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26 Jul 2014, 08:27
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This topic here: in-the-sequence-x0-x1-x2-xn-each-term-from-x1-to-xk-126564.html
got me thinking about what does "n" actually mean? Before I was 100% sure that "n" denotes the rank that the element holds in a set. But in the above example the 11th element had an "n" of 10. Not 11, but 10. So no, "n" does not denote the rank that the element holds in a set. So what does "n" denote then?
So I decided to play around with numbers.
Example: imagine an AP with difference (d)=3 and the first term being 2
If the sequence begins with "n"=0
Rank order: first-second-third
"n": 0-1-2
Actual numbers: 2-5-8
If the sequence begins with n=1
Rank order: first-second-third
"n": 1-2-3
Actual numbers: 5-8-11
If the sequence begins with n=4
Rank order: first-second-third
"n": 4-5-6
Actual numbers: 14-17-20
So what I take away from this is:
The formula Last term - first term = (n-1) times difference is not correct. Proof:
Let's try it for the first sequence:
For third term: 8-2 = (2-1)*3 = doesn't work
Let's try it for the second sequence:
For third term: 11-5 = (3-1)*3 = WORKS
Let's try it for the third sequence:
For third term: 20-14 = (6-1)*3 = doesn't work
Ok so now let's change the formula to Last term - first term = (RANK-1) times difference
Then it works for all three, because in all three cases rank is the same = 3.
So to summarise my questions:
1. In a set, what exactly does "n" stand for?
2. Why do we need it in a set when we could have just used RANKS instead?
3. When a sequence exercise tells us n>1 or n>9 or n>0, why do they do it? What are we supposed to take away from this?
Hope this makes sense!
got me thinking about what does "n" actually mean? Before I was 100% sure that "n" denotes the rank that the element holds in a set. But in the above example the 11th element had an "n" of 10. Not 11, but 10. So no, "n" does not denote the rank that the element holds in a set. So what does "n" denote then?
So I decided to play around with numbers.
Example: imagine an AP with difference (d)=3 and the first term being 2
If the sequence begins with "n"=0
Rank order: first-second-third
"n": 0-1-2
Actual numbers: 2-5-8
If the sequence begins with n=1
Rank order: first-second-third
"n": 1-2-3
Actual numbers: 5-8-11
If the sequence begins with n=4
Rank order: first-second-third
"n": 4-5-6
Actual numbers: 14-17-20
So what I take away from this is:
The formula Last term - first term = (n-1) times difference is not correct. Proof:
Let's try it for the first sequence:
For third term: 8-2 = (2-1)*3 = doesn't work
Let's try it for the second sequence:
For third term: 11-5 = (3-1)*3 = WORKS
Let's try it for the third sequence:
For third term: 20-14 = (6-1)*3 = doesn't work
Ok so now let's change the formula to Last term - first term = (RANK-1) times difference
Then it works for all three, because in all three cases rank is the same = 3.
So to summarise my questions:
1. In a set, what exactly does "n" stand for?
2. Why do we need it in a set when we could have just used RANKS instead?
3. When a sequence exercise tells us n>1 or n>9 or n>0, why do they do it? What are we supposed to take away from this?
Hope this makes sense!
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Hello imoi,
The value of \(n\) is just a place holder for a number of the sequence. In the example your provided via your link, since the start of the sequence started when \(n=0\), then the \(n=10\) is the 11th term. From a sequence, it is important to know where it started or stopped to determine the value of n, i.e. if the first term was \(n=1\), then the final answer would have been \(n=11\).
I am not sure what you are trying to prove for with your example. In terms of \(\frac{(last term - first term)}{difference} = number of steps\) between the two values.
Example 1: \(\frac{(8-2)}{3}= 2\) -> two steps between the two values, if you knew that the first term is \(n=0\), then the 3rd term is \(n=2\). This would apply for all your examples.
Hope this helps. If not, can you please reword your statement so its a bit clearer?
The value of \(n\) is just a place holder for a number of the sequence. In the example your provided via your link, since the start of the sequence started when \(n=0\), then the \(n=10\) is the 11th term. From a sequence, it is important to know where it started or stopped to determine the value of n, i.e. if the first term was \(n=1\), then the final answer would have been \(n=11\).
I am not sure what you are trying to prove for with your example. In terms of \(\frac{(last term - first term)}{difference} = number of steps\) between the two values.
Example 1: \(\frac{(8-2)}{3}= 2\) -> two steps between the two values, if you knew that the first term is \(n=0\), then the 3rd term is \(n=2\). This would apply for all your examples.
Hope this helps. If not, can you please reword your statement so its a bit clearer?
26 Jul 2014, 09:03
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wunsun
Thanks a million for your reply. I was worried that I didn't communicate my issue too effectively.
I definitely understand what you are saying about knowing where the sequence starts. I just don't get the idea of the "zeroth" term - why does it even exist?
Also for some problems, such as this one:
the-sequence-a1-a2-a3-a4-a5-is-such-that-an-a-n-1-5-for-166830.html
we are given more severe constrains (above or equal to 2 and below or equal to 5). How does that impact my solution to this problem? It doesn't really, does it? So what's the point of writing that?
Sorry if I am retarded haha
