arithmetic sequence sum
- Thread starter Makito
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CLEARLY stated:
We have an arithmetic sequence.
Sum of first 2 terms = 13
4th term = -1
Go here :
Intro to arithmetic sequences | Algebra (video) | Khan Academy
Sal introduces arithmetic sequences and their main features, the initial term and the common difference. He gives various examples of such sequences, defined explicitly and recursively.
Come back if you have questions.
Although you never explicitly stated such in the body of your message, I can infer from your title that \(a_n\) is meant to be an arithmetic sequence. With that in mind, we can use what you (should) already know about arithmetic sequences: They have some starting value and then each successive term is some common difference more/less than the previous term.
We're given that \(a_4 = -1\) but we're not told a common difference. Hmm... well, the most important principle in all of algebra is that if we don't know a value, we can give that value a name so as to make talking about it and working with it easier. Let \(d\) be the common difference of the sequence \(a_n\). Can you see why that means we can say that \(a_4 = a_3 + d \implies -1 = a_3 + d\)? How can you rearrange that equation such that we have \(a_3 = (\text{something})\)? Using this result, what can you say about \(a_2\)? And then what can you say about \(a_1\)? Finally, what do you get when you plug these values into the first given equation that \(a_1 + a_2 = 13\)? Where does that lead you?
We're given that \(a_4 = -1\) but we're not told a common difference. Hmm... well, the most important principle in all of algebra is that if we don't know a value, we can give that value a name so as to make talking about it and working with it easier. Let \(d\) be the common difference of the sequence \(a_n\). Can you see why that means we can say that \(a_4 = a_3 + d \implies -1 = a_3 + d\)? How can you rearrange that equation such that we have \(a_3 = (\text{something})\)? Using this result, what can you say about \(a_2\)? And then what can you say about \(a_1\)? Finally, what do you get when you plug these values into the first given equation that \(a_1 + a_2 = 13\)? Where does that lead you?
Steven G
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a1 + a2 is 13 translates into (a1)+ (a1+d) = (2a1+d) = 13
Now a4 =a1 + 3d = -1
Now solve 2a1+d = 13 and a1 + 3d = -1 for a1 and d
Can you finish from here?
Last edited:
Steven G
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No no, only Subhotosh can recommend such videos. You can get into great trouble doing this. No need to worry as I won't report you.CLEARLY stated:
We have an arithmetic sequence.
Sum of first 2 terms = 13
4th term = -1
Go here :
Intro to arithmetic sequences | Algebra (video) | Khan Academy
Sal introduces arithmetic sequences and their main features, the initial term and the common difference. He gives various examples of such sequences, defined explicitly and recursively.www.khanacademy.org
Come back if you have questions.
BTW would you be interested in purchasing a bridge I have for sale, it's called the Brooklyn Bridge
HOKAY: a dollar down and a dollar a week;
I'll have my people contact your people...
pka
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People who need a calculator do not know. You Makito want to know.
Arithmetic series have only two essential parts: a first term and a common different.
Suppose that the first term is \(\displaystyle a_1\) and a common different is \(\displaystyle d\).
Then the sequence is \(\displaystyle a_n=a_1+ [(n-1)d]\)
Now the sum \(\displaystyle {S_n} = \sum\limits_{k = 1}^n {{a_1}(k - 1)d} = {a_1}d\sum\limits_{k = 1}^n {(k - 1)} = {a_1}d\frac{{(n - 1)n}}{2} \)
If \(\displaystyle a_1+a_2=13\) then \(\displaystyle a_1+(a_1+d)=13 \text{ or }2a_1+d=13\)
AND \(\displaystyle a_4=a_1+(4-1)d=-1\)
Now Show us what you can do with his help!