Arithmetic progression used to determine geometric progression

[Hivoyer]

Homework Statement

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

Homework Equations

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'?

 

[Office_Shredder]

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

 

[Hivoyer]

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

 

[Curious3141]

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

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 $q^6$?