Solving Arithmetic and Series Problems in IB Math Methods

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The discussion focuses on solving two arithmetic and series problems in IB Math Methods. The first problem involves finding the first term of an arithmetic sequence with a common difference of 2 and a sum of 120, leading to the conclusion that the first term is 8, derived from the equation Sn = 120. The second problem concerns the series defined by 4 + 10 + 16 + 22 + ... + (6n - 2), with confusion about whether the series is finite or infinite, and how to express the sum as a function of n. Participants clarify that -7.5 is not a valid solution for the number of terms, as it cannot represent a negative count. The conversation emphasizes the importance of correctly interpreting the series and applying mathematical principles to derive valid solutions.
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I've been working on these problems for a long time trying to prove to myself that I'm not stupid (though I am convinced I am because I don't feel like "IB material" at the moment), but this will be impossible since I have other things to do. So, my two questions are these:

1. An arithmetic sequence has a common difference of 2 and a sum of 120. The first term is numerically equal to the number of terms. FInd all possible values of hte first.

I know that the answer is 8 because it is a multiple choice question and through elimination, the others aren't possible, but I don't understand how to prove this mathematically.

2. Determine Sn of the series 4 + 10 + 16 + 22+...+ (6n - 2).

I tried to solve this one, but when I did, the n's canceled out, making it "all real numbers" and since (6n - 2) is the last term, the series isn't infinite.

PLEASE HELP ME! Thanks!
 
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1) The series is: [sum] [ x + 2*y] from y = 0 to x-1

To solve an arithmetic series it is 1/2 times the size of the series times the quantity of the first term plus the second

Sn = (1/2) * n * ( a1 + a2 )

the first term you know is the number of terms in the series and the last term is equal to n+2*(n-1) based on the equation.

So..

Sn = (1/2) * n * ( n + [ n + 2*(n-1) ] )

Then you solve Sn = 120

And you get n = 8 and -7.5

2) I'm unsure what your asking

the series is
[sum] 6x-2 from x=1 to ?

your saying the series isn't infinite, so what is it?

if the series goes from x=1 to infinity then the series diverges
 
Originally posted by dduardo


Then you solve Sn = 120

And you get n = 8 and -7.5

2) I'm unsure what your asking

the series is
[sum] 6x-2 from x=1 to ?

your saying the series isn't infinite, so what is it?

if the series goes from x=1 to infinity then the series diverges

Thanks for the help on the first question. I was assuming the series was finite because (6x - 2) is the last term in the series. Another question: I don't understand how -7.5 can be relevant because it says that n is equal to the number of terms, but there can't be -7.5 terms... can there?
 
I think that what the question for is the sum as a function of n.

Sn = F(n)

What you would do is write out the summation, and then use the rules that you know to reduce it into a formula.

For example:

[sum]c from i=0 to n = cn (c is a constant
[sum]i from i=0 to n = 1/2(i+1)*i
 
Astronomer107: -7.5 can occur in a quasi-hyperdimesion manifold, that is forumalated by using the...

No, you are right, -7.5 isn't an answer. (Only in my imagination)

Also, what are the choices for question 2.

Is there an answer: n*(3*n+1)
 
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