The first term of a finite geometric series is 6 and the last term is 4374. The sum of all terms is 6558. Find the common ratio and the number of terms in the series.
So i get that T_{1= 6.}So, a=6. Sn= 6558. how to find r?
If $\displaystyle t_1 = a$, $\displaystyle t_2 = ar$, $\displaystyle t_3 = ar^2$, ..., $\displaystyle t_n = ar^{n-1}$ and $\displaystyle S_n = t_1 + t_2 + \dots + t_n$, then, according to the standard formula for the sum of geometric progression, we get
$\displaystyle S_n = \frac{a(r^n-1)}{r-1}=\frac{ar^n-a}{r-1}=\frac{rt_n-t_1}{r-1}$
from where
$\displaystyle r=\frac{S_n-t_1}{S_n-t_n}=\frac{S_n-t_1}{S_{n-1}}$.
Not using any memorized formulas, just thinking about it:
Given that 6r^n = 4374, divide through by 6 to get r^n = 729. Note that 729 = 27^2, or 9^3, or 3^6; so the answer must be that either r=27 and n=2, r=9 and n=3, or r=3 and n=6. See which one results in a series whose sum of terms 6 + 6r + 6r^2 + ...6r^n = 6558.