# Progressions

• Dec 23rd 2009, 05:31 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, 05:42 AM
I-Think
G.P.: $\Sigma{ar^k}$

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

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

Part b
6th term: $ar^5$
With value of a and r, insert for answer
• Dec 23rd 2009, 05:44 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) $6 + 6r + 6r^2 = 126$

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

b) 6th term is $6r^5$
• Dec 23rd 2009, 05:46 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 $S_{n}=a_{1}\cdot \frac{1-r^{n}}{1-r}$

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

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

Quote:

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