A) \[a=3,\,\,b=12\]
B) \[a=12,\,b=3\]
C) \[a=2,\,b=32\]
D) \[a=4,\,\,b=16\]
Correct Answer: C
Solution :
\[\alpha +\beta =3;\alpha \beta =a;\] \[\gamma +\delta =+12;\gamma \delta =b\] \[\alpha ,\beta ,\gamma ,\delta \]are in increasing G.P. \[\beta =\alpha x,\gamma =\alpha {{x}^{2}},\delta =\alpha {{x}^{3}}\] \[\alpha +\beta =\alpha +\alpha x=3=\alpha (1+x)\] ?(1) \[\gamma +\delta \alpha {{x}^{2}}+\alpha {{x}^{3}}=12=\alpha {{x}^{2}}(1+x)\] ?(2) Dividing \[\frac{3}{12}=\frac{\alpha (1+x)}{\alpha {{x}^{2}}(1+x)}\]or \[\frac{1}{4}=\frac{1}{{{x}^{2}}}\]or x = 2 \[\Rightarrow \] \[\beta =2\alpha \]and \[\alpha +2\alpha =3\Rightarrow \alpha =1\]and \[\beta =2\] \[\therefore \] \[a=\alpha \beta =2\] \[\gamma =\alpha {{x}^{2}}=1\times {{2}^{2}}=4;\delta =\alpha {{x}^{3}}=1\times {{2}^{3}}=8\] \[\therefore b=\gamma \delta =4\times 8=32\]
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