SOLUTION: Hello, We are working on Arithmetic Sequences. I am stuck. Can someone please help. I need to find a1 and d. a15 = 129, a8=22(a1) Thanks.

Algebra ->  Sequences-and-series -> SOLUTION: Hello, We are working on Arithmetic Sequences. I am stuck. Can someone please help. I need to find a1 and d. a15 = 129, a8=22(a1) Thanks.      Log On


   

Question 936448: Hello,
We are working on Arithmetic Sequences. I am stuck.
Can someone please help. I need to find a1 and d.
a15 = 129, a8=22(a1)
Thanks.
Found 3 solutions by Fombitz, Edwin McCravy, MathTherapy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a%5Bn%5D=a%5B1%5D%2Bd%28n-1%29
.
.
a%5B15%5D=a%5B1%5D%2B14d=129
a%5B8%5D=a%5B1%5D%2B7d=22
Subtract the two values,
a%5B1%5D%2B14d-a%5B1%5D-7d=129-22
7d=107
d=107%2F7
Then use either equation,
a%5B1%5D%2B7%28107%2F7%29=22
a%5B1%5D%2B107=22
a%5B1%5D=-85

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
a15 = 129, a8=22(a1) 

   ,   ,   ,   ,   ,   ,   ,  22,   ,   ,   ,   ,   ,   ,129 

Let d be the common difference.  Then since 129 is the 7th term after
22, one would have to add the common difference d seven times to get 129,
ao

22+7d=129
   7d=107
    d=107%2F7
    d=15%262%2F7

Now since 22 is the 7th term before the first term we would have to subtract
this common difference d seven times to get the first term,  

so the first term = 22-7%28107%2F7%29=22-107=-85 

First term = -85 and common difference = 15%262%2F7.

FYI, the whole sequence is:

-85,-69%265%2F7,-54%263%2F7,-39%261%2F7,-23%266%2F7,-8%264%2F7+++,6%265%2F7,+22,37%262%2F7,52%264%2F7,67%266%2F7,83%261%2F7+++,98%263%2F7,113%265%2F7,129

Edwin

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
We are working on Arithmetic Sequences. I am stuck.
Can someone please help. I need to find a1 and d.
a15 = 129, a8=22(a1)
Thanks.
If what you're saying is that: a%5B15%5D+=+129, and a%5B8%5D+=+22%28a%5B1%5D%29, then:

1st term, or highlight_green%28a%5B1%5D+=+3%29, and common difference, or highlight_green%28d+=+9%29



Since you confirmed that it is in fact: a%5B15%5D+=+129, and a%5B8%5D+=+22%28a%5B1%5D%29, then the solution follows:

a%5Bn%5D+=+a%5B1%5D+%2B+%28n+-+1%29d+

a%5B15%5D+=+a%5B1%5D+%2B+%2815+-+1%29d

a%5B15%5D+=+a%5B1%5D+%2B+14d

129+=+a%5B1%5D+%2B+14d ------ Substituting 129 for a%5B15%5D 

a%5B1%5D+=+129+-+14d ------- eq (i)



a%5Bn%5D+=+a%5B1%5D+%2B+%28n+-+1%29d 

a%5B8%5D+=+a%5B1%5D+%2B+%288+-+1%29d

a%5B8%5D+=+a%5B1%5D+%2B+7d

22%28a%5B1%5D%29+=+a%5B1%5D+%2B+7d ------ Substituting 22%28a%5B1%5D%29 for a%5B8%5D 

22a%5B1%5D+-+a%5B1%5D+=+7d 

21a%5B1%5D+=+7d ------- eq (ii)

21(129 – 14d) = 7d --------- Substituting 129 – 14d for a%5B1%5D in eq (ii)

2,709 – 294d = 7d

7d + 294d = 2,709

301d = 2,709

d, or common difference = 2709%2F301, or highlight_green%289%29



21a%5B1%5D+=+7%289%29 -------- Substituting 9 for d in eq (ii)

21a%5B1%5D+=+63

1st term, or a%5B1%5D+=+63%2F21, or highlight_green%283%29

You can do the check!! 

===================

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