Particles of masses m, 2m, 3m, ….. nm grams are placed on the same line at distances l, 2l, 3l, …. nl cm from a fixed point. The distance of center of mass of the particles from the fixed point in centimeter is

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3+...............+{\mathrm{m}}_{\mathrm{n}}{\mathrm{x}}_{\mathrm{n}}}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+.........+{\mathrm{m}}_{\mathrm{n}}}$

            $=\frac{\mathrm{m}\left(l\right)+2\mathrm{m}\left(2\mathrm{l}\right)+3\mathrm{m}\left(3l\right)+...............+n\mathrm{m}\left(\mathrm{nl}\right)}{\mathrm{m}+2\mathrm{m}+3\mathrm{m}+.........+n\mathrm{m}}$

            $=\frac{\mathrm{ml}(1^2+2^2+3^2+........+{\mathrm{n}}^2)}{\mathrm{m}(1+2+3+.......+\mathrm{n})}$

            $=\mathrm{l}\left(\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\right)\times \frac{2}{\mathrm{n}\left(\mathrm{n}+1\right)}$

            $=\frac{\left(2\mathrm{n}+1\right)}{3}l$