A) 1 : 2 : 3
B) 2: 3 : 1
C) 1 : 3 : 2
D) 3 : 2 : 1
Correct Answer: A
Solution :
a, b, c are in A.P. Þ a + c = 2b ....(i) Also, \[b-a,\,c-b,\,a\] are in G.P. Þ \[{{(c-b)}^{2}}=(b-a)a\] Þ \[(b-a)\,(c-b)\,=(b-a)\,a\]\[(\because \,\,c-b=b-a\] as a, b, c are in A.P.) Þ \[c-b=a\] \[(\because \,\,a\ne b)\] Þ \[b=c-a\] ?..(ii) From (i) and (ii), \[a=\frac{b}{2}\text{ and }c=\frac{3b}{2}\] \ \[a:b:c::\frac{b}{2}:b:\frac{3b}{2}\] Þ \[a:b:c::1:2:3\].
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