Arithmetic Progression Problem (1 Viewer)

[kevinsta]

Find the nth term of the arithmetic sequence 7,14,21,28
Dosent tell me what term to find so i don't know what to do.

 

[photastic]

Umm, Tn=a+(n-1)d, from info, a=7,d=7
Hence, Tn=7+(n-1)7 =7n???

 

[Drongoski]

I think kevinsta has difficulty understanding what 'n-th term' means. Or am I wrong, kevin?

 

[kevinsta]

Umm, Tn=a+(n-1)d, from info, a=7,d=7
Hence, Tn=7+(n-1)7 =7n???

Really? just 7n? what does 7n mean

 

[braintic]

Really? just 7n? what does 7n mean

It means:
The 1st term is 7(1)=7
The 2nd term is 7(2)=14
The 3rd term is 7(3)=21
.
.
.
The nth term is 7(n)

(which you can confirm by looking at the question again)


In fact, this question should not require the use of any formulae - you should be able to just 'see' it.

 

[kevinsta]

I think kevinsta has difficulty understanding what 'n-th term' means. Or am I wrong, kevin?

yea i dont understand what the n-th term means.. someone told me it means anyterm but i still dont get it

 

[Squar3root]

yea i dont understand what the n-th term means.. someone told me it means anyterm but i still dont get it

the nth term is some arbitrary number in the set of N. So n can be 1,2,3,..., n-1, n

 

[GoldyOrNugget]

Squar3root's answer is confused.

Here, n is a pronumeral representing a positive integer. The nth term of the arithmetic progression is the term that comes in the nth position, expressed in terms of the pronumeral n. For example, the 1st term (i.e. the term when n=1) is 7. The 2nd term (n=2) is 14. We notice that a term at position n in the sequence is found by multiplying 7 by the value of n. So the nth term is said to be 7 times n or simply 7n.

 

[Squar3root]

Squar3root's answer is confused.

Here, n is a pronumeral representing a positive integer. The nth term of the arithmetic progression is the term that comes in the nth position, expressed in terms of the pronumeral n. For example, the 1st term (i.e. the term when n=1) is 7. The 2nd term (n=2) is 14. We notice that a term at position n in the sequence is found by multiplying 7 by the value of n. So the nth term is said to be 7 times n or simply 7n.

that's basically what I said

 

[GoldyOrNugget]

that's basically what I said

It's really not.

the nth term is some arbitrary number in the set of N

What is N? The only valid interpretation I have is that N the set of natural numbers, which is wrong. In general, the nth term of an arithmetic progression doesn't have to be a natural number or even an integer. If you don't mean N to be N the set of natural numbers, then what do you mean?

. So n can be 1,2,3,..., n-1, n

Two problems here. "n can be 1,2,3,..., n-1, n" is nonsensical. You can say "n can be 1,2,3,..., k-1,k" if k is independent of n, but you can't replace k with n. The second problem is that if N is the set of natural numbers as per the previous sentence, then there's no upper bound the value of n that can be chosen. It would be more correct to say "n can be 1,2,3,...", but this does not clarify the answer for OP.

 

[Squar3root]

It's really not.



What is N? The only valid interpretation I have is that N the set of natural numbers, which is wrong. In general, the nth term of an arithmetic progression doesn't have to be a natural number or even an integer. If you don't mean N to be N the set of natural numbers, then what do you mean?



Two problems here. "n can be 1,2,3,..., n-1, n" is nonsensical. You can say "n can be 1,2,3,..., k-1,k" if k is independent of n, but you can't replace k with n. The second problem is that if N is the set of natural numbers as per the previous sentence, then there's no upper bound the value of n that can be chosen. It would be more correct to say "n can be 1,2,3,...", but this does not clarify the answer for OP.

Sorry about the formatting. Usually I have nice signposting but I am on my phone

By N yes I do mean natural numbers. You said that n can be 'anything' but it really can't (actually it cannot be). You can't have the -7th term in a sequence much alike you can't have the sqrt(86)th term. So yeah n has to be a number over the natural field. RL example: count to ten, a normal person would say 1,2,3,....., 9,10 and not 1, 1.1, 1.12, Pi, sqrt(7), ..., 9.9,10

n is just some random letter I chose to be consistent with the OP, I could have chosen a,d,z whatever

 

[GoldyOrNugget]

I know that n has to be a natural number, but you didn't say that. You said "the nth term is some arbitrary number in the set of N". This is wrong. The nth term of an arithmetic progression can be negative, fractional, etc. as in the progression -1, -0.5, 0, 1.5. ...

The set of natural numbers is not a field.

There's no problem with the choice of the letter n. The problem is with the conflation of "nth term" and "n", as well as the suggestion that n is chosen from some set of integers with an upper bound. n can be any positive integer, so it's incorrect to give an upper bound on this set at all.

 

[Squar3root]

I know that n has to be a natural number, but you didn't say that. You said "the nth term is some arbitrary number in the set of N". This is wrong. The nth term of an arithmetic progression can be negative, fractional, etc. as in the progression -1, -0.5, 0, 1.5. ...

The set of natural numbers is not a field.

There's no problem with the choice of the letter n. The problem is with the conflation of "nth term" and "n", as well as the suggestion that n is chosen from some set of integers with an upper bound. n can be any positive integer, so it's incorrect to give an upper bound on this set at all.

"the nth term is some arbitrary number in the set of N" - ohh i see where I went wrong (stupid english). what i mean is that the nth term can be anything but the n in 7n can only be over the set of N

I thought N was a field like the complex field (C), real field (R) and rational field (Q). I remember wolfgang or one of the tutors saying "... the solution for x is (this) over the field of N"