# hard gcse maths question (9-1) (Harder than A Level?) Watch

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I'm stuck on the last question of my homework paper and even my Chinese friend who's a maths genius can't do it:

22) Here are the first five terms of an arithmetic sequence.

7 13 19 25 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

This is what I have written down so far:

6n+1

(v is the closest i can get to a down arrow)

nv1 = 7 nv2 = 13

7^2 = 49

13^2 = 169

169 - 49 = 120

120/24 = 5

xvn+1 = xvn + 6

xvn+2 = xvn+1 + 6

(xvn+2 + 6)^2 - (xvn+1 + 6)^2 = 24m

(xvn+2)^2 + 36 - (xvn+1)^2 = 24m

(xvn+2)^2 - (xvn+1)^2 + 72 = 24m

[(xvn+2)^2 - (xvn+1)^2 + 72]/m = 24

Any help would really be appreciated.

Thank you!

22) Here are the first five terms of an arithmetic sequence.

7 13 19 25 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

This is what I have written down so far:

6n+1

(v is the closest i can get to a down arrow)

nv1 = 7 nv2 = 13

7^2 = 49

13^2 = 169

169 - 49 = 120

120/24 = 5

xvn+1 = xvn + 6

xvn+2 = xvn+1 + 6

(xvn+2 + 6)^2 - (xvn+1 + 6)^2 = 24m

(xvn+2)^2 + 36 - (xvn+1)^2 = 24m

(xvn+2)^2 - (xvn+1)^2 + 72 = 24m

[(xvn+2)^2 - (xvn+1)^2 + 72]/m = 24

Any help would really be appreciated.

Thank you!

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#2

(Original post by

I'm stuck on the last question of my homework paper and even my Chinese friend who's a maths genius can't do it:

22) Here are the first five terms of an arithmetic sequence.

7 13 19 25 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

**Flkeavy**)I'm stuck on the last question of my homework paper and even my Chinese friend who's a maths genius can't do it:

22) Here are the first five terms of an arithmetic sequence.

7 13 19 25 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

SInce the nth term is , this means that any terms in this sequence must have this form. So if we're considering any two terms in this sequence then you could represent them as and .

So the difference between the squares of these two numbers will be

Can you try to continue from here? If you're still stuck, please post all your working/thoughts.

EDIT : there’s a much quicker way to do this as explained later in the thread.

Last edited by Sir Cumference; 9 months ago

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#3

i got to 36nsquared + 12n - 36Msquared - 12m

how do u prove this is a multiple of 24

how do u prove this is a multiple of 24

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#4

(Original post by

i got to 36nsquared + 12n - 36Msquared - 12m

how do u prove this is a multiple of 24

**WarningPlanet02**)i got to 36nsquared + 12n - 36Msquared - 12m

how do u prove this is a multiple of 24

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#5

Thanks a lot worked it out but in an exam would never get the multiple of 12 part then prove its even.

Will we get this sort of question in the exam

Will we get this sort of question in the exam

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#6

(Original post by

Thanks a lot worked it out but in an exam would never get the multiple of 12 part then prove its even.

Will we get this sort of question in the exam

**WarningPlanet02**)Thanks a lot worked it out but in an exam would never get the multiple of 12 part then prove its even.

Will we get this sort of question in the exam

It's really tough though so if this was in the real thing it would probably be the hardest question in the 3 papers.

By the way, you could have dealt with by using difference of two squares, which saves you having to expand.

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#7

(Original post by

One term = 6n+1

Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m

Factorise and take 12 our

12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.

**danielwinstanley**)One term = 6n+1

Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m

Factorise and take 12 our

12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.

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#8

Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?

**Here are the first five terms of an arithmetic sequence.****7, 13, 19, 25, 31****Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.**From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?

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#9

**danielwinstanley**)

One term = 6n+1

Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m

Factorise and take 12 our

12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.

A number being a multiple of 12 does not always imply the number is a multiple of 24 - example; 12 is not a multiple of 24.

(Original post by

Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?

**notnek**)Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

**Here are the first five terms of an arithmetic sequence.****7, 13, 19, 25, 31****Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.**From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?

You get down to then you start to consider odd/even parities of to prove that you can always take out a factor of 2 in any case. So if are both even or odd, then is even thus a 2 can be taken out. If are of opposing parities then is odd but is even thus you can take out a factor of 2.

A mere expansion and factorisation won't prove it's always a multiple of 24 due to the reasons I outlined in the reply above - so consideration of parity of brackets would be required.

Of course, you can always say that but then you'd need to show the numerator is even somehow - and this requires the same approach with parities as above.

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#10

**notnek**)

Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

**Here are the first five terms of an arithmetic sequence.**

**7, 13, 19, 25, 31**

**Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.**

From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?

Once you get to 12(3n^2+n-3m^2-m), group terms as follows:

12((3n^2+n)-(3m^2+m)).

The 2 brackets are even, because if n is odd, let n=2k+1. We get 3n^2+n = 12k^2+14k+4 =

2(6k^2+7k+2). If n is even, let n=2k. We get 3n^2+n = 12k^2+2k = 2(6k^2+k)

So 3n^+n and 3m^2+m are even, so let 3n^2 = 2p and 3m^2+m = 2q. Their difference is 2(p-q), which is even.

