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What is the sum of an 8-term geometric series if the first term is 14 and the last term is -3,919,104?
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well you need to find the common ratio. to do this you need the information for the 8th term a term in a geometric series is \[a_{n} = a \times r^{n -1}\] so you know \[-3919104 = 14 \times r ^7\] solve for r. then the sum of a geometric series is \[S_{8} = \frac{14( 1 - r^8)}{1 - r}\] hope this helps.
−3919104=14×r^7 or r^7 = -279936 or r^7 = -(2*3)^7 hence r=-6 use as campbell_st has formulated
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