In grade 8 we did work on number patterns. A number pattern is a list of numbers that follow a fixed rule so that one can predict or work out what the value will be of a number anywhere in the pattern. For example: 3; 5; 7; 9;…We can see easily that there is a constant difference of 2 between each of the numbers. In other words, they go up by two each time. Here’s another example:
15; 12; 9; 6;… What is the difference in this sequence? That’s right they come down by 3 each time so the difference is -3.
Remember we call each number in the sequence a term and we write it like this T1; T2 etc. meaning the first, second and so on terms. So, the tenth term would be T10.
We call the pattern a sequence
In grade 8 we learned that when there is a constant difference between terms we can use a formula to describe the rule about the sequence: Tn = d(n) + c where n is the number of the term e.g.first, tenth, 30th and so on; d is the constant difference and c is the number that must be added to or subtracted from d to obtain the first term.
Look at the two examples above:
3; 5; 7; 9;… Tn = 3, d = 2 so c = Tn – d = 3 – 2 = 1
Let’s see if that works in the formula to find the value of T3, shall we?
Tn = d(n) + c
T3 = 2(3) + 1
= 6 + 1 = 7 The third term is indeed 7 so it works!
Now you should be able to use this formula to easily solve problems with number sequences that have a constant difference.
Each doe the problems assigned to you and post your solutions in the comments box below. Then look at each others problems and solutions and either say why you agree with the answers or post a better solution. Complete this work on Monday 13 April.
Sinovuyo
Write down the next three numbers in the following sequences:
(a) 6; 9; 12; 15;…
(b) 1/2; 1; 3/2; 2;…
What is the constant difference in each sequence?
What is the rule for determining any term in each pattern?
What is the value of the 28th term in each pattern?
What term in the first sequence has the value 27?
Malakhiwe
Write down the next three numbers in the following sequences:
(a) -4; -8; -12; …
(b) 1; 2/3; 1/3; 0;…
What is the constant difference in each sequence?
What is the rule for determining any term in each pattern?
What is the value of the 28th term in each pattern?
What term in the first sequence has the value -32?
You may want to refresh your memories and look at the wiki and blog post for g8 mathematics
Okay Sinovuyo and Malakhiwe you were supposed to have posted your work here by yesterday. Where is it, guys? I’m waiting…
Mathematics revision
A-18,21,24
B-5\2; 3; 7\2
1-The constant in A is add 3
2-The constant difference in B is add half
3N for A
Half N for B
28th term for A = 84
28th term for B = 14
The 9th term in the first sequence has the value of 27
The rule in each case is: 3n +3 (remember T1 is 6 not 1 so you have to “make up” the difference between the first term and the constant difference)
If we used your rule (3n) then T1 would be 3(1) = 3 and not 6. so the rule is 3n + 3 then T1= 3(1) +3 = 6
In the second series T1 is 5/2 not 1 so the rule would have to be 1/2(n) + c. So what is c? (The difference between 1 and 5/2 i.e. 3/2)
So the rule is 1/2n + 3/2
NOW if we use the rule then…
T28 = 3(28) +3 = 87 (not 84 because remember we started with 6 not 1!
and 3(n) + 3 = 27
3n = 27-3 = 24
n =24/3 = 8 so it is the 8th term (not the 9th
Well done for posting SIno! We’ll wait for MK’s input then I’ll give feedback.
1
a)-16,-20,-24
b)-1/3,-2/3,-1
2).
a)You subtract 4 from the previous term
b)you subtract 1/3 from the previous term
3TN=d(n)+c
4.
a)=-112
b)=
5
a)=-128
The rule in each case is: 3n +3 (remember T1 is 6 not 1 so you have to “make up” the difference between the first term and the constant difference)
If we used your rule (3n) then T1 would be 3(1) = 3 and not 6. so the rule is 3n + 3 then T1= 3(1) +3 = 6
In the second series T1 is 5/2 not 1 so the rule would have to be 1/2(n) + c. So what is c? (The difference between 1 and 5/2 i.e. 3/2)
So the rule is 1/2n + 3/2
NOW if we use the rule then…
T28 = 3(28) +3 = 87 (not 84 because remember we started with 6 not 1!
and 3(n) + 3 = 27
3n = 27-3 = 24
n =24/3 = 8 so it is the 8th term (not the 9th
1
a)-16,-20,-24 (Correct)
b)-1/3,-2/3,-1 (Correct)
2).
a)You subtract 4 from the previous term (Correct d=-4
b)you subtract 1/3 from the previous term (Correct d=-1/3)
3TN=d(n)+c (Yes, but do you know what this means?)
4.
a)=-112 (T28= -4(28) + 0 [c= (-4) – (-4) = 0]
= -112+0
= -112 Correct
b)= T28= -1/3(28) + 4/3
= – 28/3 + 4/3
= -24/3
= -8
5
a)=-128 No – see below
Tn = -4(n) + 0 =-32
=> -4n = -32
=> n = -32/-4 = 8 So 8th term is -32
Okay MK and Sino, well done on your first post now you need to check out each other’s solutions and decide if you agree that they are correct and why or post the corrections you think are needed then I will follow up with some final feedback and further exercises. Please post your responses today.
Uhm sir can you please explain the work to me because what I did there were kind of rough calculations.