A) \[3\]
B) \[4\]
C) \[5\]
D) \[6\]
Correct Answer: B
Solution :
[b] \[xyz=\frac{15}{2}\] if a, x, y, z, b are in A.P. Common difference \[=d=\frac{b-a}{n-1}=\frac{b-a}{4}\] \[\Rightarrow \,\,\,\,x=\frac{3a+b}{4},\,\,y=\frac{a+b}{2}\]and \[z=\frac{3b+a}{4}\] \[\Rightarrow \,\,\,\,\,\,\,xyz=\frac{15}{2}\] \[\left( \frac{3a+b}{4} \right)\,\left( \frac{a+b}{2} \right)\left( \frac{3b+a}{4} \right)=\frac{15}{2}\] ?.(1) Also, \[xyz=\frac{18}{5}\]if a, x, y, z, b are in H.P. \[\Rightarrow \,\,\,\,\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}\] are m A.P. \[\Rightarrow \,\,\,\,x=\frac{4ab}{3b+4},\,\,y=\frac{2ab}{a+b}\] and \[z=\frac{4ab}{3a+b}\] \[\Rightarrow \,\,\,\,xyz=\frac{18}{5}\] \[\left( \frac{4ab}{3b+a} \right)\,\left( \frac{2ab}{a+b} \right)\left( \frac{4ab}{3a+b} \right)=\frac{18}{5}\] ? (2) Multiplying equations (1) and (2), we get \[{{a}^{3}}{{b}^{3}}=\frac{18}{5}\times \frac{15}{2}=27\] \[\Rightarrow \,\,\,\,ab=3\] \[\Rightarrow \,\,\,\,\,a=1,3\] and \[\Rightarrow \,\,\,\,\,b=3,1\]
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