A) 2 and 8
B) 1 and 9
C) 3 and 7
D) None
Correct Answer: B
Solution :
[b] As x, y, z, are A.M. of a and b |
\[\therefore \,\,\,x+y+z=3\left( \frac{a+b}{2} \right)\] |
\[\therefore \,\,\,\,\,15=\frac{3}{2}(a+b)\Rightarrow a+b=10\] ??. (1) |
Again \[\frac{1}{x},\frac{1}{y},\frac{1}{z}\] are A.M. of \[\frac{1}{a}\]and \[\frac{1}{b}\] |
\[\therefore \,\,\,\,\,\,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}\,\left( \frac{1}{a}+\frac{1}{b} \right)\] |
\[\therefore \,\,\,\,\,\,\frac{5}{3}=\frac{3}{2}.\frac{a+b}{ab}\] |
\[\Rightarrow \,\,\,\,\,\,\,\frac{10}{9}=\frac{10}{ab}\Rightarrow ab=9\] ??.. (2) |
Solving (1) and (2), we get |
\[a=9,\,\,b=1,\,\,9\] |
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