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

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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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