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Question

The sum of three numbers which are consecutive terms an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.


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    Solution

    The correct option is

    Let the first term of an A.P. is a and its common difference be d.

    a1+a2+a3=21

    a+(a+d)+(a+2d)=21

    3a+3d=21

    a+d=7 ..........(i)

    Now, according to the questiong :

    a,a+d1 and a+2d+1 are in G.P.

    (a+d1)2=a(a+2d+1)

    (7+aa1)2=a[a+2(7a)+1]

    (6)2=a(15a)

    36=15aa2

    a215a+36=0

    (a3)(a12)=0

    a=3,12

    Now, putting a = 2, 12 in equation (i), we get d = 5, -5, respectively.

    Thus, for a = 2 and d = 5, the numbers are 2, 7 and 12.

    And, for a = 12 and d = -5, the numbers are 12, 7 and 2.


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