Question

What is the sum of an 8-term geometric series if the first term is −11, the last term is 180,224, and the common ratio is −4?
@perl

Answer 1

@kc_kennylau

Answer 2

Let \(a\) be the first term, \(r\) be the common ratio and \(n\) be the number of terms.
The sequence is: \(\Large a, ar, ar^2, ar^3,\cdots,ar^{n-1}\)
The series is: \(\Large \dfrac{a(r^n-1)}{r-1}\)

Proof: http://fym.la.asu.edu/~tturner/MAT_117_online/SequenceAndSeries/geometric_sequence_17.gif

Answer 3

P.S. Series is the sum.

Answer 4

isnt the formula \[a _{1-a _{n}r}/1-r\]

Answer 5

@kc_kennylau

Answer 6

They're just the same :)

Answer 7

\[\Large\dfrac{a(r^n-1)}{r-1}=\dfrac{a(1-r^n)}{1-r}\]

Answer 8

i got it wrong i get something out of this world idk why it isnt working

Answer 9

Mind showing me what you got? :)

Answer 10

ok \[-11-(8)(-4)/1-4\]

Answer 11

right this is how it looks when i pllugg in the numbers

Answer 12

Sorry I'm back now

Answer 13

Do you mind drawing it out? I cannot comprehend your LaTeX. Sorry for any inconvenience caused.

Answer 14

:(can you just show me the proper wat and illl tell you my mistyake

Answer 15

OK :D
\[\Large S=\dfrac{-11[(-4)^8-1]}{(-4)-1}\]

Answer 16

why the exponent 8?

Answer 17

Number of terms :)

Answer 18

and -1?

Answer 19

ohh i see my mistake now i wasnt plugging it in properly ok now i will simplify

Answer 20

\[\Large a=-11,r=-4,n=8\]
\[\Large S=\frac{a(r^n-1)}{r-1}=\frac{-11[(-4)^n-1]}{-4-1}\]

Answer 21

8*

OK :)

Answer 22

-144181?

Answer 23

@kc_kennylau

Answer 24

lolz tagging me has no use xP

Answer 25

lol you keep scaring me when you dont reply this website make you insecure

Answer 26

is it right tho?

Answer 27

but I calculated 144177...
http://bit.ly/1aCX0MU

Answer 28

omggggggggggggggggggggggggggggggggggggggggggggggg

Answer 29

thats was sooooo cool how did you do tht

Answer 30

http://lmgtfy.com

Answer 31

http://lmgtfy.com/?q=2%2B2

Answer 32

gj :D

Answer 33

lol