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The sum of a geometric series is -2044. If the first term is -4 and the common ratio is 2, what is the final term in the sequence?
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Sn = a[r^n-1]/r-1
use this and find n
Wait @Yahoo!
then an = a*r^(n-1)
Why @waterineyes any mistake
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Sorry my mistake... Ha ha ha..
\[-2044 = -4[2^n -1]/(2-1)\]
511 = 2^n - 1 510 = 2^n
find n??
\(-2044 = -4. \frac{2^{n}-1}{2-1}\) \(-2044 = -4. (2^{n}-1)\) \(511 = 2^{n}-1\) \(512 = 2^{n}\) \(2^9 = 2^{n}\) 9=n
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512
sorry 512 = 2^n
now n = 9
subs in an = a * r^(n-1)
\[\large 2^n = 512 \implies 2^n = 2^9 \implies \color{green}{n = 9}\]
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an = -4 * 2^8
you can replace n here by 9 @Yahoo!
256 * -4 = -1024
lol thanks!
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