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 2 years ago
is it okay to do this...? and did i get it correct anyway? the third term of an exponential sequence is 12 and the 9th term is 768. find the values of the common ratio.
\[\large ar^2=12 \]\[\large ar^8=768\]
\[\large \frac{ar^8}{ar^2}=\frac{768}{12}\]
\[\large r^6=64 \]
\[\large r=2 \ , \ r=2\]
I forgot the formulas and decided to go with this but i don't know if it is correct or even if it's acceptable... can anyone confirm for me please?
 2 years ago
is it okay to do this...? and did i get it correct anyway? the third term of an exponential sequence is 12 and the 9th term is 768. find the values of the common ratio. \[\large ar^2=12 \]\[\large ar^8=768\] \[\large \frac{ar^8}{ar^2}=\frac{768}{12}\] \[\large r^6=64 \] \[\large r=2 \ , \ r=2\] I forgot the formulas and decided to go with this but i don't know if it is correct or even if it's acceptable... can anyone confirm for me please?

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sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.0truth is i've even forgotten if this is exactly how to do it lol. i literally have no memory of the classes i took for this topic

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0You are perfectly correct, math can be done in infinite ways :)

Mimi_x3
 2 years ago
Best ResponseYou've already chosen the best response.0it looks like a geometric progression.. i dont know what's an exponential sequence.. http://en.wikipedia.org/wiki/Exponential_sheaf_sequence

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.3\[a_3=12\] \[a_9=768\]  \[ar^2=12\] \[ar^8=768\] \[\frac{12}{r^2}r^8=768\] \[r^6=64\] \[r= \pm 2\]

sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.0same thing :) different name

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.3GP formula \[\Huge T_n=a_1r^{n1}\]
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