Question

If m is the A.M. of two distinct real numbers $$\;l\;$$ and $$n$$ $$\left( {l,\;n > 1} \right)\;$$ and $${G_1},\;{G_2}\;$$ and $${G_3}$$​ are three geometric means between $$\;l\;$$ and $$n$$ , then $${\left( {{G_1}} \right)^4} + 2{\left( {{G_2}} \right)^4} + {\left( {{G_3}} \right)^{4\;}}$$ equals

Answer 1

Hint: Given that m is the Arithmetic mean between $$\;l\;$$ and $$n$$ we need to three condition on $${G_1},\;{G_2}\;$$ and $${G_3}$$​ after the conditions equation will form and after solving the equations we can get the answer.

Complete step-by-step answer:
Given m is the Arithmetic mean between $$\;l\;$$ and $$n$$
$$ \Rightarrow m = 2l + n$$ ​...(1)
And given $${G_1},\;{G_2}\;$$ and $${G_3}$$​ ​ are Geometric mean between $$\;l\;$$ and $$n$$
${G_{1}} = {l^{\dfrac{3}{4}}}{n^{\dfrac{1}{4}}} \\
{G_{2}} = {l^{\dfrac{1}{2}}}{n^{\dfrac{1}{2}}}$
and
$${G_{3}} = {l^{\dfrac{1}{4}}}{n^{\dfrac{3}{4}}}$$
Hence,
$${
 {G_1}^{4} + 2{G_2}^{4} + {G_3}^{4} = {l^3}n + 2{l^2}{n^2} + l{n^3} \\
= ln{\left( {l + n} \right)^2} \\
= 4l{m^2}n } $$
Hence If m is the A.M. of two distinct real numbers $$\;l\;$$ and $$n$$ $$\left( {l,\;n > 1} \right)\;$$ and $${G_1},\;{G_2}\;$$ and $${G_3}$$​ are three geometric means between $$\;l\;$$ and $$n$$ , then $${\left( {{G_1}} \right)^4} + 2{\left( {{G_2}} \right)^4} + {\left( {{G_3}} \right)^{4\;}}$$ equals $$4l{m^2}n$$

Note: Here in this type of geometric mean problem, we need to form conditions and solve then to get the answer. In mathematics, the geometric mean is a mean or average, which indicates the central tendency or typical value of a set of numbers by using the product of their values (as opposed to the arithmetic mean which uses their sum).