\(\cdot\) Weights are suspended at a distance \(=1 \mathrm{cm}, 2 \mathrm{cm}, 3 \mathrm{cm}, \ldots \ldots .100 \mathrm{cm}\)

respectively to the masses.

For the system to be in equilibrium the point must lie in the center of mass.

So center of mass of the system can be calculated as

\(\mathrm{COM}=\frac {\left(1^{2}+2^{2}+3^{2} \ldots 100^{2}\right)} {(1+2+3 \ldots+100)}\)

\(COM = \left[ {\frac{{\frac{{\left( {100} \right)\left( {101} \right)\left( {201} \right)}}{6}}}{{\frac{{\left( {100} \right)\left( {101} \right)}}{2}}}} \right]\)

\(\mathrm{COM}=\frac {201}{3}=67 \mathrm{cm}\)

Hence at \(67 \mathrm{cm}\) from the origin or \(66 \mathrm{cm}\) from the first particle, at that point system will be supported to get equilibrium.