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Q)

The sum of three decreasing numbers in AP is 27. If -1, -1, and 3 are added to them respectively the resulting series is in GP. The numbers are

$(a)\;17,9,1\qquad(b)\;16,9,2\qquad(c)\;13,9,5\qquad(d)\;5,9,13$

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A)
Answer : (a) 17,9,1
Explanation : Let the 3 numbers in AP be a+d,a and a-d (as the numbers are decreasing)
Sum is a+d+a+a-d=3a=27 , a=9
As the three numbers are in GP when -1,-1 and 3 are added,
$(9+d-1)\;(9-d+3)=(9-1)^2$
$(8+d)\;(12-d)=64$
$96+4d-d^2=64$
$d^2-4d-32=0$
$(d+4)(d-8)=0$
$d=-4,8$
Putting the value of d , The series is 9-4 ,9, 9+4
5,9,13 but this is increasing
Therefore d=8 , series is 9+8 , 9 , 9-8
Answer is 17 ,9 , 1.
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