1 / 43

Arithmetic Series Problems Solved with Examples

Learn how to solve arithmetic series problems with step-by-step examples for finding terms, sums, and common differences. Explore various scenarios and formulas to master arithmetic series concepts.

eweyant
Download Presentation

Arithmetic Series Problems Solved with Examples

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1. The first term of an arithmetic series is − 1 and the second term is 5. (i) Find d, the common difference. –1, 5 d = Tn – Tn– 1 +6 = 5 – (–1) d = 6

  2. 1. The first term of an arithmetic series is − 1 and the second term is 5. (ii) Find T8, the eighth term of the series. Tn = a + (n – 1)d a = –1, d = 6 = –1 + (n – 1)6 = –1 + 6n – 6 Tn = 6n − 7 T8 = 6(8) – 7 = 48 – 7 T8 = 41

  3. 1. The first term of an arithmetic series is − 1 and the second term is 5. (iii) The kth term of the series is 113. Find k. Tn = 6n − 7 113 = 6k – 7 120 = 6k 20 = k The 20th term has a value of 113. Therefore, k = 20.

  4. 1. The first term of an arithmetic series is − 1 and the second term is 5. (iv) Find S18, the sum of the first 18 terms of the series. a = –1, d = 6, n = 18 = 9 [–2 + (17)6] = 9 [–2 + 102] = 9 [100] S18 = 900

  5. 2. In an arithmetic sequence, the first term is 3 and the fifth term is 19. Find: (i) d, the common difference T1 = 3 = aT5 = 19 Tn = a + (n – 1)d T5 = 3 + (5 – 1)d = 19 3 + 4d = 19 4d = 16 d = 4 The common difference is +4.

  6. 2. In an arithmetic sequence, the first term is 3 and the fifth term is 19. Find: (ii) T12, the 12th term of the sequence Tn = a + (n – 1)da = 3, d = 4, n = 12 T12 = 3 + (12 – 1)4 = 3 + (11)4 = 3 + 44 T12 = 47

  7. 2. In an arithmetic sequence, the first term is 3 and the fifth term is 19. Find: (iii) S25, the sum of the first 25 terms. a = 3, d = 4, n = 25 = 12·5 [6 + (24)4] = 12·5 [6 + 96] = 12·5 [102] S25 = 1275

  8. 3. The first term of an arithmetic series is 5 and the eighth term is 33. Find: (i) T3, the third term of the series Tn = a + (n – 1)d T1 = 5 = a T8 = 33 T3 = 5 + (3 – 1)4 Tn = a + (n – 1)d = 5 + (2)4 T8 = 5 + (8 – 1)d = 33 = 5 + 8 5 + 7d = 33 T3 = 13 7d = 28 d = 4

  9. 3. The first term of an arithmetic series is 5 and the eighth term is 33. Find: (ii) S28, the sum of the first 28 terms. a = 5, d = 4, n = 28 = 14 [10 + (27)4] = 14 [10 + 108] = 14 [118] S28 = 1,652

  10. 4. The fourth term of an arithmetic series is 2 and the 12th term is 34. (i) Find the first term, a, and the common difference, d. Tn = a + (n – 1)d T4 = a + (4 – 1)d 2 = a + 3d Tn = a + (n – 1)d T12 = a + (12 – 1)d 34 = a + 11d

  11. 4. The fourth term of an arithmetic series is 2 and the 12th term is 34. (i) Find the first term, a, and the common difference, d. Solve these equations simultaneously: a + 3d = 2 (× − 1) a + 11d = 34 – a – 3d = – 2 a + 3d = 2 a + 11d = 34 a + 3(4) = 2 8d = 32 a + 12 = 2 d = 4 a = – 10 The first term, a, is – 10. The common difference, d, is 4.

  12. 4. The fourth term of an arithmetic series is 2 and the 12th term is 34. (ii) Find the sum of the first 16 terms. a = – 10, d = 4, n = 16 = 8 [–20 + (15)4] = 8 [–20 + 60] = 8 [40] = 320

  13. 4. The fourth term of an arithmetic series is 2 and the 12th term is 34. (iii) For what values of n is the sum of the first n terms equal to 80? a = – 10, d = 4, Sn = 80 80 = n [2n – 12] 80 = 2n2 – 12n 0 = 2n2 – 12n – 80 (2)

  14. 4. The fourth term of an arithmetic series is 2 and the 12th term is 34. (iii) For what values of n is the sum of the first n terms equal to 80? 0 = n2 – 6n – 40 0 = (n – 10)(n + 4) n – 10 = 0 and n + 4 = 0 n = 10 and n = –4 If n = 10 or n = – 4 then Sn = 80. However, in the context of a question, n = – 4 is rejected as term numbers cannot be negative. Therefore, the first 10 patterns will sum to 80.

