SERIES OF COMMON DIFFERENCES. A Series of common Differences, usually termed Arithmetical Progression, is a series in which the difference of any two consecutive terms, taken in order, is the same; as, in the series The first term is 1, and the common difference is 1. This is an increasing series, as each successive term is larger than the previous one. is an increasing series, and the common difference is 2. is an increasing series, and the common difference is 3. is a decreasing series, with a common difference of 2. REM.-In an increasing series, the common difference is found by subtracting any term, except the last, from the following term. In a decreasing term it is found by subtracting any term, except the first, from the preceding term. PROBLEM. The sum of a series of equal differences is equal to one-half the product of the number of terms and the sum of the first and last terms. As, the sum of ten terms of the series, 1, 2, 3, etc., is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. 11, 11, 11, 11, 11, 11, 11, 11, 11, 11. Sum of all the terms of both series = 11 x 10 = 110, and sum of 1 series, படிம் In these series a represents the first term and d the common difference. COR.-Any term in an increasing series is equal to the first term plus the difference taken once less than the number of terms, and in a decreasing series any term is equal to the first term minus the common difference taken as often as the number of the term minus 1, and any term may be regarded as the last term. Therefore, the expression for the last term in the former is 1 = a + (n − 1) d. The expression for the last term in the latter is 1 = a – (n − 1) d. ( The last term is equal to the first plus or minus the product of the difference and the number of terms less one. EXAMPLES. 1. Find the ninth term in the series 1, 3, 5, 7, etc. FORMULA.-1 = a + (n − 1) d = 1 + (8 x 2) = 1 8 = 17 = ninth term. 2. Find the tenth term of the series 2, 4, 6, 8, etc. Ans. 20. 3. Find the the fifth term of 12, 10, 8, 6, etc. l= 12 4, Ans. 8 = = . 4. Find the sum of six terms of the series 1, 3, 5, 7, etc. 1 = 11; f 1) d = 15 – 12 = 3. S= (15 + 3) = 63. 6. Find the sum of 100 terms of the series 1, 3, 5, 7, 9, etc. S = 10000. y. Find the sum of a decreasing series, whose first term is 12 and the common difference 4. S = 150. 8. Insert three terms of a series between 3 and 4. , , 1. a 9. The first term of a series is 1 and the number of terms 23; what must be the common difference in order that the sum may be 1493 ? Com. diff. FORMULA 2. i = a + (n − 1) d. Subtract a from both members of Formula 2 and divide by (n − 1), I-a = (n − 1)d. 7 d= 1 = 1 Ans. a n The common difference is equal to the last term minus the first divided by the number of terms less one. REM.When terms are to be inserted, find the common difference and then apply it. 10. What is the sum of an increasing series whose first term is , the common difference 1, and the number of the terms 20 ? Sum = 105. 11. What is the sum of an increasing series whose first term is 2, common difference 3, and number of terms 12 ? S = 222. SERIES OF EQUAL RATIOS. 1 2 A Series of Equal Ratios, improperly called Geometrical Progression, is one in which the ratios of any two consecutive terms taken in the same order is . equal; as, 1, 2, 4, 8, 16, is an increasing series, and any term divided by the one immediately preceding it gives a quotient of 2. Observe the series, 1, 2, 4, 8, 16, 32. Write it literally, ag ar, ara, ars, ari, arb. a = 1st term, and r = ratio. Each term is equal to the product of the 1st term and the ratio raised to the power of the number of terms less FORMULA (1). nth term = am-1. 5 6 one. PROBLEM. (1) To find the sum of a series of equal ratios. S = a +ar+arl + ar3+ art +ars, and multiply both terms by r, Sr = ar+ar+ar+ar+ar+are. Subtract the 1st equation from the 2d, Sr - S = arh a = a (76 - 1). Factor the terms, (r — 1) S = a (p16 — 1). (2) |