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 Post subject: Finding Postion of the Nth TermPosted: Thu Jan 07, 2010 2:38 pm

Joined: Sat Jan 19, 2008 4:27 pm
Posts: 156
The formula for finding the Nth term of a sequence is
N = dN + (a-d)

Anyone know what's the formula for finding the position in a sequence of a term within that sequence?

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 Post subject: Posted: Thu Jan 07, 2010 3:08 pm

Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1245
I don't understand the question... do you have any sub- or super-scripts missing?

But I would guess... re-arrange

gives

(N+(d-a))/d

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 Post subject: Finding Postion of the Nth TermPosted: Thu Jan 07, 2010 3:22 pm

Joined: Sat Jan 19, 2008 4:27 pm
Posts: 156

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 Post subject: Posted: Thu Jan 07, 2010 3:32 pm

Joined: Thu Nov 22, 2007 10:30 am
Posts: 531
Location: Warwickshire
For some reason, I feel the need to work out N:

N = dN + (a-d)

Take dN away from both sides:
N- dN = (a-d)

Which is the same as:
N(1-d) = (a-d)

Divide both sides by (1-d)
N = (a-d)/(1-d)

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 Post subject: Posted: Thu Jan 07, 2010 8:10 pm

Joined: Sat Jan 19, 2008 4:27 pm
Posts: 156
No this doesn't work - how do they expect 11 year old to do this!

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 Post subject: Posted: Thu Jan 07, 2010 9:20 pm

Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1245
I feel we don't have the whole question....

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 Post subject: Posted: Thu Jan 07, 2010 10:30 pm

Joined: Sun Aug 17, 2008 11:05 am
Posts: 3
The formula N = dN + (a-d) isnâ€™t really correct â€“

N on the left hand side is the value of the Nth term, and N on the right hand side is the term number, so two different letters are required. d is the common difference, and a is the value of the first term

For example, take the sequence 3, 5, 7, 9, 11, â€¦â€¦â€¦â€¦.

Then, if we let the value of the Nth term be y, we would have
y=dN +(a-d)
say we want to find the value of the 8th term in the sequence.
Then,
Y =2 x8 +(3-2)
=16 +1
=17

Now for the original question: the formula for finding the position in a sequence of a term within that sequence?

Say weâ€™re told that 21 is a number in the above sequence, but we need to find the position of it in that sequence; we can re-arrange the same formula â€“ we know the value of y (21) and we need to find N.
So,
21 = 2N +(3-2)
Subtract (3-2) from each side:
21 â€“(3-2) =2N
21-1 =2N
N=10 i.e the 10th term in the sequence is the number 21

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 Post subject: Posted: Thu Jan 07, 2010 11:31 pm

Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1245
heather wrote:
...

For example, take the sequence 3, 5, 7, 9, 11, â€¦â€¦â€¦â€¦.

...

Say weâ€™re told that 21 is a number in the above sequence, but we need to find the position of it in that sequence; we can re-arrange the same formula â€“ we know the value of y (21) and we need to find N.
So,
21 = 2N +(3-2)
Subtract (3-2) from each side:
21 â€“(3-2) =2N
21-1 =2N
N=10 i.e the 10th term in the sequence is the number 21

So I was right... If a=3, d=2, and term being considered is 21...

(N+(d-a))/d

(21+(2-3))/2

= 10

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 Post subject: Posted: Fri Jan 08, 2010 4:33 pm

Joined: Sat Jan 19, 2008 4:27 pm
Posts: 156
Thank you for this explaination.

Slump.. I could get your formula to work in simple sequences like the one here but not ones where the sequence started say at 37 - but then maybe that was my mistake

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