Explanation

(c)
Let $A_{j},\ H_{j}$ where $j = 1,\ 2,\ 3,\ldots,9$ denote the 9 AM’s and HM’s between 2 and 3.
Then $2,\ A_{1},\ A_{2},\ A_{3},\ldots,A_{9},$ 3 are in AP, let $d$ be the common difference of this AP, then
$3 = 2 + 10d\ \Rightarrow d = \frac{1}{10}$
If $A$ denotes the $j$th arithmetic mean, then
$A = 2 + jd = 2 + \left( \frac{j}{10} \right)\ \ \ \ \ \ \left( \because d = \frac{1}{10} \right)\ldots$(i)
Again, $2,\ H_{1},\ H_{2},\ldots,H_{9},\ 3$ will be in HP.
$\Rightarrow \frac{1}{2},\frac{1}{H_{1}},\frac{1}{H_{2}},\ldots\frac{1}{H_{9}},\frac{1}{3}$ will be in AP
Let $d$ be the common difference of this AP, then
$\frac{1}{3} = \frac{1}{2} + 10d\ \Rightarrow d = \frac{- 1}{60}$
If $H$ be the $j$th harmonic mean, then
$\frac{1}{H} = \frac{1}{2} + jd = \frac{1}{2} - \frac{j}{60}$ …(ii)
$\therefore\ A + \frac{6}{H} = 2 + \frac{j}{10} + 6\left( \frac{1}{2} - \frac{j}{60} \right)$ [from Eqs.(i) and (ii)]
$= 5 + \frac{j}{10} - \frac{j}{10} = 5$