Question 1202469: The 3rd term of a geometric sequence is 36, and the 6th term is 9/2. What is the recursive formula for the sequence?
Found 3 solutions by mananth, ikleyn, greenestamps: Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! The 3rd term of a geometric sequence is 36, and the 6th term is 9/2. What is the recursive formula for the sequence
tn = ar^(n-1)
t3= a*r^(3-1)
t3= ar^2=36
t6 = a*r(6-1) = ar^5= 9/2
t3/t6 = ar^2/ar^5 = 36/(9/2)
1/r^3= 36/9 *2
1/r^3= 8
r^3= 1/8
r=1/2
ar^2=36
plug r
a*(1/2)^2 =36
a/4 =36
a= 144
tn = 144*(1/2)^n-1
Answer by ikleyn(52788) (Show Source): Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!
The 3rd term, 36, is the first term, multiplied by the common ratio 2 times: .
The 6th term, 9/2, is the first term, multiplied by the common ratio 5 times: .
Divide the formulas for the 6th and 3rd terms to calculate the common ratio:


A RECURSIVE formula tells how to get each term from the preceding term; for a geometric sequence the rule is "multiply by the common ratio". Since in this problem the common ratio is 1/2, the recursive formula is

To complete the definition of the recursive formula for the sequence, we need to specify the first term. Since the 3rd term is 36 and the common ratio is 1/2,



ANSWER: a(1)=144; for n>1, a(n)=(1/2)*a(n-1)
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