Number patterns with a constant difference – Gr 9

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You have shown real understanding of how number patterns work. Let us recap what we know:
1. Numbers following each other often have a regular pattern and we call these number series or number patterns.
2. The numbers in a series are called terms and are written as T1; T2 and so on and in general as Tn.
3. One type of pattern which occurs often is one where there is a constant difference between terms.
4. The general rule for patterns with a constant difference is Tn = d(n) + c

HOW TO USE THE GENERAL RULE FOR PATTERNS WITH CONSTANT DIFFERENCES
a. Determine the constant difference (d)
b. Subtract d from the first term to work out the value of c

Here is an example:
-5; -2; 1; 4; …
a. The difference between each term is 3 so d=3
b. -5 -3 = -8 so c=-8
Therefore the rule is Tn = 3(n) -8

Now to work out the value of a particular term e.g. what is the 60th term of this pattern?
T60 = 3(60) -8 = 120 -8 =

    112

Or to work out which term has a given value e.g. what term is equal to 55?
3(n) – 8 = 55
3(n) = 55 + 8
n = 63/3
n = 21 (The 21st term is 55)

WHAT YOU HAVE TO DO NOW
Post your solutions for the following problems tomorrow (Friday 23 April)
(i) State the constant difference in each pattern and the rule for each sequence.
(ii) Calculate the 45th term for each pattern
(iii) In the first sequence which term is -73?
(a) -2; -5; -8; …
(b) 2; 11/4; 7/2;…
(c) 15; 10; 5;…

WHAT COMES NEXT?
Coming shortly – Number patterns with a constant ratio – Watch this page!

3 thoughts on “Number patterns with a constant difference – Gr 9”

  1. i.a)d=3
    -2-3=-5 so c=-5
    Tn=3(n)-5
    B)d=3/4
    2-3/4=5/4 so c =5/4
    Tn=3/4(n)5/4
    C)d=5
    15-5=10 so c =10
    Tn5(n)10
    ii)a)T45=-3(45)-5=130-5
    =123
    b)T45=3/4(45)-5/4=-37/30-5/4
    =-1/60
    c)T45=5(45)-10=215-10
    =205
    iii)

    • (i) a) d = -3 (not 3) watch the sign to get from -2 to -5 you subtract 3
      b) d= 3/4 Correct
      c) d= -5 Again from 15 to 10 you SUBTRACT 5 so it is -5 not 5!
      (ii) T45 = d(45) + c so
      (a) T45 = -3(45) + (-1) [from -2 to -3 you subtract 1]
      = -135 -1
      = -136
      (b) T45 = 3/4(45) + 5/4 [from 3/4 to 2 = from 3/4 to 8/4 you ADD
      5/4]
      = 135/4 +5/4
      = 140/4
      = 35
      (c) T45 = -5(45) – 20 [from 15 to -5 you SUBTRACT 20]
      = -225 -20
      = -245
      (iii) -3(n) -1 = -73
      => -3(n) = -73 +1
      => n = -72/3
      = 24 (24th term)

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