Question

What is the sum of a 7-term geometric series if the first term is -11, the last term is -45,056, and the common ratio is -4?

Answer 1

I'm sorry, but I have to ask again: Do you know the formula?

Answer 2

is it the same formula lol..

Answer 3

not like arithmetic ... O_O

Answer 4

this seems interesting... let me try it on paper, I'll let you know IF i can do it ;)

Answer 5

haha kk thanx

Answer 6

uff... this is not easy (for me) ... using logs It seems it doesn't work. Maybe I made a mistake somewhere... Let me think :(

Answer 7

... I can't, is just impossible. Do you have any advanced calculator?
to find "n" here..
\[\LARGE \log_4(n)+1-n=\log_4(11)\]

Answer 8

no :(

Answer 9

.... well I guess I can't do it. I have to confess that this is a cute problem , but no matter how much I beg her, she just won't give up :) lol

Answer 10

haha

Answer 11

call someone maybe they can help :S ... and I think that one with logs there above, I just broke some rules lol. now I see it, my math sucks anyway.

Answer 12

Shouldn't we use the formula:\[a \frac{1-r^n}{1-r}\]

Answer 13

yes - you dont need to know the last term

Answer 14

if you don't know the last term what n is going to be ? :O

Answer 15

the formula is:
\[\LARGE Sn=a_1\cdot {1-r^n\over 1-r }\]

Answer 16

S7 = -11 * (1 - (-4)^7)
---------
1-(-4)

Answer 17

Geometric series shall be in the form of
a, ar, ar^2 , ar^3,....ar^6 (these are first 7 terms)
a=-11
ar^6=-45056
taking the ratio
ar^6/a=-45056/-11
r^6=4096 (as given in the question common ratio r=4 will satisfy this). Just to make sure everything is in geometric

sum of first 7 terms gives (I am not using striaght formula, I shall derive one that helps you to verify)
y=a(1+r+r^2+r^3+...+r^6)
y/a=1+r+r^2+r^3+...+r^6 --- (1)
now mulitplying with r on both sides we get
ry/a=r +r^2+r^3+...+r^7 --- (2)
subtracting (1) from (2)
ry/a-y/a=r^7-1 (remaining terms cancel each other)
y(r-1)/a=r^7-1
y=a(r^7-1)/(r-1) [r should not be equal to 1]

in general sum of n terms in geometric series is
a(r^(n+1)-1)/(r-1) [considering r>1]
a(1-r^(n+1))/(1-r) [considering r<1]

sum of 7 terms with a=-11, r=4 gives
-11*(4^7-1)/(4-1)
-11*(16384-1)/3
-11*(16383/3)
-11*5461
-60071

Answer 18

The common ratio is -4, not 4.

Answer 19

Mine yielded -36047, but I am sure I made a mistake.

Answer 20

ok theses are the 2 questions and the possible answer choices if it helps you guys--->What is the sum of a 7-term geometric series if the first term is -11, the last term is -45,056, and the common ratio is -4? (3 points)
A) -143,231
B) -36,047
C) 144,177
D) 716,144
21.
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? (3 points)
A) -143,231
B) -36,047
C) 144,177
D) 716,144

Answer 21

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWWWWWWWWWWWWWWWWWWWWWWWWWW :@:@:@ I SWEAR I GONN' BREAK MY PC RIGHT NOW :@ AAAHHH... (angry ) how come I didn't know it :(

Answer 22

I got B from the first.

Answer 23

I didn't read the question properly, and I thought that we have to find a_n then
we have to find S_n and that "n" I just couldn't find . :(

Answer 24

i just want to say thank you guys so much for all the help and im srry if its making u mad :/

Answer 25

[-11 * 16385] / 5
= - 36047

Answer 26

And I got C for the second.

Answer 27

thanks bmp :)

Answer 28

\[\Large S_8=-11\cdot {1-(-4)^8 \over 1-(-4)} \]
pff.. Peace of cake .. Lord I'm stupid :(

Answer 29

No problem, mate :-)

Answer 30

Yeah, its my bad. common ratio is -4 instead of 4 :)
That makes the same as bmp answer :)