  15. 5. The following diagram shows the second and fourth patterns in a sequence. Each pattern is made from toothpicks. (i) Draw the first and third steps of the pattern. Pattern 3 Pattern 1

  16. 5. The following diagram shows the second and fourth patterns in a sequence. Each pattern is made from toothpicks. (ii) Find an expression for Tn, the nth term of the sequence. 3, 5, 7, 9 d = Tn – Tn– 1 Tn = a + (n – 1)d a = 3 = 5 – 3 = 3 + (n – 1)2 d = 2 = 3 + 2n – 2 Tn = 2n + 1

  17. 5. The following diagram shows the second and fourth patterns in a sequence. Each pattern is made from toothpicks. (iii) How many toothpicks will be needed to make the 22nd pattern? Tn = 2n + 1 T22 = 2(22) + 1 = 44 + 1 T22 = 45

  18. 5. The following diagram shows the second and fourth patterns in a sequence. Each pattern is made from toothpicks. (iv) Eamon wants to make the first 15 patterns of this sequence. How many toothpicks will he need in total? a = 3, d = 2, n = 15 = 7·5 [6 + (14)2] = 7·5 [6 + 28] S15 = 255 = 7·5 [34]

  19. 6. The sum of the firstn terms of an arithmetic series is given by Sn= n2 − 12n. (i) Use S1 and S2 to find the first term and the second term of the series. Sn = n2 – 12n S1 = 12 – 12(1) S2 = 22 – 12(2) = 1 – 12 = 4 – 24 S1 = – 11 S2 = – 20 T1 = S1 = – 11 T2 = S2 – S1 = –20 – (– 11) = –20 + 11 T2 = – 9

  20. 6. The sum of the firstn terms of an arithmetic series is given by Sn= n2 − 12n. (ii) Find d, the common difference. T1 = – 11, T2 = – 9 d = Tn – Tn– 1 = – 9 – (– 11) = – 9 + 11 d = 2

  21. 6. The sum of the firstn terms of an arithmetic series is given by Sn= n2 − 12n. (iii) Find an expression for Tn, the nth term of the series. Tn = a + (n – 1)da = – 11, d = 2 = – 11 + (n – 1)2 = – 11 + 2n – 2 Tn = 2n – 13

  22. 6. The sum of the firstn terms of an arithmetic series is given by Sn= n2 − 12n. (iv) Find the values of n for which Sn= 28. a = – 11, d = 2, Sn = 28 0 = (n – 14)(n + 2) n – 14 = 0 and n + 2 = 0 n = 14 and n = –2 28 = n [–12 + n] n = –2 is rejected as pattern numbers cannot be negative. 28 = –12n + n2  The first 14 patterns sum to 0. 0 = n2 – 12n – 28

  23. 6. The sum of the firstn terms of an arithmetic series is given by Sn= n2 − 12n. (v) After how many terms of the series will Snequal zero? a = − 11, d = 2, Sn = 0 0 = n (n – 12) n = 0 and n – 12 = 0 n = 12 0 = n [–12 + n] n = 0 is rejected as pattern numbers cannot be zero 0 = –12n + n2  The first 12 patterns sum to 0. 0 = n2 – 12n

  24. 7. The third term of an arithmetic series is 2 and the ninth term is − 16. (i) Find the first term, a, and the common difference, d. a + 2d = 2 (× – 1) Tn = a + (n – 1)d T3 = a + (3 – 1)d a + 8d = – 16 2 = a + 2d – a – 2d = – 2 a + 8d = – 16 T9 = a + (9 – 1)d 6d = – 18 –16 = a + 8d d = – 3

  25. 7. The third term of an arithmetic series is 2 and the ninth term is − 16. (i) Find the first term, a, and the common difference, d. Let d = – 3: a + 2d = 2 a + 2(– 3) = 2 a – 6 = 2 a = 8 The first term a, is 8. The common difference, d, is – 3.

  26. 7. The third term of an arithmetic series is 2 and the ninth term is − 16. (ii) Find the sum of the first 15 terms. a = 8, d = –3, n = 15 = 7·5 [16 + (14)(– 3)] = 7·5 [16 – 42] = 7·5 [–26] S15 = –195

  27. 7. The third term of an arithmetic series is 2 and the ninth term is − 16. (iii) For what value of n is the sum of the first n terms equal to − 55? a = 8, d = –3, Sn = –55 3n2 – 19n – 110 = 0 (3n + 11)(n – 10) = 0 –110 = n [19 – 3n] 3n + 11 = 0 and n – 10 = 0 –110 = 19n – 3n2 3n = – 11 and n = 10

  28. 7. The third term of an arithmetic series is 2 and the ninth term is − 16. (iii) For what value of n is the sum of the first n terms equal to − 55? is rejected as pattern numbers cannot be negative. n = 10 The first 10 patterns sum to – 55.

