometrical progression may be finite, will easily appear from the following illustration: Take a straight line, AK, and bisect it in B; bisect BK in C; CK in D, and so on continually; then will A B C D K AK-AB+BC+CD+, &c., in infinitum, =AB+AB+AB &c., in infinitum, =2AB, which agrees with the example. |