Number patterns

Advanced Math 7 · Unit 9 · sequences-patterns-7 · Teacher edition

sequences-patterns-7
Name ______________________________Date ______________
  1. 1.

    What is the 1010th term of the arithmetic sequence 4,11,18,25,4, 11, 18, 25, \ldots?

    Answer: ______________

  2. 2.

    Write a formula for the nnth term TnT_n of the sequence 7,12,17,22,7, 12, 17, 22, \ldots in terms of nn. Give it in the form an+ban + b.

    Answer: ______________

  3. 3.

    A pattern of dots grows as shown. Each figure is a triangle of dots with one more row than the figure before.

    Figure1234
    Dots361015

    How many dots are in Figure 88?

    Answer: ______________

  4. 4.

    Which formula gives the nnth term of 3,7,11,15,3, 7, 11, 15, \ldots?

    1. (A)

      Tn=4n+3T_n = 4n + 3

    2. (B)

      Tn=3n+4T_n = 3n + 4

    3. (C)

      Tn=n+4T_n = n + 4

    4. (D)

      Tn=4n1T_n = 4n - 1

  5. 5.

    How many terms of the arithmetic sequence 7,10,13,16,7, 10, 13, 16, \ldots are less than 100100?

    1. (A)

      3030

    2. (B)

      3131

    3. (C)

      3232

    4. (D)

      3333

    5. (E)

      3434

Answer key — Number patterns

  1. 1.
    67sequences-patterns-7-01

    T10=4+(101)×7=4+63=67T_{10} = 4 + (10 - 1) \times 7 = 4 + 63 = \mathbf{67}. (Adding dd ten times, 4+70=744 + 70 = 74, gives the 1111th term.)

  2. 2.
    5n+25n+2sequences-patterns-7-02

    Tn=7+5(n1)=7+5n5=5n+2T_n = 7 + 5(n - 1) = 7 + 5n - 5 = \mathbf{5n + 2}. Check: n=17n = 1 \to 7, n=212n = 2 \to 12, n=317n = 3 \to 17. ✓ (n+5n + 5 is not a formula for TnT_n; it only restates the step.)

  3. 3.
    45sequences-patterns-7-03

    Figure 88 is the 99th triangular number: 9×102=45\frac{9 \times 10}{2} = \mathbf{45}. By differences: 15+6=2115 + 6 = 21, +7=28+7 = 28, +8=36+8 = 36, +9=45+9 = 45. ✓

  4. 4.
    (D)

    Tn=4n1T_n = 4n - 1

    sequences-patterns-7-04

    d=4d = 4 gives Tn=4n+bT_n = 4n + b, and T1=3T_1 = 3 forces b=1b = -1. So Tn=4n1\mathbf{T_n = 4n - 1}: 3,7,11,153, 7, 11, 15. ✓ (n+4n + 4 gives 5,6,75, 6, 7 — the “add the difference” trap; 4n+34n + 3 uses a+dna + dn and gives 7,11,157, 11, 15, one term early.)

  5. 5.
    (B)

    3131

    sequences-patterns-7-05

    3n+4<100n<323n + 4 < 100 \Rightarrow n < 32, so n=1,2,,31n = 1, 2, \ldots, 31: 31\mathbf{31} terms. Check: T31=97<100T_{31} = 97 < 100 and T32=100T_{32} = 100, which is not less than 100100. (3232 counts the term equal to 100100.)