A) 9, 18, 27, 36, 45
B) 8, 24, 72, 216, 648
C) 4, 22, 38, 46
D) None of these
Correct Answer: B
Solution :
[b] Let \[b=ar,\,\,c=a{{r}^{2}}\] Given that \[12=\frac{2ab}{a+b}=\frac{2a.ar}{a+ar}=\frac{2ar}{1+r}\] or \[ar=6\,\,(1+r)\] ?..(1) Also, \[36=\frac{2bc}{b+c}=\frac{2.ar.a{{r}^{2}}}{ar+a{{r}^{2}}}=\frac{2a{{r}^{2}}}{1+r}\] \[\Rightarrow \,\,\,\,a{{r}^{2}}=18(1+r)\] ?..(2) Dividing (ii) by (i), we have \[\frac{a{{r}^{2}}}{ar}=\frac{18(1+r)}{6(1+r)}\Rightarrow r=3\] \[\therefore \] From (i), \[a\times 3=6(1+3)\Rightarrow a=\frac{6\times 4}{3}=8\] \[\therefore \] First five numbers are \[8,24,72,216,648.\]
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