marcybaby
Find the general term an for the geometric sequence a1=-3 and a2=6?
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marcybaby
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\[a_{1}=-3 and a_{2}=6\]
RadEn
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use the formula :
an = a1 r^(n-1)
with a1 is given -3 and to get the ratio(r), defined r = a2/a1 = 6/-3 = -2
therefore,
an = a1 * r^(n-1)
an = -3 * (-2)^(n-1)
|dw:1367598379663:dw|
marcybaby
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so that's the answer? @RadEn I'm confused
nincompoop
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it is not the answer, rather hint as how to solve the problem.
marcybaby
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ahhh!! i'm so lost and confused!! @nincompoop
waterineyes
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Can you find here the common ratio "r" ??
marcybaby
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-2?
waterineyes
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Yep..
waterineyes
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So, general term can be found as:
\[\large \color{green}{a_n = a \cdot r^{n-1}}\]
waterineyes
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a = -3 and r = -2..
Plug in and find..
marcybaby
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6?
waterineyes
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How?
waterineyes
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Use that formula,
marcybaby
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-3(-2)^n-1?
waterineyes
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Yep, this is the general term you require...
marcybaby
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okay, but I am stuck here. I dont what is the next step
marcybaby
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i get 6^n-1?
waterineyes
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Hey, this is the required answer, you can't solve it further..
marcybaby
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so this cant get be simplified any more?
waterineyes
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No..
waterineyes
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\[\large a_n = -3\cdot (-2)^{n-1}\]
marcybaby
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okay... how come tho? 6^n-1 looks like an ideal answer to me.
waterineyes
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You cannot multiply \(-3\) with \((-2)^{n-1}\)
marcybaby
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ohhhhhhhhhhh!!!!!!!!!! duh!! now i get it! my brain hurts been studing all week.
waterineyes
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It is okay..
marcybaby
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thank you for your time and patience
waterineyes
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You are welcome dear..