Proving a term is in a quadratic sequence
Hi, could anyone explain why n=-12,n=11 proves that 136 is the 11th term in the sequence? I can't find anything on this from the MathsGenie or MathsWatch videos and don't remember coming across it, so I'd appreciate an answer.
Taken from here:
https://www.mathsgenie.co.uk/resources/8-quadratic-sequencesans.pdf
Taken from here:
https://www.mathsgenie.co.uk/resources/8-quadratic-sequencesans.pdf
(edited 10 months ago)
Original post by The_Architect
Hi, could anyone explain why n=-12,n=11 proves that 136 is the 11th term in the sequence? I can't find anything on this from the MathsGenie or MathsWatch videos and don't remember coming across it, so I'd appreciate an answer.
Taken from here:
https://www.mathsgenie.co.uk/resources/8-quadratic-sequencesans.pdf
Taken from here:
https://www.mathsgenie.co.uk/resources/8-quadratic-sequencesans.pdf
In the first part of the question you show that the nth term of the sequence is given by the formula n^2 + n + 4. If 136 is a term in the sequence, then that formula must produce the value 136 for some value of n.
So you need to solve the equation n^2 +n + 4 = 136 which can be rearranged as a standard quadratic with 0 on the RHS. The equation has 2 integer solutions, one positive and one negative. It's usual to label terms in a sequence with positive integers, so you reject the negative solution for n.
Reply 2
Original post by davros
In the first part of the question you show that the nth term of the sequence is given by the formula n^2 + n + 4. If 136 is a term in the sequence, then that formula must produce the value 136 for some value of n.
So you need to solve the equation n^2 +n + 4 = 136 which can be rearranged as a standard quadratic with 0 on the RHS. The equation has 2 integer solutions, one positive and one negative. It's usual to label terms in a sequence with positive integers, so you reject the negative solution for n.
So you need to solve the equation n^2 +n + 4 = 136 which can be rearranged as a standard quadratic with 0 on the RHS. The equation has 2 integer solutions, one positive and one negative. It's usual to label terms in a sequence with positive integers, so you reject the negative solution for n.
So we can safely assume that the equation will always have only one positive?
Original post by The_Architect
So we can safely assume that the equation will always have only one positive?
If you're talking about the terms in a sequence, then you would normally label them 1st term, 2nd term, 3rd term, etc , so a negative term wouldn't make much sense. (You could technically define negative terms in a sequence but it's not normal terminology and I wouldn't expect it to come up in school exams. So if your 'n' is labelling a term in a sequence and you have to solve an equation for n you expect it to be positive.)
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