A) 5, 9, 13
B) 15, 9, 3
C) 13, 9, 5
D) 17, 9, 1
Correct Answer: D
Solution :
Let the three terms of the series is \[a+d,\,\,a,\,\,a-d.\] \\[a+d+a+a-d=27\] Þ \[3a=27\]Þ \[a=9\] Now, \[(a+d-1),(a-1),(a-d+3)\]are in G.P. \[\Rightarrow \]\[{{(a-1)}^{2}}=(a+d-1)(a-d+3)\] \[\Rightarrow 64=(8+d)(12-d)\]\[\Rightarrow 64=-{{d}^{2}}+4d+96\] \[\Rightarrow {{d}^{2}}-4d-32=0\]\[\Rightarrow {{d}^{2}}-8d+4d-32=0\] Þ \[(d-8)(d+4)=0\], \[\therefore d=-4,\,8\] Series is 5, 9, 13 (for d = - 4) and 17, 9, 1 (for d = 8) \ Decreasing A.P. is 17, 9, 1.
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