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Arithmetic

Arithmetic. Sequences & Series jeff.bivin@lz95.org. Last Updated: October 11, 2005. Arithmetic Progression. n th term. 5, 8, 11, 14, 17, 20, … 3n+2, … -4, 1, 6, 11, 16, … 5n – 9, . . . 11, 7, 3, -1, -5, … -4n + 15, . . . n th term. Jeff Bivin -- LZHS.

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Arithmetic

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  1. Arithmetic Sequences & Series jeff.bivin@lz95.org Last Updated: October 11, 2005

  2. Arithmetic Progression nth term 5, 8, 11, 14, 17, 20, … 3n+2, … -4, 1, 6, 11, 16, … 5n – 9, . . . 11, 7, 3, -1, -5, … -4n + 15, . . . nth term Jeff Bivin -- LZHS

  3. nth term of arithmetic sequence Tn = a + d(n – 1) a = First term d = common difference n = number of terms. Common difference = the difference between two consecutive terms in a sequence. d = Tn – Tn-1 Jeff Bivin -- LZHS

  4. Find the nth term of the following AP. 8, 11, 14, 17, 20, … First term is 8 Common difference is 3 Tn = a + d(n – 1) Tn = 8 + 3(n – 1) Tn = 8 + 3n – 3 Tn = 3n + 5 Jeff Bivin -- LZHS

  5. Finding the nth term 6, 1, 8, 15, 22, … First term is -6 common difference is 7 Tn = a + d(n – 1) Tn = -6 + 7(n – 1) Tn = -6 + 7n – 7 Tn = 7n - 13 Jeff Bivin -- LZHS

  6. Finding the nth term 23, 19, 15, 11, 7, … First term is 23 common difference is -4 Tn = a + d(n – 1) Tn = 23 + -4(n – 1) Tn = 23 -4n + 4 Tn = -4n + 27 Jeff Bivin -- LZHS

  7. Finding the 956th term a1 = 156 d = -16 n = 956 156, 140, 124, 108, . . . Tn = a + d(n – 1) T956 = 156 + -16(956 – 1) T956 = 156 - 16(955) T956 = 156 - 15280 T956 = -15124 Jeff Bivin -- LZHS

  8. Finding the 100th term a = 5 d = 6 n = 100 5, 11, 17, 23, 29, . . . Tn = a + d(n – 1) T100 = 5 + 6(100 – 1) T100 = 5 + 6(99) T100 = 5 + 594 T100 = 599 Jeff Bivin -- LZHS

  9. Finding the number of terms in the AP a = 10 d = -2 Tn = -24 10, 8, 6, 4, 2, . . .-24 Tn = a + d(n – 1) -24 = 10 -2(n – 1) -34 = -2(n – 1) 17 = n-1 n = 18 Jeff Bivin -- LZHS

  10. The 5th term of an AP is 13 and the 13th term is -19. Find the first term & the common difference. T5 = a + 4d = 13……..(1) T13= a + 12d = -19……….(2) (2) – (1): 8d = -19 - 13 8d = - 32 d = -4 Substitute d = -4 into (1): a + 4(-4) = 13 a – 16 = 13 a = 29 Jeff Bivin -- LZHS

  11. Problem solving In a race, a swimmer takes 35 seconds to swim the first 100 m, 39 seconds to swim the second 100m and 43 seconds to swim the third 100 m. If he continues to swim in this manner, how long does he take to finish the 10th lap of 100m.

  12. Summing it up Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an Sn = an + (an - d) + (an - 2d) + …+ a1 Jeff Bivin -- LZHS

  13. 1 + 4 + 7 + 10 + 13 + 16 + 19 a1 = 1 an = 19 n = 7 Jeff Bivin -- LZHS

  14. 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 a1 = 4 an = 24 n = 11 Jeff Bivin -- LZHS

  15. Find the sum of the integers from 1 to 100 a1 = 1 an = 100 n = 100 Jeff Bivin -- LZHS

  16. Find the sum of the multiples of 3 between 9 and 1344 Sn = 9 + 12 + 15 + . . . + 1344 a1 = 9 an = 1344 d = 3 Jeff Bivin -- LZHS

  17. Find the sum of the multiples of 7 between 25 and 989 Sn = 28 + 35 + 42 + . . . + 987 a1 = 28 an = 987 d = 7 Jeff Bivin -- LZHS

  18. Evaluate Sn = 16 + 19 + 22 + . . . + 82 a1 = 16 an = 82 d = 3 n = 23 Jeff Bivin -- LZHS

  19. Evaluate Sn = -29 - 31 - 33 + . . . - 199 a1 = -29 an = -199 d = -2 n = 86 Jeff Bivin -- LZHS

  20. Find the sum of the multiples of 11 that are 4 digits in length Sn = 10 01+ 1012 + 1023 + ... + 9999 a1 = 1001 an = 9999 d = 11 Jeff Bivin -- LZHS

  21. Review -- Arithmetic Sum of n terms nth term Jeff Bivin -- LZHS

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