please help me on how to solve this question
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.
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
$\displaystyle a+ar+ar^2=126$
$\displaystyle 6r^2+6r+6=126$
$\displaystyle r^2+r+1=21$
Solve this quadratic equation for your answer
Part b
6th term:$\displaystyle ar^5$
With value of a and r, insert for answer
For a geometric progression,the first term is 6 and the sum of the 3 terms is 126.FindThe nth partial sum is $\displaystyle S_{n}=a_{1}\cdot \frac{1-r^{n}}{1-r}$a)the possible values of the common ratio
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.
Just use $\displaystyle a_{n}=a_{1}r^{n-1}$ once you find r.b)the 6th term of the geometric progression.