If $$m$$ is the A.M. of two distinct real numbers $$l$$ and $$n ( l, n > 1)$$   and $${{{G}}_1}{{,}}{{{G}}_2}$$  and $${{{G}}_3}$$ are three geometric means between $$l$$ and $$n,$$ then $${{G}}_1^4 + {{2G}}_2^4{{ + }}{{G}}_3^4$$    equals:

Solution :
$$\eqalign{ & m = \frac{{l + n}}{2}{\text{ and common ratio of G}}{\text{.P}}{\text{. }} = r = {\left( {\frac{n}{l}} \right)^{\frac{1}{4}}} \cr & \therefore \,\,{{{G}}_1} = {l^{\frac{3}{4}}}{n^{\frac{1}{4}}},{{{G}}_2} = {l^{\frac{1}{2}}}{n^{\frac{1}{2}}},{{{G}}_3} = {l^{\frac{1}{4}}}{n^{\frac{3}{4}}} \cr & \,\,\,\,{{G}}_1^4 + 2{{G}}_2^4 + {{G}}_3^4 = {l^3}n + 2{l^2}{n^2} + l{n^3} \cr & = ln{\left( {l + n} \right)^2} \cr & = \,ln \times 2{m^2} \cr & = 4l{m^2}n \cr} $$