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84. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    84. Let a, ar and ar2 be the three nos. which is in G.P.

    Then, a + ar + ar2 = 56

    a ( 1 + r + r2) =56  -I

    Given, that a1, ar 7, ar2 - 21 from an AP we have,

    (ar−7)−(a−1)=(ar2−21)−(ar−7)

    ⇒ar−7−a+1=ar2−21−ar+7

    ⇒ar−a−6=ar2−ar−14

    ⇒ar2−ar−ar+a−a−14−6

    ⇒ar2−2ar+a=8

    ⇒ a(r2−2r+1)=8 ………………. II

    Now, dividing equation I by II we get,

    ⇒a(1+n+n2)a(r2−2r+1)=568

    ⇒1+r+r2=7(n2−2n+1)

    ⇒1+r+r2=7r2−14n+7

    ⇒7r2−14n+7−1−x−r2=0

    ⇒6r2−15r+6=0

    ⇒2r2−5r+2=0 (dividing by 3 throughout)

    ⇒2r2−4r−r+2=0

    ⇒2r(r−2)−(r−2)=0

    ⇒(r−2)(2r−1)=0

    ⇒r−2=02r−1=0

    ⇒r=2r=12

    So, when r = 2, putting in equation I,

    ⇒a(1+2+22)=56

    ⇒a(1+2+4)=56

    ⇒a(7)=56

    ⇒a=567=8

    The numbers are 8, 8× 2, 8× 22 = 8, 16, 32.

    And When r=12 putting in equation I,

    a(1+12+122)=56

    ⇒a(1+12+14)=56

    ⇒a(4+2+14)=56

    ⇒a×74=56

    ⇒a=56×47=32

    So, the numbers are 32,32×12,32×(12)2⇒32,16,8

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