Consider the chemical reaction scheme shown in the given figure....

Reaction of acetic acid with ammonia followed by heating yields acetamide.

$\text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2$ (X)

Acetamide (X) undergoes Hoffmann bromamide degradation upon reaction with $\text{Br}_2$ and $\text{NaOH}$ to form a primary amine with one carbon atom less, which is methanamine.

$\text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2, \text{NaOH}} \text{CH}_3\text{NH}_2$ (Y)

Acetamide (X) undergoes reduction with $\text{LiAlH}_4$ to form a primary amine with the same number of carbon atoms, which is ethanamine.

$\text{CH}_3\text{CONH}_2 \xrightarrow{\text{(i) LiAlH}_4, \text{ether} \text{ (ii) H}_2\text{O}} \text{CH}_3\text{CH}_2\text{NH}_2$ (Z)

Y is methanamine ($\text{CH}_3\text{NH}_2$) and Z is ethanamine ($\text{CH}_3\text{CH}_2\text{NH}_2$).

Both are primary amines and differ by a $-\text{CH}_2-$ unit. Therefore, they are successive members of the same homologous series.

Both Y and Z will give a positive carbylamine test since they are primary amines.

Answer: $Y$ and $Z$ are successive members of the same homologous series.