Understanding Nth Term in Number Sequences

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The discussion focuses on finding the Nth term of a number sequence, specifically 8, 11, 14. The Nth term is calculated as 3N + 5, which correctly gives 17 for the 4th term. The formula a + (n-1)d is also applied, where 'a' is the first term and 'd' is the common difference, yielding the same result. The relationship between the two formulas is clarified by explaining that the second formula adjusts the indexing so that the first term corresponds to n=1. Understanding this adjustment helps connect the two methods for calculating terms in a sequence.
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Hi all

I'm studying sequences and series, the problem I am having is this:-

Given the sequence 8 11 14

Find the Nth Term.

I have worked out the nth term to be 3N+5, so if I wanted to find the 4th term it would be 3*4+5 = 17.

The problem is that I have come across the following formula:-

a+(n-1)d where a = 1st term, N is the Nth term, d common difference.

If I plug in the values into the above formula I get:-
8+(4-1)*3 = 17

I get the same result?? the problem I am having is that I do not understand how the two are related?

Can someone explain??
 
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tomtomtom1 said:
a+(n-1)d where a = 1st term, N is the Nth term, d common difference.
Try expanding the term in parentheses and writing the result in the form A*n+B.
 
Our sequence is
a, a+d, a+2d, ...

which is equivalent to

a+0d, a+1d, a+2d, ...

So what we have is:
1st = a+0d
2nd = a+1d
3rd = a+2d
...
nth = a+(n-1)d

So in order to use a formula that's of this form, we need to make our 1st term (n=1) have a value of 0, n=2 have a value of 1... our nth value have a value of (n-1).

Hence a+(n-1)d does just this.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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