Given that t_{7}=x+56,t_{8}=5x+41 and t_{9}=6x+44 are consecutive terms in an arithmetic sequence, find the following values: x = The difference between terms is: The first term of this sequence is:
ok this is arithmetic sequence, do you remember the thing called the common difference?
mhm
so you know the difference of t8 and t7 will have the same difference as t9 and t8
you have an equation right there
replace the phrase will have the same with an equal sign
and you should be able to translate the difference between m and n as m-n
so it's like \[t_7=x+56\] = \[t_8 = 6x + 44\] ? i got lost
i want you to translate "the difference of t8 and t7 is the same as the difference of t9 and t8" First of all, the same as means equals so we can use the equal symbol = so we now can write it as "the difference of t8 and t7 = the difference of t9 and t8" difference of m and n means m-n so looking at the difference of t8 and t7 this means t8-t7 and looking at the difference of t9 and t8 this means t9-t8 now there should be an equal sign between these two expressions because that it what is says in the initial sentence above so basically you have t8-t7=t9-t8
replace t7,t8,and t9 will what they are in terms of x then solve
don't forget parenthesis
if you want i will show you where you need them (t8)-(t7)=(t9)-(t8)
so just insert those things in terms of x in there and subtract now
then solve
oh, so i just plug the equations in according to the terms...hm
let me know what you get for x or you can show your work so i can see where you went wrong in solve for x or show me where you are getting lost
4635749.268
wowzers that is definitely not correct
lol ik..
oh wtf i typed in the wrong thing
-3.78
ok we have the equation (t8)-(t7)=(t9)-(t8) you were given: t_{7}=x+56,t_{8}=5x+41 and t_{9}=6x+44 so i'm going to replace t8 with 5x+41 and i'm going to replace t7 with x+56 and i'm going to replace t9 with 6x+44 like so (t8)-(t7)=(t9)-(t8) (5x+41)-(x+56)=(6x+44)-(5x+41)
did you get this far?
yes
so i typed the left side into the calc and got -12.14893616
ok distribute first if it helps you like i'm talk about where you have negative in front of the parenthesis
and then i got 3.21276596 on the right side
for example i will do one side (only distributing) 5x+41-x-56=6x+44-(5x+41) you try to take care of the distributing on the other side
6x + 44 - 5x - 41
great so we have the following equation 5x+41-x-56=6x+44-5x-41 now i'm going to regroup the terms on the left hand side so i have my like terms together on that particular side like so 5x-x+41-56=6x+44-5x-41 you can regroup the terms on the other side
do you know what i mean?
4x - 13 = x + 3
is 41-56 equal to -13? or -15?
whoops -15
great so we have 4x-15=x+3
now we want to get our x terms on one side and our constant terms on the other since we are trying to in fact isolate x
ah crap, that's the part im always iffy about
ok i will begin the process we have 4x-15=x+3 I'm going to go ahead and subtract x on both sides giving me (4x-15)-x=(x+3)-x I did this because that means that there will be no x on one side 4x-15-x=x+3-x 4x-x-15=x-x+3 4x-x-15=3 so we have now 4x-x-15=3
i'm sure you can do the 4x-x part, right?
4x - 15 = 3 4x-15+15 = 3 + 15 4x = 18 x = 4.5
4x-x is not 4x
let me rewrite 4x-x 4x-x is the same as saying 4x-1x
and 4-1 is not 4 but 3
eek i was thinking 5
so you should get a different x now
x = 3.6
hmmm ok 3x-15=3 is right to say then you added 15 on both sides 3x-15+15=3+15 3x+0=3+15 3x=18 so you are saying x is ?
6...lol omg
yes so we have answered the first question
now the second question is asking for the common difference
just plug it in
and then count up the differences, im assuming
so it wants to know what t8-t7 equals (note: this number will be the exact number as doing t9-t8 because this is a common difference number)
so d = 9
great
ok now we need to find the 1st term
a + d (n-1)
you know exactly what formula we will be using we know that \[t_n=t_1+d \times (n-1)\] we know the 7th, 8th, and 9th term so we can use any one of them so if we choose to use the 7th terms then we let n=7 and solve for the value that is t1
wherever you see an n replace it with 7
you will notice you will need to know what t7 is equal to
54 as t_7?
so x=6 and t7=x+56 so t7 cannot be 54
oh psht 62 xD
yeppie ok we had the equation \[t_n=t_1+d \times (n-1)\] and you replace all the n's with 7 (you could have used even 8 or 9 since knew the values for also t8 and t9) but anyways replacing n's with 7's we have \[t_7=t_1+d \times (7-1)\] you said t7 was 62 so we have \[62=t_1+d \times (7-1)\] also d represents the common difference which was 9 so we can also input that \[62=t_1+9 \times (7-1)\]
this is your last equation to solve
i recommend doing order of operations on the left hand side where you can first
62=54
where is t1?
i think you meant 62=t1+54
one more step to go you can do this you need to find out what the first term in the sequence is we need to isolate t1 to find that
how do you get t1 by itself there is a plus 54 so we need to _________________
8 = t1
yep
HAHA YEAH
you did very well math is like a puzzle sometimes try to enjoy them
omg thank you so much; this was pretty fun xD
i have to admit i enjoyed it too i always enjoyed trying to figure out where things go when it comes to math
do you tutor outside of open study? you'd be a really great tutor to have tbh
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