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20 June, 09:52

In a G. P., the third term exceeds the first term by 16.

If the sum of the third term and the fourth term is 72,

find the common ratios.

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  1. 20 June, 11:55
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    Step-by-step explanation:

    In a G. P, the nth term is given as

    Un=ar^ (n-1)

    Where

    a is first term

    n is nth term

    And r is common ratio

    So in the question given above,

    The third term exceed the first term by 16

    i. e U3=U1+16

    Where U1=a. First term

    U3=a+16

    Given also that, the sum if the third term and fourth term is 72.

    Then U3+U4=72.

    We are told to find common ratio (r)

    U3=ar^3-1

    U3=ar^2

    Also, U4=ar^3

    U3+U4=72

    ar^2+ar^3=72

    ar^2 (1+r) = 72. equation 1

    Also for

    U3=a+16

    ar^2=a+16

    ar^2-a=16

    a (r^2-1) = 16. From (x^2-y^2) = (x+y) (x-y)

    Then,

    a (r-1) (r+1) = 16. Equation 2

    Divide equation 2 by equation 1

    a (r-1) (r+1) / ar^2 (1+r) = 16/72

    Then a cancel a and (1+r) cancel (1+r)

    So,

    (r-1) / r^2=2/9

    Cross multiply

    9 (r-1) = 2r^2

    2r^2-9r+9=0

    Solving the quadratic equation

    2r^2-6r-3r+9=0

    2r (r-3) - 3 (r-3) = 0

    (r-3) (2r-3) = 0

    r-3=0. Or. 2r-3 = 0

    Then r=3 or r=3/2
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