a1=2 and an=3(an-1)^2?? A 2,12,27 b 3,12,27 c 2,12,432 d 3,12,432
@rvc can you help me with this??
@Nnesha can you help???
so you have a_2 find a_2 by replace n with 2
ok :) THANKYOU!!!
\[a_n=3(a_{n-1})^2 \\ a_2=3(a_{2-1})^2 \\ a_2=3(a_1)^2\\ a_2=3(2)^2 \\a_2=3(4) \\ a_2=12\] I will go ahead and give this one since all choices have a_2 is 12 then find a_3 by taking that same thing involving n and replacing n with 3
is that \(a_n=3a_{n-1}^2\)
do you think its d or b
you didn't state what are you looking for by the way is a_2?
the first 3 terms
all i see is a1 given to you and the the equation that generates the term of this sequence
from you choices it appears you are only looking for one term? are you sure that you are in need of 3 terms?
yes
then why the need of multiple choice answers?
I would ask you to recheck, did the question say find the 3th term of the sequence or find the first three terms?
\[a_1,a_2,a_3 \text{ are the first three terms } \] it sounds like they are asking you to state a_1 and a_2 anyways even though they are both given... you need to only really find the third term \[a_n=3 \cdot a_{n-1}^2\] have you replaced n with three yet ?
hmm where a_2 i don't see it
all the choices have a_2 is 12 technically it is given
which represents the first 3 terms of the sequence? that the question with the prob
hmm it seems to me a though those given number are just one term? wouldn't you agree
a_1 is the first term a_2 is the second term a_3 is the third term there are three terms in each choice
hmm the choices say a) b) c) d) where do you see a1, a2,a3?
He said find the first three terms
Those terms are called a_1,a_2,a_3
a_1 and a_2 are already given
that is the first and second term
we just need to find the third term
ohhh ok i got it its B you take the a2 into the equation and the 3 4 and you get B thankyou both for helping me :D:D
for the problem is miss worded and not stated correctly! but ate any rate finding the first three term just do as @freckles said
first of all @aswemayknow why are you saying the first term is 3 a_1 is given as 2 wasn't it? also can is you replace the n's with 3's and do the computation
Thankyou :D:D
oh my bad! i realized my mistake heheh each choice is given 3 terms
\[a_n=3 \cdot a_{n-1}^2 \] put n as 3 \[a_3=3 \cdot a_{3-1} ^2 \] 3-1 is 2 what is a_2 ?
hehe B is not correct pal ssecond term is 12 first term is 2 given to you
B and D cannot be true since the first term is a1=2 for A and C since they both have a2=12 we need a3 to see which is true
use what @freckles did \(a_3=3a_2^2=2(12)^2\)
\[3(12)^2 \]
yep 3 I'm sloppy today no focus heheh
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