Two springs of spring constants ${\mathrm{k}}_1 \mathrm{and} {\mathrm{k}}_2$ are joined in series. The effective spring constant of the-combination is given by:

Let us consider two springs of spring constants ${\mathrm{k}}_1 \mathrm{and} {\mathrm{k}}_2$ joined in series as shown in the figure.

Under a force F, they will stretch by ${\mathrm{y}}_1 \mathrm{and} {\mathrm{y}}_2$.
So $\mathrm{y} = {\mathrm{y}}_1+{\mathrm{y}}_2$
or $\frac{\mathrm{F}}{\mathrm{k} } = \frac{{\mathrm{F}}_1}{{\mathrm{k}}_1}+\frac{{\mathrm{F}}_2}{{\mathrm{k}}_2}$
but as springs are massless, so force on them must be the same, i,e., ${\mathrm{F}}_1 = {\mathrm{F}}_2 = \mathrm{F}$
So, $\frac{1}{\mathrm{k}} = \frac{1}{{\mathrm{k}}_1}+\frac{1}{{\mathrm{k}}_2 } \mathrm{or} \mathrm{k} = \frac{{\mathrm{k}}_1{\mathrm{k}}_2}{{\mathrm{k}}_1+{\mathrm{k}}_2}$

Aliter: In series combination,
$\frac{1}{{\mathrm{k}}_{\mathrm{S}}} = \frac{1}{{\mathrm{k}}_1}+\frac{1}{{\mathrm{k}}_2} = \frac{{\mathrm{k}}_2+{\mathrm{k}}_1}{{\mathrm{k}}_1{\mathrm{k}}_2 } \Rightarrow {\mathrm{k}}_{\mathrm{S}} = \frac{{\mathrm{k}}_1{\mathrm{k}}_2}{{\mathrm{k}}_1+{\mathrm{k}}_2}$