Asked by Anonymous
The geometric mean between the first two terms in a geometric sequence is 32. if the third term is 4 find the first term
Answers
Answered by
Reiny
let the terms be a, ar, 4
4a = 32
a = 8 <---- done here
Bonus marks:
ar^2 = 4
8r^2 = 4
r^2 = 1/2
r = ± 1/√2
terms would be
8 , 8/√2, 8/√2^2
8 , 8/√2, 4
looks good,
a = 8
4a = 32
a = 8 <---- done here
Bonus marks:
ar^2 = 4
8r^2 = 4
r^2 = 1/2
r = ± 1/√2
terms would be
8 , 8/√2, 8/√2^2
8 , 8/√2, 4
looks good,
a = 8
Answered by
Reiny
Forget the previous post, I didn't read it carefully
let the 3 terms be a, ar , ar^2
geometric mean of first two terms = √(a(ar)) = 32
a √r = 32 --> a = 32/√r
also ar^2 = 4 --> a = 4/r^2
32/√r = 4/r^2
32r^2 = 4√r
8r^2 = √r
64r^3 = r
r^2 = 1/64
r = ± 1/8
from a = 4/r^2
a = 4(1/64) = 256
let the 3 terms be a, ar , ar^2
geometric mean of first two terms = √(a(ar)) = 32
a √r = 32 --> a = 32/√r
also ar^2 = 4 --> a = 4/r^2
32/√r = 4/r^2
32r^2 = 4√r
8r^2 = √r
64r^3 = r
r^2 = 1/64
r = ± 1/8
from a = 4/r^2
a = 4(1/64) = 256
Answered by
Jia
Ty
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.