Explanation

(d)
Since, $a,b,c$ are in AP.
(A)$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in HP $\Rightarrow$ $\frac{k}{a},\frac{k}{b},\frac{k}{c}$ are in HP.
(B)$a + k,\ b + k,\ c + k$ are in AP.
(C )$ka,kb,kc$ are in AP.
(D)$a^{2},b^{2},c^{2}$ are in AP.
Then, $b^{2} - a^{2} = c^{2} - b^{2}$
$\therefore(b - a)(b + a) = (c - b)(c + b) = (b - a)(c + b)$
$\left\lbrack \begin{array}{r}
\because a,b,c\ are\ in\ AP \\
\therefore b - a = c - b
\end{array} \right\rbrack$
$\Rightarrow \ b + a = c + b$
$\Rightarrow \ a = c$, which is not true