SOLUTION: An arithmetic sequence a1,a2,......a100 has a sum of 15 000. Finf the first term and the common difference if the sum of the terms in the sequence a3, a6, a9,... a99 is 5016.

Algebra ->  Sequences-and-series -> SOLUTION: An arithmetic sequence a1,a2,......a100 has a sum of 15 000. Finf the first term and the common difference if the sum of the terms in the sequence a3, a6, a9,... a99 is 5016.      Log On


   

Question 1121963: An arithmetic sequence a1,a2,......a100 has a sum of 15 000. Finf the first term and the common difference if the sum of the terms in the sequence a3, a6, a9,... a99 is 5016.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52812) About Me  (Show Source):
You can put this solution on YOUR website!
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Use the formula for the sum of the first n terms of an arithmetic progression


S%5Bn%5D = a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+ellipsis+%2B+a%5Bn%5D = %28a%5B1%5D+%2B+%28n-1%29%2Ad%2F2%29%2An.


Since an arithmetic sequence a1,a2,......a100 has the sum of 15000, you have first equation


%28a%5B1%5D+%2B+%2899%2F2%29%2Ad%29%2A100 = 15000,   or


2a%5B1%5D+%2B+99d = 300.



The second progression is also an arithmetic progression with the first term of a%5B1%5D%2B2d  and the common difference of 3d.


Therefore, applying the same formula for the second progression (which has 33 terms), you get


%28a%5B1%5D%2B2d+%2B+%2832%2F2%29%2A%283d%29%29%2A33 = 5016,   or


a%5B1%5D+%2B+50d = 152.


Thus you have this system of 2 equations in 2 unknowns


2a%5B1%5D+%2B+99d = 300.    (1)

a%5B1%5D+%2B+50d = 152.     (2)


Multiply eq(2) by 2 (both sides) and then subtract the eq(1) from it. You will get


d = 304 - 300 = 4.


Then from (2)  a%5B1%5D = 152 - 50*4 = 152 - 200 = - 48.


Answer.   a%5B1%5D = -48;  d = 4.

Solved.

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    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - Mathematical induction and arithmetic progressions
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions


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The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".


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Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The sum of the first 100 terms is 15000, so the average of the terms is 15000/100 = 150.

In particular, the average of the first and last terms is 150; and the sum of the 50th and 51st terms is 150.

The sub-sequence has 33 terms with a sum of 5016, so the average is 5016/33 = 152.

In particular, the 17th (middle) term alone is 152.

The 17th term of the sub-sequence is the 51st term of the full sequence.

So we know the average of the 50th and 51st terms is 150; and we know that the 51st term is 152. That means the 50th term is 148.

With the 50th term 148 and the 51st term 152, the common difference is 4.

With a common difference of 4 and 51st term 152, the first term is 152-50(4) = 152-200 = -48.

Answer: first term -48; common difference 4.

CHECK:
a1+a2+...+a100 = (-48)+(-44)+...+(348) = 100((-48)+348)/2 = 100(150) = 15000
a3+a6+...+a99 = (-40)+(-28)+...+344) = 33((-40)+344)/2 = 33(152) = 5016