A) 7,14
B) 10,14
C) 12, 7
D) 14,12
Correct Answer: A
Solution :
Clearly, X is a binomial variate. with parametres n and \[\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\] such that \[3x+2y-2z+15=0,\] \[{{u}_{1}}\] Now, P (X = 4), P (X = 5) and P (X = 6) are in AP. \[{{u}_{2}}\]\[{{u}_{1}}\]and\[{{u}_{2}}\]are in AP. \[{{u}_{2}}\]\[{{u}_{1}}\] \[{{u}_{1}}\]\[{{u}_{2}}\] \[{{u}_{2}}\]\[{{u}_{2}}\] \[\frac{13}{30}\]\[\frac{23}{30}\] \[\frac{19}{30}\]\[\frac{11}{30}\] \[{{D}_{k}}=\left| \begin{matrix} a & {{2}^{k}} & {{2}^{16}}-1 \\ b & 3({{4}^{k}}) & 2({{4}^{16}}-1) \\ c & 7({{8}^{k}}) & 4({{8}^{16}}-1) \\ \end{matrix} \right|,\]\[\sum\limits_{k=1}^{16}{{{D}_{k}}}\]\[{{x}^{2}}-4x+4{{y}^{2}}=12\]\[\frac{\sqrt{3}}{2}\] \[\frac{2}{\sqrt{3}}\] n = 7 or 14
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