CHAPTER VIII SERIES AND LIMITS ARITHMETICAL SERIES 167. The numbers 5, 9, 13, 17, 21 appear upon examination to have been selected according to some law and arranged in a definite order. Each number after the first is greater than the one immediately preceding by 4. Such a regulated succession of numbers is called series. When, as here, the increase is the same throughout, the series receives the special name of arithmetical series or arithmetical progression. An arithmetical series is a succession of numbers in which each number after the first minus the preceding one always gives the same difference. This difference is called the common difference. Instead of increasing, the numbers in the series may decrease, so that the first number is the largest and the common difference is negative. Arithmetical series are frequently encountered in the study of mathematics, hence it is desirable to develop certain formulas relating to such series. EXERCISES 168. State which of the following series are arithmetical series: THE LAST TERM 169. If a stands for the first number or term of an arithmetical series, and d for its common difference, then the series may be written in general terms thus, Let a, a +d, a + 2 d, a + 3d, a + 4 d, etc. The stand for the last term of the arithmetical series. value of the last term evidently depends upon three things; namely, the value of a, the value of d, and also the number of terms in the series. Denote the number of terms by n. It is to be observed that the coefficient of d in the second term is 1, in the third term is 2, in the fourth term is 3, in the fifth term is 4. If n denotes the number of terms, what must be the coefficient of d in the last term? From what we have observed it must be one less than the number of the term, that is, n-1. Hence we have the formula for the nth term, 170. 1. Find the 10th term in the arithmetical series, 2, 7, 12, 17, ... Hence the 10th term is 47, as may be verified by writing down all the terms to the 10th term. .... 2. Find the 15th term of the series -2, -4, -6, -8, .• •. 3. Find the 12th term of the series 1, 11, 2, 21, 3, 4. Find the 20th term of the series 5, 5 + 2 x, 5 + 4 x, 5+6x,.... 5. Find the 24th term of the series √2, √2 + 1, √2 + 2, √2 +3,.... .... 6. Find the (n - 1)th term of the series 3, 6, 9, 12, 15, Find the expression for any given term (the nth term) of the following series of numbers: 13. Find the (n − 2)th term of the series 5, 1, -3, −7, .... 14. A bullet is fired vertically upward so that at the end of the first second it has a velocity of 200 ft. per sec., at the end of the second second a velocity of 168 ft. per sec., at the end of the third second a velocity of 136 ft. per. sec., and so on. Compute the velocity at the end of the sixth second, at the end of the tenth second. Interpret the second answer. 15. A body falling from rest falls 16 ft. during the first second, 48 ft. during the second, 80 ft. during the third, 112 ft. during the fourth, and so on. How far will it fall during the ninth second? ARITHMETICAL MEANS 171. The arithmetical means between two numbers are numbers which, together with the two given numbers as first and last terms, form an arithmetical series. If the two given numbers are 5 and 50, then 14, 23, 32, 41 are four arithmetical means, because 5, 14, 23, 32, 41, 50 is an arithmetical series. EXERCISES 172. 1. Insert six arithmetical means between 7 and 63. The six terms to be found, and the given numbers 7 and 63, will make 8 terms. Hence the required series is 7, 15, 23, 31, 39, 47, 55, 63. 2. Insert 5 arithmetical means between 6 and 72. 3. Insert 4 arithmetical means between 7 and 23. 4. Insert 8 arithmetical means between 7 and 81. SUM OF AN ARITHMETICAL SERIES 173. The sum of the terms of an arithmetical series can always be found by writing down all the terms, and adding them. But, if the number of terms is great, this operation is quite laborious. We proceed to derive a formula by which the sum of a large number of terms may be computed with less labor. Observing that, if the last term is 7, the term immediately preceding may be written 7-d, the term before this 72 d, and so on, we may indicate the sum of the series thus, S=a+(a + d) + (a + 2d) + ... + (1 − 2 d) + (1 − d)+l. (1) Reverse the order of the terms in the right side of (1), S = 1 + (1 − a) + (1 − 2 d) + . . . +(a + 2d)+(a + d)+a. d) (2) +(a+1)+(a+1)+(a+1). (3) In (3) there are as many parentheses (a + ) as there are terms in the series; hence, M 2 S = n(a + 1), S 8 = 22 (a + 1). (B) |