Originally Posted by

**HallsofIvy** What was that referring to? Since your original problem had NO a, b, or n, I don't see how it applies.

Seriously, this is just basic algebra- except for the definition of "geometric sequence" and fifth root, really just arithmetic.

Skeeter and I told you that if the first term in a geometric series is 'a' and "constant ratio" is 'r' then the sixth term is $\displaystyle ar^5$. Here you are told that the first term is 81, so a= 81, and the six term is 954 so $\displaystyle 81r^5= 954$. Then, as skeeter said, $\displaystyle r^5= \frac{954}{81}= \frac{106}{9}$, by cancelling a "9" in both numerator and denominator. To get rid of that 5th power, take the fifth root: $\displaystyle r= \sqrt[5]{\frac{106}{9}= 11.777777... = 1.638$ approximately. Not anything near "17.159"- that's larger than 106/9 already! The fifth root of a number larger than 1 will be much less that the number.

Now that you know that r= 1.638, the four values you need to find are (81)(1.638), $\displaystyle (81)(1.638)^2$, $\displaystyle (81)(1.638)^3$, and $\displaystyle (81)(1.638)^4$.