Will give medal and become fan!!! Identify the 31st term of an arithmetic sequence where a1 = 26 and a22 = -226. Answers -274 -284 -334 -346
@bohotness @freckles @ganeshie8 @vera_ewing
@XoALIEoX
@StudyGurl14
@Holly00d1248
\[d=\frac{a_{22}-a_1}{22-1}=\]
d represents the common difference by the way
\[a_n=a_1+(n-1)d \text{ can be used to find } a_{31} \text{ once we find } a_1 \text{ and } d\] So that is why I'm asking you to compute that above to find d
@freckles I dont understand?
which equation or formula do you now understand?
not*
@freckles I understand the equation but I can figure out how to work it
\[d=\frac{a_{22}-a_1}{22-1}= \] the bottom equals 22-1=? you are given the top numbers...a_(22)=-226 and a_1=26 so I mean a_(22)-a_1=-226-26=? do find 22-1 all you do is think of having 22 somethings and 1 gets taken away how many somethings do you have left?
to find -226-26 since both of them are negative you add 226 and 26 and then put a - on the sum when done
a21/21?
the bottom is right but how do you just assume the difference of the 22nd term and the 1st is going to give you the 21st term ?
I think you would be better off by doing what I said
again you are given the 22nd term is -226 and the 1st term is 26
-252 which isnt in the answer choice
enter those numbers into the top in the respective spot
lol I didn't say that was an choise I said we were trying to find d. \[d=\frac{a_{22}-a_1}{22-1}=\frac{-226-26}{21}=\frac{-252}{21}\]
so now we have: \[a_n=a_1+(n-1)d \\ \text{ where} a_1=26 \text{ and } d=\frac{-252}{21} \\ a_n=26+(n-1) \frac{-252}{21}\] now to find the 31st term replace n with 31
idk how to do it!!!!!
-12
a1 = 26 a22 = –226 a22=a1+21d Substitute! -226=26+21d 21d=-252 d= -12 Now lets find the 31st term. a31=a1+30d a31=26+30 * (-12) a31=26 + -360 a31= -334
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