A) 1
B) 3
C) 5
D) 6
Correct Answer: C
Solution :
Let \[{{A}_{j}},\ {{H}_{j}},\] where \[j=1,\ 2,\ 3,.......9\] denote the 9 A.M.?s and H.M.?s between 2 and 3. Then \[2,{{A}_{1}},{{A}_{2}}........{{A}_{9}},3\] are in A.P. Let \[d\] be the common difference of this A.P. Then \[-d=\frac{{{a}_{1}}-{{a}_{2n}}}{2n-1}\]\[\Rightarrow \]\[d=\frac{1}{10}\] If\[A\] denotes the \[{{J}^{th}}\] arithmetic mean, then \[A=2+jd=2+\left( \frac{j}{10} \right)\] Again \[2,\ {{H}_{1}},\ {{H}_{2}}.......,{{H}_{9}},\ 3\] will be in H.P. \[i.e.\], \[\frac{1}{2},\frac{1}{{{H}_{1}}},\frac{1}{{{H}_{2}}},......\frac{1}{{{H}_{9}}},\ \frac{1}{3}\] will be in A.P. Let \[D\] be the common difference of this A.P. Then \[\frac{1}{3}=\frac{1}{2}+10D\]\[\Rightarrow \]\[D=-\frac{1}{60}\] If \[H\] be the \[{{J}^{th}}\] harmonic mean, then \[\frac{1}{H}=\frac{1}{2}+jD=\frac{1}{2}-\frac{j}{60}\] \[\therefore \]\[A+\frac{6}{H}=2+\frac{j}{10}+6\left( \frac{1}{2}-\frac{j}{60} \right)=5+\frac{j}{10}-\frac{j}{10}=5\]. Aliter: As we know \[{{A}_{m}}=2+\frac{m(3-2)}{9+1}=2+\frac{m}{10}\] and \[\frac{1}{{{H}_{m}}}=\frac{1}{2}+\frac{m(2-3)}{2\times 3(9+1)}=\frac{1}{2}-\frac{m}{60}\] \[\therefore \]\[{{A}_{m}}+6\times \frac{1}{{{H}_{m}}}\ \ i.e.\ \ A+\frac{6}{H}=2+\frac{m}{10}+3-\frac{m}{10}=5\].
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