What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

1 Answer
George C.
Nov 17, 2015

#976560#

Explanation:

The common ratio is #root(7)(781250/10) = root(7)(78125) = 5#

#sum_(n=1)^8 10*5^(n-1)#

#=1/4(5-1)sum_(n=1)^8 10*5^(n-1)#

#=1/4(5 sum_(n=1)^8 10*5^(n-1) - sum_(n=1)^8 10*5^(n-1))#

#=1/4(sum_(n=2)^9 10*5^(n-1) - sum_(n=1)^8 10*5^(n-1))#

#=1/4(10*5^8 + color(red)(cancel(color(black)(sum_(n=2)^8 10*5^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^8 10*5^(n-1)))) - 10)#

#=1/4(10*5^8-10) = 5/2(5^8-1) = 5/2*390624 = 976560#

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