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Q4 - Arithmetic and Geometric series

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Let a1, a2, a3 denote the first three terms of a geometrical sequence, for which a1 + a2 + a3 = 26.

    a1 + 3, a2 + 4, a3 - 3 are the first three terms of an arithmetical sequence.

    Find the first term and the common quotient (ratio) of the geometrical sequence!

    2. Relevant equations

    Arithmetical sequences:

    an = a1 + (n-1)d
    Sn = (n/2)(a1 + an)

    Geometric sequences:

    an = a1rn-1
    Sn = a1 * ((r-1n)/(r-1))

    3. The attempt at a solution

    By taking the sum of the given arithmetical sequence:

    (a1 + a2 + a3) + 3 + 4 -3 = 26 +4 = 30

    We know that:

    Sn = (n/2)(a1 + an)
    30 = (3/2)(a1 + 3 + a3 - 3) > (3/2)(a1 + a3)
    30 = (3/2)(20)

    Thus a1 + a3 = 20

    Because of the given

    a1 + a2 + a3 = 26

    This implies

    (a1 + a3) + a2 = 26
    20 + a2 = 26
    a2 = 6

    Now, I found the second term of each sequence, however, I do not know how to go forward from this point. The four functions given for arithmetic and geometric sequences do not seem to help me any further because there are always two unknown variables!
     
  2. jcsd
  3. Aug 24, 2010 #2

    ehild

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    You do not need the sum of the arithmetic sequence, but it might be useful that the middle term is the arithmetic mean of the neighbours.

    ehild
     
  4. Aug 25, 2010 #3
    I might be missing some important understandig about the mean, but I cannot resolve the problem any further with the help of this hint. I have been breaking my head over it a while longer, but do not see where this fits in.

    Could you perhaps provide me with another hint? Thanks!
     
  5. Aug 25, 2010 #4

    eumyang

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    In other words, the middle term of the arithmetic sequence, a2 + 4, is the average of the two other terms a1 + 3 and a3 - 3.


    69
     
  6. Aug 25, 2010 #5

    ehild

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    You know that the first three terms of the geometric sequence is a1, a1r, a1r2, and their sum is 26:

    a1+a1r+a1r2=26. *

    There is the arithmetical sequence, with first term

    b1=a1+3,

    second term

    b2=a2+4=a1r+4,

    and third term

    b3=a3-3=a1r2-3.

    By definition,

    b2-b1=b3-b2, which is equivalent by

    b3+b1=2b2

    so

    a1+3+a1r2-3=2(a1r+4). **

    You have two unknowns (a1 and r) and two equations. Solve.
     
  7. Aug 29, 2010 #6
    Now I got the following problem. I tried to use the logical steps you provided in the previous problem but I think I am missing something.

    Problem
    The sum of the first three terms of a geometrical sequence is 26 and their product is 216. Find the sum of the first four terms!

    so we have

    a + ar + ar2 = 26
    a * ar * ar2 = 216

    where a is the first term

    we can get r in terms of a by putting

    a = 26 * ((1-r)/1-r2))

    Now it is possible to solve the equation

    a3r3 = 216 by putting the value we found for a

    However, this results in a very long and tiresome caclulation ending up with the following:

    2170r6 - 6591r5 + 6510r4 - 2197r3 - 81r2 + 27 = 0

    I suppose I could solve this equation with synthetic division and then checking for which values the original equations holld, but I feel I got on the wrong track real soon here. Help greatly appreciated.

    Cheers,
     
  8. Aug 29, 2010 #7

    ehild

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    You made a mistake: a = 26 * ((1-r)/1-r^3)). But it is better to keep the first equation in the original form and start with the second one.

    The problem is not that terrible, as you can take the cubic root of the second equation: ar=6

    ehild
     
    Last edited: Aug 29, 2010
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