# Homework Help: Arithmetic progression used to determine geometric progression

1. Jun 1, 2013

### Hivoyer

1. The problem statement, all variables and given/known data
an arithmetic progression(a1-a9) has 9 numbers.
a1 equals 1
The combination(S) of all of the numbers of the arithmetic progression is 369

a geometric progression(b1-b9) also has 9 numbers.
b1 equals a1(1)
b9 equals a9(unknown)

find b7

2. Relevant equations

3. The attempt at a solution

basically I use Sn = ((2*a1 + (n-1)*d)/2)*n
and I get 369 = 9 + 36*d; d = 10
then I find a9:
a9 = a1 + 8*d
a9 = 1 + 80 = 81; and I know b9 equals a9, so b9 = 81
then with the formula for the geometric progression I do:
bn = b1*q^(n-1)
b9 = 1*q^8
81 = q^8
9 = q^7
3 = q^6; which should be b7, however in the book's answers, it's not '3', but '27'.How is that even possible if b1 is said to be '1'?

2. Jun 1, 2013

### Office_Shredder

Staff Emeritus
How did you go from 81 = q8 to 9 = q7 to 3 = q6?

It looks like one side you were taking the square root of, and the other side you were dividing by q, which is definitely not the same operation

3. Jun 1, 2013

### Hivoyer

oh yeah, sorry about that, so it seems that squaring them isn't the way to proceed anyway, can you offer a tip?

4. Jun 2, 2013

### Curious3141

You should review the laws of exponents. $(a^m)^n = a^{mn}$, for instance.

So $q^8 = 81$. What's $q^4$? What's $q^2$? And therefore what's $q^6$?