A) 1
B) 2
C) 3
D) None of these
Correct Answer: C
Solution :
Let the numbers be\[a-d,\ a,\ a+d\]. Then \[{{(a-d)}^{2}},\ {{a}^{2}},\ {{(a+d)}^{2}}\] are in G.P. \[\therefore \]\[{{a}^{4}}={{(a-d)}^{2}}{{(a+d)}^{2}}\]\[\Rightarrow \]\[{{d}^{4}}-2{{a}^{2}}{{d}^{2}}=0\] \[\Rightarrow \]\[d=0,\ \pm \sqrt{2}a\] Hence \[d\] has three values.
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