consider the following geometric sequences and find the terms indicated:
*The 1st term is 440 and the 12th is 880. find S6
You should know that $\displaystyle t_n = ar^{n-1}$.
So $\displaystyle a = t_1 = 440$ and $\displaystyle t_{12} = ar^{11} = 880$.
This means $\displaystyle 440r^{11} = 880$
$\displaystyle r^{11} = 2$
$\displaystyle r =2^{\frac{1}{11}}$.
This means $\displaystyle t_n = 440(2^{\frac{1}{11}})^{n-1}$.
You should also know that $\displaystyle S_n = \frac{a(r^n-1)}{r-1}$
So $\displaystyle S_6 = \frac{440[(2^{\frac{1}{11}})^6-1]}{2^{\frac{1}{11}} - 1}$
$\displaystyle = \frac{440(2^{\frac{6}{11}} - 1)}{2^{\frac{1}{11}} - 1}$.
Use a calculator to simplify.