Discussion Forum

If m is the AM of two distinct real numbers l and n (l,n>1) and G₁, G₂ and G₃  are three geometric means between  l and n , then $G_{1}^{4}+2G_{2}^{4}+G^{4}_{3}$  equals 

Answer:

Explanation:

Given, m is the AM of l and n 

$\therefore$    l+n=2m          .......(i)

 and G₁, G₂, G₃  are geometric means between l

 and n

$\therefore$ l, G₁, G₂, G_3, n are in GP.

 Let r be the common ration of this GP

    $\therefore$        $G_{1}=lr^{}$

                              $G_{2}=lr^{2}$

                               $G_{3}=lr^{3}$

                                $n=lr^{4}$

$\Rightarrow$     $r=\left(\frac{n}{l}\right)^{1/4}$

Now,  $G_{1}^{4}+2G_{2}^{4}+G^{4}_{3}= (lr)^{4}+2(lr^{2})^{4}+(lr^{3})^{4}$

            =  $l^{4}\times r^{4 }(1+2r^{4}+r^{8})$

          =  $l^{4}\times r^{4 }(r^{4}+1)^{2}$

      $=l^{4}\times\frac{n}{l}\left( \frac{n+l}{l}\right)^{2}$

          $=ln\times4m^{2}=4lm^{2}n$