A geometric sequence is
3/4, 9, 108, 1296, 15,552, 186,624, ... . Which is the general term of the sequence?

A)tn = 3/32(12)n - 1, where n ∈N and n ≥1
B)tn = 3/16(12)n - 1, where n ∈N and n ≥1
C)tn = 3/8(12)n - 1, where n ∈N and n ≥1
D)tn = 3/4(12)n - 1, where n ∈N and n ≥1

Question

A geometric sequence is 
3/4, 9, 108, 1296, 15,552, 186,624, ... . Which is the general term of the sequence?

A)tn = 3/32(12)n - 1, where n ∈N and n ≥1	
B)tn = 3/16(12)n - 1, where n ∈N and n ≥1	
C)tn = 3/8(12)n - 1, where n ∈N and n ≥1	
D)tn = 3/4(12)n - 1, where n ∈N and n ≥1

anonymous · Answer

a = 3/4
ar = 9

r = $$9\div3/4$$ = 12

tn = ar^(n-1) general formula

tn = (3/4) 12^(n-1)
Answer is D

anonymous · Answer

D