a1=64, a^k+1 =1/2a^k
I guess that is
a1=64, a^(k+1) = (1/2)a^k
If I guess it right, then,
a^(k+1) = (1/2)a^k
(a^k)(a) = (1/2)a^k
a = 1/2
What now?
a1 = 64 and a = 1/2
What is this "a"? So what if it is equal to 1/2?
I thank all of you again for helping me, and having the patience to withstand my idiocy. I understand the concept, but maybe it's the wording of the question that is throwing me off, so I am wondering again if any of you could help me out with another question. Thank you all!!
Directions: Write the first five terms of the geometric sequence. Find the common ratio and write the nth term of the sequence as a function of n.
a1=64, a^k+1 =1/2a^k
Edit: Okay I looked at the question, and with the help of my sister, I figured it out. So I guess I can leave the question up just in case anyone is interested in doing it, however I stress again that it is very kind of all of you who have been able to help me. Also I'll leave this thread up in case I have questions with some of the other problems, or for others who are struggling as well in this portion of Precalculus. Thanks!
Hello, Norman!
I believe those "exponents" are actually subscripts.
Write the first five terms of the geometric sequence.
Find the common ratio and write the term of the sequence as a function of
. .
We are told that the first term is 64,
. . and each term is one-half of the preceding term.
The common ratio is: .
The first five terms are: .
The general term is: .