Thus we have 12(3n^2+n-3m^2-m)= 12(2z)=24z, where z is an element of the set of integers.

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#11

**danielwinstanley**)

One term = 6n+1

Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m

Factorise and take 12 our

12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.

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#12

(Original post by

A number being a multiple of 24 implies the number is a multiple of 12.

A number being a multiple of 12 does not always imply the number is a multiple of 24 - example; 12 is not a multiple of 24.

I can't see it for GCSE level.

You get down to then you start to consider odd/even parities of to prove that you can always take out a factor of 2 in any case. So if are both even or odd, then is even thus a 2 can be taken out. If are of opposing parities then is odd but is even thus you can take out a factor of 2.

A mere expansion and factorisation won't prove it's always a multiple of 24 due to the reasons I outlines in the reply above - so consideration of parity of brackets would be required.

Of course, you can always say that but then you'd need to show the numerator is even somehow - and this requires the same approach with parities as above.

**RDKGames**)A number being a multiple of 24 implies the number is a multiple of 12.

A number being a multiple of 12 does not always imply the number is a multiple of 24 - example; 12 is not a multiple of 24.

I can't see it for GCSE level.

You get down to then you start to consider odd/even parities of to prove that you can always take out a factor of 2 in any case. So if are both even or odd, then is even thus a 2 can be taken out. If are of opposing parities then is odd but is even thus you can take out a factor of 2.

A mere expansion and factorisation won't prove it's always a multiple of 24 due to the reasons I outlines in the reply above - so consideration of parity of brackets would be required.

Of course, you can always say that but then you'd need to show the numerator is even somehow - and this requires the same approach with parities as above.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.

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#13

(Original post by

Thanks, yes the parity argument is what I ended up doing. I supose it isn't really beyond GCSE but the percentage of GCSE students that would even consider this method would be tiny.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.

**notnek**)Thanks, yes the parity argument is what I ended up doing. I supose it isn't really beyond GCSE but the percentage of GCSE students that would even consider this method would be tiny.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.

Specimen Set 1, Question 22 of the last paper, by the way,

Spoiler:

I don't see how the concluding statement is worth 1 mark??

Show

I don't see how the concluding statement is worth 1 mark??

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#14

**notnek**)

Thanks, yes the parity argument is what I ended up doing. I supose it isn't really beyond GCSE but the percentage of GCSE students that would even consider this method would be tiny.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.

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#15

Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics - Specimen Papers Set 1 - September 2015 © Pearson Education Limited 2015 166 Specimen Papers Set 1 - September 2015 © Pearson Education Limited 2015

Q22

Q22

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#16

(Original post by

Would a proof considering quadratic residues mod 24 be accepted at GCSE?

**A02**)Would a proof considering quadratic residues mod 24 be accepted at GCSE?

But it isn't taught at GCSE or A Level.

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#17

(Original post by

Would a proof considering quadratic residues mod 24 be accepted at GCSE?

**A02**)Would a proof considering quadratic residues mod 24 be accepted at GCSE?

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#18

(Original post by

Most candidates would likely prove that the proposition is true in one given case, true for consecutive terms, or skip the question entirely, so the number of full, complete solutions, would be ridiculously small.

Specimen Set 1, Question 22 of the last paper, by the way,

**_gcx**)Most candidates would likely prove that the proposition is true in one given case, true for consecutive terms, or skip the question entirely, so the number of full, complete solutions, would be ridiculously small.

Specimen Set 1, Question 22 of the last paper, by the way,

Spoiler:

I don't see how the concluding statement is worth 1 mark??

Show

I don't see how the concluding statement is worth 1 mark??

Proof is a grade 8/9 topic so to add in a requirement of an even/odd argument to complete the proof is just unfair. And as you said, 1 mark for it is ridiculous!

It's possible that the question was made and then only afterwards they realised that the solution was harder than they thought (but that's a very cynical view of Edexcel exam writers).

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#19

(Original post by

Thanks for that. I'm sure I looked through this paper twice but didn't see the question

Proof is a grade 8/9 topic so to add in a requirement of an even/odd argument to complete the proof is just unfair. And as you said, 1 mark for it is ridiculous!

It's possible that the question was made and then only afterwards they realised that the solution was harder than they thought (but that's a very cynical view of Edexcel exam writers).

**notnek**)Thanks for that. I'm sure I looked through this paper twice but didn't see the question

Proof is a grade 8/9 topic so to add in a requirement of an even/odd argument to complete the proof is just unfair. And as you said, 1 mark for it is ridiculous!

It's possible that the question was made and then only afterwards they realised that the solution was harder than they thought (but that's a very cynical view of Edexcel exam writers).

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#20

(Original post by

I can't imagine more than a handful of students nationwide getting full marks on a question like that. This question is probably more difficult than quite a few of the current further maths proof by induction to show a sequence is divisible by something style questions.

**A02**)I can't imagine more than a handful of students nationwide getting full marks on a question like that. This question is probably more difficult than quite a few of the current further maths proof by induction to show a sequence is divisible by something style questions.

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