Write the explicit formula that represents the geometric sequence -2, 8, -32, 128 i need help on this problem
First find your r.
\[\frac{8}{-2} =~?\]\[\frac{-32}{8}=~?\]
...etc.
basically: \(\large\color{slate}{ r=(a_n ) \div (a_{n-1}) }\)
arithmetic sequence? or geometric sequence? @SolomonZelman
geometric
what do you mean, I am right!
Ok :) You had written arithmetic, haha
oh lol
so which one is right ?
\(\large\color{slate}{ a_n=a_1\times r^{n-1} }\)
do you see the common ratio?
Yeah... to find a common ratio, divide the preceding term by the one before it.
Make sure you first have a common ratio between all your terms.
i just dont know how to even start on this problem. like im horrible in math
Find your ratio first. \(\color{blue}{\text{Originally Posted by}}\) @Jhannybean \[\frac{8}{-2} =~?\]\[\frac{-32}{8}=~?\] \(\color{blue}{\text{End of Quote}}\)
use anything for "r" : \(\large\color{slate}{ r=(8) \div (-2) }\) \(\large\color{slate}{ r=(-32) \div (8) }\) \(\large\color{slate}{ r=(128) \div (-32) }\) see how I am dividing a term, by the one before to find r? this is denoted as: \(\large\color{slate}{ r=(a_n) \div (a_{n-1}) }\)
just wondering if anyone explained the common ratio here?
common ratio, if anyone didn't know: is a number by which you multiply a term, to find the term right after
The common ratio is the term that you find that is the same for each pair in the sequence.
That is why we are asking you to find the common rati between each pair of terms.
@Jhannybean I see you did
Elizabeth, lost?
Perhaps as Einstein said "make it simple, but no simpler"
first need to find the common ratio
Elizabeth, do you want me to explain everything very thoroughly?
@SolomonZelman is the ideal person to explain
that is debatable... I wouldn't necessarily say that, whether I hold like that or not. Although I will certainly attempt my best!
Elizabeth, say something, please
So @elizabeth710 ? Are you still here?
yes im here
okay, will start from this: Do you understand what the notations I will write in blue mean? \(\large\color{blue}{ a_1 }\) , \(\large\color{blue}{ a_2 }\) , \(\large\color{blue}{ a_3 }\) \(\large\color{blue}{ r }\) , \(\large\color{blue}{ a_n }\)
sadly no :/
What do you mean by your second line?
got disconnected-:(
Okay, so say you have the following sequence: \(\large\color{brown}{ 1,~3,~9,~27~... }\)
So \(\large\color{brown}{ 1 }\) is the first term, denoted as \(\large\color{brown}{ a_1 }\) . \(\large\color{brown}{ 3 }\) is the 2nd term, denoted as \(\large\color{brown}{ a_2 }\) . \(\large\color{brown}{ 9 }\) is the 3rd term, denoted as \(\large\color{brown}{ a_3 }\) . \(\large\color{brown}{ 27 }\) is the 4th term, denoted as \(\large\color{brown}{ a_4 }\) .
and on... see ? So, can you explain to me, what would \(\large\color{brown}{" a_5" }\) mean?
Elizabeth, if you are not getting what I am saying, please say so...
okay, between each term of a geometric sequence is what we call a "common ratio"
the reason we call it a common ratio is becuase if you divide the next term, by the previous term, you will have a ratio which is "common"
i.e divide 8 by -2 = -4, then divide -32 by 8 = -4
This is what @SolomonZelman is trying to explain to you
i got a_n=-2×-4
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