algebra

if 1, 2, 7 and 20, respectively, are added to the first terms of an arithmetic progression, a geometric progression of four terms is obtained. find the first term and common difference of the arithmetic progression

the answers are both 3 .. but i don't know the solution, please help me, thank you

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asked by Sam
  1. let the first 4 terms of the AP be
    a-d, a , a+d, and a+2d

    new sequence is
    a-d+1, a+2, a+d + 7, and a+2d+20

    so (a+2)/(a-d+1) = (a+d+7)/(a+2)
    (a+2)^2 = (a-d+1)(a+d+7)
    a^2+4a+4 = a^2+ad+7a-ad-d^2-7d+7a+d+7
    -4a = -d^2 - 6d + 3
    4a = d^2 + 6d - 3 **

    (a+d+7)/(a+2) = (a+2d+20)/(a+d+7)
    (a+d+7)^2 = (a+2)(a+2d+20)
    a^2 + d^2 + 49 + 2ad +14a + 14d
    = a^2 + 2ad + 22a + 4d + 40

    8a = d^2 + 10d + 9 ***

    double **, then subtract ***
    0 = d^2 + 2d - 15 = 0
    (d + 5)(d-3) = 0
    d = -5, or d = 3

    if d = 3 in **
    4a = 3^2 + 18 - 3 = 24
    a = 6
    the original AP according to my definition , was 3, 6, 9, 12
    check: if I add the numbers as stated, I get 4, 8, 16, and 32 , which is a GP

    if d = -5, in **
    4a = 25 -30 - 3 = -8
    a = -2
    the original AP is 3, -2, -7, -12
    adding the numbers as stated will give me:
    4, 0, 0, 8 , but we can't have a 0 in a GP

    so we have to go with my first part of the solutions, which yielded the AP
    3, 6, 9, 12
    Making the first term 3, and the common difference as 3

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    posted by Reiny

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