A) Greater than or equal to 10
B) Less than or equal to 10
C) Only equal to 10
D) None of these
Correct Answer: D
Solution :
We have \[\frac{1}{a},\ \frac{1}{b},\ \frac{1}{c}\] are in A.P. Let \[\frac{1}{a}=p-q,\ \frac{1}{b}=p\] and \[\frac{1}{c}=p+q\], where \[p,\ q>0\]and \[p>\ q\]. Now, substitute these values in \[\frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b}\] then it reduces to \[10+\frac{14{{q}^{2}}}{{{p}^{2}}-{{q}^{2}}}\] which is obviously greater than\[10(\text{as}\ p>q>0)\]. Trick: Put\[a=1,\ b=\frac{1}{2},\ c=\frac{1}{3}\]. The expression has the value\[\frac{3+1}{2-\frac{1}{2}}+\frac{1+1}{\frac{2}{3}-\frac{1}{2}}=\frac{8}{3}+12>10\].
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