- #1
ThomasMagnus
- 138
- 0
A finite geometric sequence has t1 = 0.1024 and t2 = 0.256. How many terms does this sequence have if its middle term has a value of 156.25?
My Solution
Common Ratio: T2/T1=(.256)/(.1024)=2.5
What term # is the middle term?
tn=ar^n-1
a=0.1024
r=2.5
tn=156.25
(156.25)=(0.1024)(2.5)^n-1
1525=(2.5)^n-1
[Log(1525)/Log(2.5)]+1=n
n=8+1=9
n is the middle term so final term should be 2n. 18 terms in the sequence
From the way I have seen other people do this question, they get the answer "17 terms". Why am I 1 term off? Can you help me with what I am doing wrong. Or am I doing it correctly :)
Thanks!
My Solution
Common Ratio: T2/T1=(.256)/(.1024)=2.5
What term # is the middle term?
tn=ar^n-1
a=0.1024
r=2.5
tn=156.25
(156.25)=(0.1024)(2.5)^n-1
1525=(2.5)^n-1
[Log(1525)/Log(2.5)]+1=n
n=8+1=9
n is the middle term so final term should be 2n. 18 terms in the sequence
From the way I have seen other people do this question, they get the answer "17 terms". Why am I 1 term off? Can you help me with what I am doing wrong. Or am I doing it correctly :)
Thanks!