A) 2, 4, 8
B) 4, 8, 16
C) 3, 6, 12
D) None of these
Correct Answer: B
Solution :
Let three terms of a G.P. are \[\frac{a}{r},\ a,\ ar\] So \[\frac{a}{r}.\ a.\ ar=512\]\[\Rightarrow \]\[{{a}^{3}}={{8}^{3}}\]\[\Rightarrow \]\[a=8\] From second condition, we get \[\frac{a}{r}+8,\ a+6,\ ar\] will be in A.P. \[\Rightarrow \] \[2(a+6)=\frac{a}{r}+8+ar\]\[\Rightarrow \]\[28=8\left\{ \frac{1}{r}+1+r \right\}\] \[\Rightarrow \] \[\frac{1}{r}+r+1=\frac{7}{2}\]\[\Rightarrow \]\[\frac{1}{r}+r-\frac{5}{2}=0\] \[\Rightarrow \] \[{{r}^{2}}-\frac{5}{2}r+1=0\]\[\Rightarrow \]\[2{{r}^{2}}-5r+2=0\] \[\Rightarrow \] \[(2r-1)(r-2)=0\]\[\Rightarrow \]\[r=\frac{1}{2},\ r=2\] \[(\because \ r>1)\] \[\Rightarrow \] \[r=2\]. Hence required numbers are \[4,\ 8,\ 16\]. Trick: Check for (a) \[2+8,\ 4+6,\ 8\] are not in A.P. (b) \[4+8,\ 8+6,\ 16\ i.e.\ 12,\ 14,\ 16\] are in A.P.
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