What is the sum of an 8–term geometric series if the first term is 12 and the last term is –3,359,232?

Question

lgbasallote · Answer

let A1 = 12 
A8 = -3,359,232
n= 8

$$\huge  A _{n} = A _{1}r ^{n-1}$$
$$\huge  -3,359,232 =12r ^{8-1}$$
$$\huge  -3,359,232 =12r ^{7}$$
$$\huge  \frac {-3,359,232}{12} =\frac {12r ^{7}}{12}$$
$$\huge  279,936=r ^{7}$$
$$\huge  6 = r$$

now can you do the S_n from here?

anonymous · Answer

no :(

lgbasallote · Answer

okay...gimme a few to write it..

lgbasallote · Answer

do you know the formula for the sum of geometric sequence?

anonymous · Answer

is it an=a1r(n-1) ?

lgbasallote · Answer

noo that's for finding a certain term in the series..it's what i used in the above see? the equation is 

$$\huge S _{n} = \frac {a _{1}(1 - r ^{n})}{1-r}$$

lgbasallote · Answer

so nowww...we substitute..
A1 = 12
r = 6
n = 8

can you do it now?

anonymous · Answer

is the answer 4,031,076

lgbasallote · Answer

yes you're right..sorry bout the confusion :D