# Geometric Sequences

• September 10th 2007, 03:32 PM
Norman Smith
Geometric Sequences
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!
• September 11th 2007, 12:42 AM
ticbol
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?
• September 11th 2007, 04:44 AM
Soroban
Hello, Norman!

I believe those "exponents" are actually subscripts.

Quote:

Write the first five terms of the geometric sequence.
Find the common ratio and write the $n^{th}$ term of the sequence as a function of $n.$

. . $a_1\:=\:64,\;a_{k+1} \:=\:\frac{1}{2}a_k$

We are told that the first term is 64,
. . and each term is one-half of the preceding term.

The common ratio is: . $r \,=\,\frac{1}{2}$

The first five terms are: . $64,\,32,\,16,\,8,\,4$

The general term is: . $a_n \;=\;64\cdot\left(\frac{1}{2}\right)^{n-1} \;=\;\frac{128}{2^n}$