(D) Let $A = \{1, 2, 3, 4, 5, 6\}$.
$A$ relation $R$ is defined on set $A$ as: $R = \{(a, b) : b = a + 1\}$.
$\therefore R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$.
For $R$ to be reflexive,$(a, a)$ must be in $R$ for all $a \in A$. However,$(1, 1) \notin R$,$(2, 2) \notin R$,etc.
$\therefore R$ is not reflexive.
For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a)$ must be in $R$. Here,$(1, 2) \in R$,but $(2, 1) \notin R$.
$\therefore R$ is not symmetric.
For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c)$ must be in $R$. Here,$(1, 2) \in R$ and $(2, 3) \in R$,but $(1, 3) \notin R$.
$\therefore R$ is not transitive.
Hence,$R$ is neither reflexive,nor symmetric,nor transitive.