katiebugg
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^n1]/r1

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use this and find n

waterineyes
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Wait @Yahoo!

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then an = a*r^(n1)

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Why @waterineyes any mistake

waterineyes
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Sorry my mistake...
Ha ha ha..

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\[2044 = 4[2^n 1]/(21)\]

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511 = 2^n  1
510 = 2^n

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find n??

rsadhvika
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\(2044 = 4. \frac{2^{n}1}{21}\)
\(2044 = 4. (2^{n}1)\)
\(511 = 2^{n}1\)
\(512 = 2^{n}\)
\(2^9 = 2^{n}\)
9=n

waterineyes
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512

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sorry 512 = 2^n

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now n = 9

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subs in an = a * r^(n1)

waterineyes
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\[\large 2^n = 512 \implies 2^n = 2^9 \implies \color{green}{n = 9}\]

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an = 4 * 2^8

waterineyes
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you can replace n here by 9 @Yahoo!

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256 * 4 = 1024

katiebugg
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lol thanks!

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Welcome