Question 217318: The third term of an arithmetic sequence is 4 and the sum of the first 8 term is 36. Write down the first 8 term of the sequence.
Answer by drj(1380) (Show Source):
You can put this solution on YOUR website! The third term of an arithmetic sequence is 4 and the sum of the first 8 term is 36. Write down the first 8 term of the sequence.
Step 1. An arithmetic sequence goes from one term to the next by always adding (or subtracting) the same value
Step 2. Let x be the number to add or subtract in the sequence.
Step 3. Let a be the first term, let a+x be the second term, let a+2x be the third term, a+3x be the fourth term, and a+(n-1)x be the nth term
Step 4. Let a+7x be the 8th term. Then the sum of the first and eighth term is a+a+7x=2a+7x. The sum of the second term and seventh term is the a+x+a+6x=2a+6x. So four the first eight terms, there will be four pairs of 2a+6x which will equal to 36 or
Also Equation B since the 3th term is 4.
Step 5. Then we have a system of equations given as Equations A and B. The following steps will solve the equation by substitution.
|Solved by pluggable solver: SOLVE linear system by SUBSTITUTION|
We'll use substitution. After moving 7*x to the right, we get:
, or . Substitute that
into another equation:
and simplify: So, we know that x=0.333333333333333. Since , a=3.33333333333333.
So a= and x=
So sequence is: , , , , , , , .
Check: The third term is and the sum is
Step 6. ANSWER: The sequence is , , , , , , , .
I hope the above steps and explanation were helpful.
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