The second and the fifth terms of a geometric sequence are 15 and 405, respectively. Is 32,805 a term of this sequence?

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Answer 1 · 2015-11-15T14:52:54

Yes

#A_n = A_1r^(n - 1)#

#A_2 = A_1r^(2 - 1)#

#=> A_2 = A_1r = 15#

#A_5 = A_1r^(5 - 1)#

#=> A_5 = A_1r^4 = 405#

#A_1rr^3 = 405#

#=> 15r^3 = 405#

#=> r^3 = 27#

#=> r = 3#

#A_2 = 15 = A_1r#
#=> A_1*3 = 15#
#=> A_1 = 5#

For 32805 to be a term of the sequence, it must satisfy the equation

#A_n = 5*3^(n - 1)#

where #n in NN#

#32805 = 5*3^(n - 1)#

#=> 6561 = 3^(n -1)#

#=> 3^8 = 3^(n - 1)#

#=> 8 = n - 1#

#=> n = 9#

Since #9 in NN#, 32805 is a term of the sequence.