Springs in Series

Consider three springs in series are attached to a fixed point on a ceiling as shown in the figure. How much force is required to stretch the springs 5 m?

Hint
The force in a spring is:
$$$F_s=k\delta $$$
where $$k$$ is the spring constant, and $$\delta$$ is the change in spring length from the un-deformed spring length.
Hint 2
The spring constant for springs in series:
$$$\frac{1}{k_{eq}}=\sum_{i}\frac{1}{k_i}$$$
First, let’s determine the equivalent spring constant for several springs in series:
$$$\frac{1}{k_{eq}}=\sum_{i}\frac{1}{k_i}$$$
Since the problem statement has three springs in series:
$$$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}=\frac{1}{1N/m}+\frac{1}{2N/m}+\frac{1}{3N/m}$$$
$$$\frac{1}{k_{eq}}=\frac{6}{6N/m}+\frac{3}{6N/m}+\frac{2}{6N/m}=\frac{11m}{6N}$$$
$$$\frac{1}{\frac{1}{k_{eq}}}=\frac{1}{\frac{11m}{6N}}$$$
$$$k_{eq}=\frac{6}{11}\:N/m$$$
A spring’s deflection and force are related by:
$$$F_s=k_{eq}\delta $$$
where $$F_s$$ is the forced applied to the spring, $$k_{eq}$$ is the equivalent spring constant, and $$\delta$$ is the change in spring length from the un-deformed spring length.
Therefore, the force required to stretch the springs 5 m is:
$$$F=\frac{6N}{11m}\times 5m=2.7\:N$$$
2.7 N