The sum of the first and third terms of a geometric sequence is 40 while the sum of its second and fourth terms is 96. How do you find the sixth term of the sequence?

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Answer 1 · 2016-01-15T19:20:04

Solve to find the initial term #a#, common ratio #r# and hence:

#a_6 = ar^5 = 1990656/4225#

The general term of a geometric sequence is:

#a_n = a r^(n-1)#

where #a# is the initial term and #r# is the common ratio.

We are given:

#40 = a_1 + a_3 = a + ar^2 = a(1+r^2)#

#96 = a_2 + a_4 = ar + ar^3 = ar(1+r^2)#

So:

#r = (ar(1+r^2))/(a(1+r^2)) = 96/40 = 12/5#

#a = 40/(1+r^2)#

#= 40/(1+(12/5)^2)#

#=40/(1+144/25)#

#=40/(169/25)#

#=(40*25)/169#

#=1000/169#

Then:

#a_6 = ar^5#

#= 1000/169*(12/5)^5#

#=(2^3*5^3*2^10*3^5)/(13^2*5^5)#

#=(2^13*3^5)/(5^2*13^2)#

#=1990656/4225#