  29. 8. The second term of an arithmetic series is − 8 and the seventh term is 12. Find the first term, a, and the common difference, d. (i) Tn = a + (n – 1)d a + d = –8 (× –1) T2 = a + (2 – 1)d a + 6d = 12 −8 = a + d –a – d = 8 a + 6d = 12 T7 = a + (7 – 1)d 5d = 20 12 = a + 6d d = 4

  30. 8. The second term of an arithmetic series is − 8 and the seventh term is 12. Find the first term, a, and the common difference, d. (i) Let d = 4: a + d = –8 a + 4 = –8 a = –12 The first term, a, is – 12. The common difference, d, is 4.

  31. 8. The second term of an arithmetic series is − 8 and the seventh term is 12. Find the sum of the first 22 terms. (ii) a = –12, d = 4, n = 22 = 11[–24 + (21)4] = 11[–24 + 84] = 11[60] S22 = 660

  32. 8. The second term of an arithmetic series is − 8 and the seventh term is 12. For what value of n is the sum of the first n terms equal to zero? (iii) a = – 12, d = 4, Sn = 0 0 = n2 – 7n 0 = n(n – 7) n = 0 and n – 7 = 0 n = 7 0 = n [–14 + 2n] n = 0 is rejected as pattern numbers cannot be 0. 0 = –14n + 2n2 0 = 2n2 – 14n  The first 7 patterns sums to 0.

  33. 9. The sum of the first nterms of an arithmetic series is given by Sn= n(2n + 1). Use Snto find the value of the fourth term in the series. Sn = n(2n + 1) T4 = S4 – S3 S4 = 4(2(4) + 1) S3 = 3(2(3) + 1) = 4(8 + 1) = 3(6 + 1) = 4(9) = 3(7) S4 = 36 S3 = 21 T4 = S4 – S3 T4 = 36 − 21 T4 = 15

  34. 10. The sum of the firstn terms of an arithmetic series is given by Sn= 3n2 + 4n. Use Snto find the value of the 12th term in the series. Sn = 3n2 + 4n T12 = S12 – S11 S12 = 3(12)2 + 4(12) S11 = 3(11)2 + 4(11) = 3(144) + 48 = 3(121) + 44 = 432 + 48 = 363 + 44 S12 = 480 S11 = 407 T12 = S12 – S11 T12 = 480 – 407 T12 = 73

  35. 11. The sum of the first nterms of an arithmetic series is given by Sn= 5n2 − 3n. Use Snto find the value of the eighth term in the series. Sn = 5n2– 3n T8 = S8–S7 S8 = 5(8)2– 3(8) S7 = 5(7)2– 3(7) = 5(64) – 24 = 5(49) – 21 = 320 – 24 = 245 – 21 S8 = 296 S7 = 224 T8 = S8–S7 T8 = 296 – 224 T8 = 72

  36. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Draw the next pattern in the sequence. (i)

  37. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Copy the table on the right and write the number of tiles needed for each of the first five patterns. (ii)

  38. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Find, in terms of n, a formula that gives the number of tiles needed to make the nth pattern. (iii) Tn = a + (n– 1)d a = 4 Tn = a + (n– 1)d d = Tn–Tn – 1 Tn = 4 + (n– 1)4 = 8 – 4 = 4 + 4n– 4 d = 4 Tn = 4n

  39. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Using your formula, or otherwise, find the number of tiles in the 12th pattern. (iv) Tn = 4n n = 12 T12 = 4(12) T12 = 48

  40. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Scarlet has 480 tiles. What is the biggest pattern in the sequence that she can make? (v) Tn = 4n = 480 n = 120 The biggest pattern with 480 tiles is the 120th pattern.

  41. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Find, in terms of n, a formula for the total number of tiles in the first n patterns (Sn). (vi)

  42. 12. Scarlet is investigating the number of tiles needed to make patterns in a sequence. The first three patterns are shown below, and the sequence continues in the same way. Sn = 2n2 + 2n = 480 2n2 + 2n– 480 = 0 n2 + n– 240 = 0 (n + 16)(n– 15) = 0 Scarlet starts at the beginning of the sequence and makes as many patterns as she can. She does not break up the earlier patterns to make the new ones. For example, after making the first two patterns, she has used up 16 tiles (4 + 12). How many patterns can she make in total with her 480 tiles? (vii) n + 16 = 0 and n– 15 = 0 n = –16 and n = 15 n = – 16 is rejected as pattern numbers cannot be negative  With a total of 480 tiles, she can form the first 15 patterns.

More Related