(N/A) Let $R_{1}, R_{2},$ and $R_{3}$ be three resistors connected in series with a battery of potential $V$ between points $A$ and $B$. The current flowing through the circuit is $I$.
The potential difference across $R_{1}, R_{2},$ and $R_{3}$ are $V_{1}, V_{2},$ and $V_{3}$ respectively. According to Ohm's law,$V_{1}=IR_{1}, V_{2}=IR_{2},$ and $V_{3}=IR_{3}$.
The total terminal voltage of the battery is the sum of individual potential drops:
$V = V_{1} + V_{2} + V_{3}$
Substituting the values:
$V = IR_{1} + IR_{2} + IR_{3}$
$V = I(R_{1} + R_{2} + R_{3})$
Dividing by $I$:
$\frac{V}{I} = R_{1} + R_{2} + R_{3}$
Since $\frac{V}{I} = R_{eq}$ is the equivalent resistance of the series combination:
$R_{eq} = R_{1} + R_{2} + R_{3}$
For $n$ resistors connected in series,the equivalent resistance is:
$R_{eq} = R_{1} + R_{2} + \ldots + R_{n}$
If $n$ resistors of equal value $R$ are connected in series,then:
$R_{eq} = nR$
In a series connection,the equivalent resistance is always greater than the largest individual resistance.