What is the 8th term of the geometric sequence if #a_3 = 108# and #a_5 = 972#?

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Answer 1 · 2016-03-12T13:26:08

#26244#

In a geometric sequence is valid the following rule

#a_(i+1)=k*a_i#, where #i in NN#

So

#a_5=k*a_4=k*(k*a_3)#
#a_5=k^2*a_3#
#972=k^2*108# => #k^2=9# => #k=3#

By the same token

#a_8=k^(8-5)*a_5#
#a_8=k^3*a_5=3^3*972=27*972# => #a_8=26244#

Answer 2 · 2016-06-02T22:37:08

#T_8 = ar^7 = 12 xx 3^7 = 26 244#

Before we can find an unknown term of a geometric sequence, we need to know the first term #(a)# and the common ratio #(r)#

Each term can be written as #T_n = ar^(n-1)#

Let's divide the two terms we have been given, their formulae and their values:

#(T_5)/(T_3) = (ar^4)/(ar^2) = 972/108#

The following happens: #(cancelar^4)/(cancelar^2) = 972/108#

Subtract the indices: #a^2 = 9 " " rArr r = 3#

In #T_3 = ar^2, if r = 3,# then #a(3^2) = 108#

#9a = 108 " "rArr a = 12#

Great! now we have #a and r# so we can find any term we want.

#T_8 = ar^7 = 12 xx 3^7 = 26 244#