geometric sequence help

• Nov 24th 2009, 07:08 PM
hovermet
geometric sequence help
Find the tenth term of a geometric sequence.

the first term is 3 the 6th term is root3/9

i stuck on the calucation for finding R

tn=ar^n-1
t6= root3/9 = 3(r)^5

(3^-1/2)/3^2=3(r)^5 <-- stuck

I checked the answer on my workbook .It says.

(3^-1/2)/3^3=r^5
3^-5=r^5

then it suddenly jumps right into r= 3^-1/2

i do not understand how they get rid of the exponent 5 in the workbook and if i factor out the 9 to 3^2, then how can i continue? will the answer be equivalent?
• Nov 24th 2009, 07:24 PM
pickslides
Using the information given

$\displaystyle \frac{\sqrt{3}}{9}= 3r^5$

Divide by 3 and take the 5th root.
• Nov 24th 2009, 07:24 PM
Jameson
Quote:

Originally Posted by hovermet
Find the tenth term of a geometric sequence.

the first term is 3 the 6th term is root3/9

i stuck on the calucation for finding R

tn=ar^n-1
t6= root3/9 = 3(r)^5

(3^-1/2)/3^2=3(r)^5 <-- stuck

I checked the answer on my workbook .It says.

(3^-1/2)/3^3=r^5
3^-5=r^5

then it suddenly jumps right into r= 3^-1/2

i do not understand how they get rid of the exponent 5 in the workbook and if i factor out the 9 to 3^2, then how can i continue? will the answer be equivalent?

I am having trouble following your notation. Please use parentheses to clear up confusion.

If you mean $\displaystyle \frac{\sqrt{3}}{9}=3r^5$ then cancel out a 3 from both sides, then take the 5th root of both sides.