GCSE - Arithmetic Sequence

it is barely conceivable that this is a GCSE question

smh

anyhow....

you can find a rule in the usual way;

the gap between the first two terms is x - 1

so (x - 1) *N + ?

letting N = 1 we can see that ? must be 2

so the n^th term is (x - 1)*N + 2

see what happens if you let 4x² - 3 equal (x - 1)*N + 2

Reply 2

OP

Original post by the bear

it is barely conceivable that this is a GCSE question

smh

anyhow....

you can find a rule in the usual way;

the gap between the first two terms is x - 1

so (x - 1) *N + ?

letting N = 1 we can see that ? must be 2

so the n^th term is (x - 1)*N + 2

see what happens if you let 4x² - 3 equal (x - 1)*N + 2

Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.

Reply 3

Original post by irrelevant kid

Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.

you would need to show a bit more detail of "4x^2-5 cannot be factorised to give an (x-1) bracket", but your idea is very good.

Reply 4

OP

Original post by the bear

you would need to show a bit more detail of "4x^2-5 cannot be factorised to give an (x-1) bracket", but your idea is very good.

Okay, thanks a lot! :smile:

Reply 5

Original post by irrelevant kid

Okay, thanks a lot! :smile:

are you sure it was a GCSE question ? :redface:

Reply 6

OP

Original post by the bear

are you sure it was a GCSE question ? :redface:

Yeah, my teacher gave me a booklet of grade 9+ questions today and this was one of them. Most of the questions weren't too bad, but this one took me a while.

Reply 7

just realized that you can find x quite easily... there is only one suitable value

:colondollar:

Reply 8

Original post by the bear

just realized that you can find x quite easily... there is only one suitable value

:colondollar:

The question is wanting to work out the value of x first. Then once obtained the value of x it wants you to substitute that value in to form a sequence which you must work out the nth term for. Now you must work out the value of 4x²−3 and show that it isn’t in the pattern of the nth term

(edited 6 years ago)

Reply 9

Original post by irrelevant kid

Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.

This is incorrect

Reply 10

Original post by Y11_Maths

This is incorrect

it is correct, but not the best method.

Reply 11

Original post by the bear

it is correct, but not the best method.

You won’t obtain any marks for this method though because it is not what the question is asking for

Reply 12

Original post by Y11_Maths

You won’t obtain any marks for this method though because it is not what the question is asking for

the question does not specifically ask you to find x.

Reply 13

Original post by the bear

the question does not specifically ask you to find x.

It is a 6 mark question so I would edge my bets. I posted how the question is meant to be answered above and I have answered this question before and with all the steps it takes it definitely wants you to find x and solve.

Reply 14

Original post by Y11_Maths

It is a 6 mark question so I would edge my bets. I posted how the question is meant to be answered above and I have answered this question before and with all the steps it takes it definitely wants you to find x and solve.

if you still think the other method does not work please explain why.

Reply 15

Original post by the bear

if you still think the other method does not work please explain why.

I didn’t say it doesn’t work I’m saying it is incorrect as you would not obtain any marks from it since that is not how the question is meant to be answered.

Reply 16

username2912782

13

Original post by irrelevant kid

Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.

Just because 4x^2-5 can't be factored to give an x-1 bracket, it doesn't mean that x-1 doesn't divide 4x^2-5.

For example consider x-1 and x+1. Clearly x+1 doesn't contain an x-1 bracket. But this doesn't necessarily mean x-1 isn't a factor.

For example, if x=3 then: x-1=2, x+1=4. Obviously 2 is a factor of 4.

Reply 17

Original post by psc---maths

Just because 4x^2-5 can't be factored to give an x-1 bracket, it doesn't mean that x-1 doesn't divide 4x^2-5.

For example consider x-1 and x+1. Clearly x+1 doesn't contain an x-1 bracket. But this doesn't necessarily mean x-1 isn't a factor.

For example, if x=3 then: x-1=2, x+1=4. Obviously 2 is a factor of 4.

Thank you

Reply 18

OP

Original post by Y11_Maths

This is incorrect

Is this right?

2x - (x+1)
Difference = x-1
5x-3-(x^2-2)
Difference = 5x-x^2-1

x-1=5x-x^2-1
x^2-4x=0
x(x-4)=0
x=4

x+1=5
2x=8
2(2x+3)/6-x=11
x^2-2=14
5x-3=17
nth term= 3n+2

4x^2-3 = 61
3n+2=61
3n=59
59/3 is not an integer so 4x^2-3 is not a term in the sequence.

Reply 19