Question

Find x such that x + 4, 3x -9, and 2x +8 are consecutive terms of an arithmetic sequence.

Answer 1

x is not a negative number, because if not, then `3x-9` has the greatest absolute value, and can't be the middle term.
x is not zero, because 4,-9, 8 is not a sequence. So it must be a positive number.

(this is what I got so far)

Answer 2

x is big enough to make `3x-9` bigger than `x+4` so
x + 4=3x -9
x +13=3x
2x=13
x=7.5

So x is bigger than 8.

The 2x+8 is bigger than 3x-9
2x+8>3x-9
17>x

So I am thinking it is 8<x<17

Okay then if it has to be an integer.... (idk why I am assuming that though)

x=10 random guess.

Answer 3

x + 4, 3x -9, 2x +8
10+4, 3(10)-9, 2(10)+8
14, 21, 28

Answer 4

so we have 3 terms, and this is an arithmetic sequence
meaning the next term is itself plus "something"

so

x + 4, 3x -9, 2x +8


so let's say that "something" is x

so

we could say that

3x - 9 = prior term + x

3x - 9 = (x+4) + x <--- solve for "x"

Answer 5

So do I combine like terms on the right side?

Answer 6

but then the x is not 10

Answer 7

well... just solve or simplify for "x" :)

Answer 8

Alright thank you both! c:

Answer 9

yw

Answer 10

it is 10
the way jdoe001 did it was correct but
he shouldn't put x as the common difference
you need to put another variable there

Answer 11

let make is n
(x+4)+n= the next term=3x-9
3x-9+n=2x+8

Answer 12

first solve for n and then solve for x ^_^

Answer 13

hmmm thought about using "y" BUT that sorta will get a bit in the way of getting "x" methinks

Answer 14

but you that x is different variable than the common difference? no?
so you need another variable y or call it whatever

Answer 15

yes hhhhmmm.... ok

Answer 16

well... I checked on the 1st two terms... and 13 checks out
it doesn't for the last one though

Answer 17

n=2x-13
and
n=-x+17

Answer 18

so x=10, nd 10 checks nicely

Answer 19

ohhh I see..... d = d and then it'd be an equation based on "x"

Answer 20

Yes^_^

Answer 21

@jherrera98 confused about what @xapproachesinfinity just said?

Answer 22

Yes haha

Answer 23

ok....lemme write it

Answer 24

what @SolomonZelman wasn't clear to me but somehow he got 10

Answer 25

he did a rundown trial and error

Answer 26

@jherrera98 one sec

Answer 27

So solving for x isn't the answer?

Answer 28

@jdoe0001

Answer 29

but he used intervals how could he know that it is integer
it could be any real?

Answer 30

yes, I can't thing logically about thinking logically. I can use my brain, but unable to explain how I use it.....

Answer 31

And I assumed it was an integer, because the question would have been very hard otherwise.

Answer 32

but since it's a arithmetic sequence you have to be adding something don't you?

Answer 33

hehe, that's not a good way to do this
it could really be a rational number hehe

Answer 34

it can be an irrational number too

Answer 35

jherrera98 what you are adding is n=7

Answer 36

that would make way harder hehe

Answer 37

What if I just equal them all to zero & find x?

Answer 38

i still need to observe this trial and error method hehe

Answer 39

No you can't. you are dealing with arithmetic sequence therefore
you need to do it using the definition of an arithmetic sequence
we know that for arithmetic sequence you add or subtract something to get the next term
we call that common difference

Answer 40

So there's many ways to solve this?

Answer 41

I can't find d so idk what to do from there

Answer 42

Don't know! the obvious one is what jdoe did

Answer 43

hmmmm not making much headway here either btw =)

Answer 44

I'm even more lost now haha

Answer 45

the only thing i would thing of is that method!
where do you lost?

Answer 46

I'm lost once the common difference isn't the same for the last equation, 2x + 8

Answer 47

see if you go this way \(\large {
\begin{array}{cccllll}
x+4&3x-9&2x+8\\
\hline\\
&(x+4)+d&(3x-9)+d
\end{array}
\\ \quad \\
\implies
\begin{cases}
3x-9=(x+4)+d\to \frac{3x-9}{x+4}=d\\
2x+8=(3x-9)+d\to \frac{2x+8}{3x-9}=d
\end{cases}
\\ \quad \\
d=d\implies \cfrac{3x-9}{x+4}=\cfrac{2x+8}{3x-9}
}\)

it ends up with \(\bf x^2-10x+7=0\ which .... is not very feasible

Answer 48

ahemm with \(\bf x^2-10x+7=0\) that is, which is not very feasible either

Answer 49

No the common difference is indeed the same
14, 21,28

Answer 50

ohhh man... I see my mistake now.. anyhow...lemme redo all that dohh =(

Answer 51

So how do I find x now? .-.

Answer 52

you did multiplication lol,

Answer 53

\(\large {
\begin{array}{cccllll}
x+4&3x-9&2x+8\\
&(x+4)+d&(3x-9)+d
\end{array}
\\ \quad \\
\implies
\begin{cases}
3x-9=(x+4)+d\to 3x-9-(x+4)=d\\
2x+8=(3x-9)+d\to 2x+8-(3x-9)=d
\end{cases}
\\ \quad \\
d=d\implies 3x-9-(x+4)=2x+8-(3x-9)
}\)

so.. there... now just a matter of solving for "x" :)

Answer 54

see it @jherrera98 ? =)

Answer 55

we have solved already for x above, it think you are not following lol

Answer 56

heheh k

Answer 57

So x=10 is the final answer? haha

Answer 58

yes, what else? that's what the question is looking for
you back to the terms and check
you get 14,21,28 which true
this sequence has 7 as the common difference

Answer 59

\(\large {
\begin{array}{cccllll}
x+4&3x-9&2x+8\\
\hline\\
&(x+4)+d&(3x-9)+d
\end{array}
\\ \quad \\
\implies
\begin{cases}
3x-9=(x+4)+d\to 3x-9-(x+4)&=d\\
2x+8=(3x-9)+d\to 2x+8-(3x-9)&=d
\end{cases}
\\ \quad \\
d=d\implies 3x-9-(x+4)=2x+8-(3x-9)
\\ \quad \\
3x-9-x-4=2x+8-3x+9\implies 2x-13=-x+17
\\ \quad \\
3x=30\implies x=\cfrac{\cancel{ 30 }}{\cancel{ 3 }}\to ?
}\)

Answer 60

Ok now I see it! Blonde moment lol Thank you all so much!

Answer 61

YW!