# GCSE Maths nth TermWatch

This discussion is closed.
#1
Help! I’m just completing an assignment and I’m stuck at the last hurdle!

I’ve got the following patterns:

n 1 2 3 4
nth?
p-nth
p 11 33 66 110
-22 -33 -44
-11 -11

And:

N 1 2 3 4
Nth?
p-nth
p 9 27 54 90
-18 -27 -36
-9 -9

I would normally be OK if the last lines of differentials were even numbers. (e.g. in the first problem above had the last line been 10 I assume the nth would be 5n2 + OR – whatever p – nth works out at. But what do you do when you have odd numbers like 11 and 9 above???

Hope someone can help.
Cheers
Eka
0
14 years ago
#2
1.
n 1 2 3 4
p 11 33 66 110

Divide the second row by 11:
1 3 6 10

These are the triangular numbers (1, 1+2, 1+2+3, etc), and the nth term is n(n+1)/2. So the nth term in the original sequence is 11n(n+1)/2. [Check: n=4 -> 11n(n+1)/2 = 11*4*5/2 = 11*2*5 = 110.]

2.
N 1 2 3 4
p 9 27 54 90

Divide the second row by 9:
1 3 6 10

The nth term in the original sequence is 9n(n+1)/2. [Check: n=3 -> 9n(n+1)/2 = 9*3*4/2 = 9*3*2 = 54.]
0
#3
(Original post by Jonny W)
1.
n 1 2 3 4
p 11 33 66 110

Divide the second row by 11:
1 3 6 10

These are the triangular numbers (1, 1+2, 1+2+3, etc), and the nth term is n(n+1)/2. So the nth term in the original sequence is 11n(n+1)/2. [Check: n=4 -> 11n(n+1)/2 = 11*4*5/2 = 11*2*5 = 110.]

2.
N 1 2 3 4
p 9 27 54 90

Divide the second row by 9:
1 3 6 10

The nth term in the original sequence is 9n(n+1)/2. [Check: n=3 -> 9n(n+1)/2 = 9*3*4/2 = 9*3*2 = 54.]

Jonny W. Thanks for the explanation … all is now clear! Much appreciated.

All the best

Eka
0
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