Explanation

(a)
We have, $x + y + \mathcal{z} = 15$, if $9,x,\ y,\ \mathcal{z,\ }a$ are in AP.
$\therefore\$ Sum = $9 + 15 + a = \frac{5}{2}(9 + a)$
$\Rightarrow \ \ 24 + a = \frac{5}{2}(9 + a)$
$\Rightarrow \ \ 48 + 2a = 45 + 5a$
$\Rightarrow \ \ 3a = 3\ \Rightarrow a = 1$ …(i)
And $\frac{1}{x} + \frac{1}{y} + \frac{1}{\mathcal{z}} + \frac{5}{3},$ if $9,x,\ y,\ \mathcal{z,\ }a$ are in HP.
$Sum = \frac{1}{9} + \frac{5}{3} + \frac{1}{a} = \frac{5}{2}\left\lbrack \frac{1}{9} + \frac{1}{a} \right\rbrack\ \Rightarrow a = 1$