# Progressions

• Dec 23rd 2009, 04:31 AM
mastermin346
Progressions

For a geometric progression,the first term is 6 and the sum of the 3 terms is 126.Find

a)the possible values of the common ratio

b)the 6th term of the geometric progression.
• Dec 23rd 2009, 04:42 AM
I-Think
G.P.:$\displaystyle \Sigma{ar^k}$

First term:6 $\displaystyle \rightarrow{ar^0=6}$
Hence $\displaystyle a=6$

I assume you mean 1st 3 terms
$\displaystyle a+ar+ar^2=126$
$\displaystyle 6r^2+6r+6=126$
$\displaystyle r^2+r+1=21$

Part b
6th term:$\displaystyle ar^5$
With value of a and r, insert for answer
• Dec 23rd 2009, 04:44 AM
skeeter
Quote:

Originally Posted by mastermin346

For a geometric progression,the first term is 6 and the sum of the 3 terms is 126.Find

a)the possible values of the common ratio

b)the 6th term of the geometric progression.

a) $\displaystyle 6 + 6r + 6r^2 = 126$

solve for r (note that there will be more than one solution)

b) 6th term is $\displaystyle 6r^5$
• Dec 23rd 2009, 04:46 AM
galactus
Quote:

For a geometric progression,the first term is 6 and the sum of the 3 terms is 126.Find
Quote:

a)the possible values of the common ratio
The nth partial sum is $\displaystyle S_{n}=a_{1}\cdot \frac{1-r^{n}}{1-r}$

So, the sum of the first 3 terms is given as $\displaystyle S_{3}=126$ and the first term is $\displaystyle a_{1}=6$

Solve $\displaystyle 126=6\cdot \frac{1-r^{3}}{1-r}\Rightarrow 126=6(r^{2}+r+1)$ for r.

Quote:

b)the 6th term of the geometric progression.
Just use $\displaystyle a_{n}=a_{1}r^{n-1}$ once you find r.