For the geometric series # 54+36+...+128/27#, how do you determine the sum?

1 Answer
Noah G
Dec 17, 2016

#s_7 = 4118/27#

Explanation:

This is a geometric series with first term #54#, common difference #36/54 = 2/3# and #n = ?#.

Before finding the sum, we must find the number of terms, #n#.

#t_n = a xx r^(n - 1)#

#128/27 = 54 xx (2/3)^(n -1)#

#128/1458 = (2/3)^(n- 1)#

#64/729 = (2/3)^(n - 1)#

#2^6/3^6 = (2/3)^( n -1)#

#(2/3)^6 = (2/3)^(n - 1)#

We can eliminate the bases now since they're equivalent.

#6 = n - 1#

#n = 7#

The sum of n terms of a geometric series is given by #s_n = (a(1 - r^n))/(1 -r)#.

#s_7 = (54(1 - (2/3)^7))/(1- (2/3))#

#s_7 = (54(1 - 128/2187))/(1/3)#

#s_7 = (54(2059/2187))/(1/3)#

#s_7 = 3(54)(2059/2187)#

#s_7 = (333,558)/(2,187) = 4118/27 ~= 152.5#

Hopefully this helps!

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