SOLUTION: Hello. Can someone please help me with this series problem. I need to find {{{a[1] }}} and {{{d }}} Given: {{{a[7]=49}}} and {{{a[3] + a[9] = 82 }}} Thank you.

Algebra ->  Sequences-and-series -> SOLUTION: Hello. Can someone please help me with this series problem. I need to find {{{a[1] }}} and {{{d }}} Given: {{{a[7]=49}}} and {{{a[3] + a[9] = 82 }}} Thank you.      Log On


   

Question 937865: Hello.
Can someone please help me with this series problem.
I need to find a%5B1%5D+ and d+
Given:
a%5B7%5D=49 and a%5B3%5D+%2B+a%5B9%5D+=+82+
Thank you.
Found 3 solutions by josgarithmetic, MathTherapy, MathLover1:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Symbolism suggests Arithmetic Sequence, with a%5B1%5D as the first term, a common difference d, and general term a%5Bn%5D=a%5B1%5D%2B%28n-1%29d.


Use the formula to rewrite your given equations.

a%5B1%5D%2B%287-1%29d=7 and a%5B1%5D%2B%283-2%29d%2Ba%5B1%5D%2B%289-1%29d=82.

Simplify and solve the system.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Hello.
Can someone please help me with this series problem.
I need to find a%5B1%5D+ and d+
Given:
a%5B7%5D=49 and a%5B3%5D+%2B+a%5B9%5D+=+82+
Thank you.
1st term, or highlight_green%28a%5B1%5D+=+1%29

Common difference, or highlight_green%28d+=+8%29      

You can do the check!! 

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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

The n%5Bth%5D term of an arithmetic sequence is:
a%5Bn%5D+=+a%5B1%5D++%2B+%28n-1%29d
where a%5B1%5D is the first term in the sequence, d is the common difference between terms and n is the number of the term to find
given:
a%5B7%5D=49 and a%5B3%5D+%2B+a%5B9%5D+=+82+
use a%5Bn%5D+=+a%5B1%5D++%2B+%28n-1%29d to find a%5B3%5D+ and a%5B9%5D+ expressed in terms of a%5B1%5D+ and d+
=> a%5B3%5D+=a%5B1%5D++%2B+2d
=> a%5B9%5D+=a%5B1%5D++%2B+8d
=> a%5B1%5D++%2B+2d%2Ba%5B1%5D++%2B+8d=82
=> 2a%5B1%5D++%2B+10d=82
=> 2a%5B1%5D++=82-+10d
=> a%5B1%5D++=41-+5d

find:
a%5B1%5D+ and d+

a%5Bn%5D+=+a%5B1%5D++%2B+%28n-1%29d if a%5B7%5D=49 we have
49+=+a%5B1%5D++%2B+%287-1%29d ....substitute 41-+5d for a%5B1%5D
49+=+41-+5d%2B+6d
49+-+41=d
highlight%28d=8%29

=> a%5B1%5D++=41-+5%2A8
=> a%5B1%5D++=41-+40
=> highlight%28a%5B1%5D++=1%29