SECTION XII. ARITHMETICAL PROGRESSION. 267. AN Arithmetical Progression is a series of numbers which increase or decrease by a Common Difference. Thus, 1, 3, 5, 7, 9 is an increasing arithmetical progression in which the common difference is 2. PROGRESSIONS. And 10, 8, 6, 4, 2 is a decreasing arithmetical progression in which the common difference is 2. The numbers composing the series are called the Terms. 268. To find the last term of an arithmetical progression. WRITTEN EXERCISES. 1. Find the 11th term of the series 3, 5, 7, 9, 11, etc. The first term is 3. The second term equals 3 plus once the common difference. 1st 2d 3d 4th Hence 11th PROCESS. 3 3 + 1 x 2 = 3 + 2 x 2 3 + 3 x 2 = 3 + 10 x 2 Hence the eleventh term equals the first term plus ten times the common difference, which equals 3 + 10 × 2 23. From this problem we derive the = The third term equals 3 plus twice the common difference. 23. RULE. The last term equals the first term increased by the common difference multiplied by the number of terms less one. 2. Find the 12th term of the series 1, 3, 5, etc. 9. Find the 60th term of the series 216, 214, 212, etc. 10. Find the 12th term of the series, §, 1, etc. 11. If a body falls 16 ft. the first second, and in each succeeding second 32 feet more than in the preceding one, how far will it fall in the 25th second? 269. To find the sum of an arithmetical progression. WRITTEN EXERCISES. 1. Find the sum of 8 terms of the series 1, 5, 9, etc. Sum 1 + 5+ 9 +13 +17 + 21 + 25 + 29 = 2 sum 2 sum = = = The sum of the series is 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29. Writing the sum of the series in a reverse order—that is, 29 + 25 + 21 +17 +13 + 9 + 5 + 1—and adding, we have twice the sum of the series equal to 30 taken 8 times, or (1 + 29) 8. Hence the sum of the series equals 1 + 29 × 8, or 120. From this problein we derive the RULE. The sum of an arithmetical progression equals half the sum of the extremes multiplied by the number of terms. 2. Find the sum of 12 terms of the series 6, 9, 12, etc. PROCESS.-12th term 6 + 11 × 3 39. 6 + 39 Sum = = × 12 = = 270, Ans. Find the sum 3. Of 16 terms of the series 2, 4, 6, etc. 9. How many times does a common clock strike in a week? 10. A traveled 15 miles the first day, 18 miles the second, 21 miles the third, and so on for 36 days; how far did he travel? 11. A boy received $5 for January, $10 for February, $15 for March, and so on; how much did he receive for the year? 12. If a body falls 16 ft. the first second, 3 times as far the second second, 5 times as far the third second, and so on, how far will it fall in 25 seconds? 13. 100 apples are placed in a row 2 yd. apart, the first being 2 yd. from a basket: a boy starts to gather them singly into the basket; how far does he travel? GEOMETRICAL PROGRESSION. 270. A Geometrical Progression is a series of numbers which increase or decrease by a common Ratio. Thus, 1, 3, 9, 27, etc. is an increasing geometrical progression, in which the ratio is 3. And 32, 16, 8, 4, etc. is a decreasing geometrical progression, in which the ratio is . The numbers composing the series are called the Terms. 271. To find the last term of a geometrical progression. WRITTEN EXERCISES. 1. Find the 7th term of the series 2, 6, 18, etc. PROCESS. The first term is 2. The second term equals 2 times the first power of the ratio. 1st 2 2d 3d 4th = 2 x 3 x 32 2 x Hence 7th 2 x 36 = = = = = 6 18 54 1458 The third term equals 2 times the second power of the ratio. Hence the seventh term equals 2 times the sixth power of the ratio, 1458. which equals 2 × 36 RULE. The last term equals the first term multiplied by the ratio raised to a power denoted by the number of terms less one. 2. Find the 6th term of the series 3, 9, 27, etc. PROCESS.-Last term 3 x 36 2187, Ans. = = 3. Find the 10th term of the series 1, 2, 4, etc. 4. Find the 8th term of the series 5, 15, 45, etc. 5. Find the 7th term of the series 3, 12, 48, etc. 6. Find the 10th term of the series 1, 1, 1, etc. 7. Find the 10th term of the series 1,,, etc. 8. Find the 9th term of the series 2, 9. A man agreed to labor at the rate of $1 for January, $2 for February, $4 for March, and so on, and to accept for his year's services as much as would be due him under this agreement for December; how much would he receive? 1 10. A boy agreed to put one cent in his savings bank on New Year's day, and on each succeeding day of the month to double the previous day's deposit; if he had carried out his agreement, what would have been the last day's deposit? 272. To find the sum of a geometrical progression. WRITTEN EXERCISES. 1. Find the sum of 5 terms of the series 2, 6, 18, etc. PROCESS. Sum = = 2 1 Multiplying the sum of the series by 3, the ratio, and subtracting the series, we have (31) times the sum of the series equal to 486 – 2, and the sum of the series equal (486 — 2) divided by (3 − 1). Observing that 162 is the last term, 3 the ratio, and 2 the first term, we have the RULE. The sum of a geometrical progression equals the last term multiplied by the ratio, minus the first term, divided by the ratio less one. 2. Find the sum of 8 terms of the series 3, 6, 12, etc. PROCESS.-Last term 3 x 27 384. 384 × 2 = - = 2 Find the sum— 3. Of 1, 2, 4, etc. to 10 terms. 6. Of 5, 15, 45, etc. to 14 terms. 7. Of 10, 20, 40, etc. to 10 terms. = 242, Ans. = 765, Ans. years, 8. If the population of a city is doubled every 5 what is its population at the end of 50 years, if it begins with 5000 inhabitants? |