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.

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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)   (Show Source): You can put this solution on YOUR website!

.
.


Subtract the two values,



Then use either equation,
Answer by Edwin McCravy(20056)   (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=
    d=

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 =  

First term = -85 and common difference = .

FYI, the whole sequence is:

,,,,,,,,,,,,,,

Edwin
Answer by MathTherapy(10552)   (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: , and , then:

1st term, or , and common difference, or 



Since you confirmed that it is in fact: , and , then the solution follows:







 ------ Substituting 129 for  

 ------- eq (i)



 





 ------ Substituting  for  

 

 ------- eq (ii)

21(129 – 14d) = 7d --------- Substituting 129 – 14d for  in eq (ii)

2,709 – 294d = 7d

7d + 294d = 2,709

301d = 2,709

d, or common difference = , or 



 -------- Substituting 9 for d in eq (ii)



1st term, or , or 

You can do the check!! 

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