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Geometric Progression

If m is the A...

Question

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 find the value of ${G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4}$ .

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Solution

Step 1: Solve for value of $m .$
Given, $m$ is the arithmetic mean of $l$ and $n .$
Then, $m = \frac{l + n}{2}$
$\Rightarrow$ $2 m = l + n$
Step 2: Solve for value of common ratio of G.P.
Now, $l, G_{1}, G_{2}, G_{3}, n$ are in G.P.
Let $r$ be the common ratio of this G.P.
$∴ G_{1} = l r; G_{2} = l r^{2}; G_{3} = l r^{3}; n = l r^{4}$
then, $r^{4} = \frac{n}{l}$
Step 3: Solve for value of ${G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4}$
${G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4} = {(l r)}^{4} + 2 {(l r^{2})}^{4} + {(l r^{3})}^{4} {G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4} = l^{4} r^{4} + 2 l^{4} r^{8} + l^{4} r^{12} {G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4} = l^{4} [r^{4} + 2 r^{8} + r^{12}]$
Substituting the values of $r^{4}$ , we get
${G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4} = l^{4} [\frac{n}{l} + 2 \frac{n^{2}}{l^{2}} + \frac{n^{3}}{l^{3}}] {G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4} = n l^{3} [1 + \frac{2 n}{l} + \frac{n^{2}}{l^{2}}]$
$= n l^{3} {[1 + \frac{n}{l}]}^{2}$
$= n l^{3} (\frac{{(l + n)}^{2}}{l^{2}})$
$= n l {(l + n)}^{2}$
$= n l {(2 m)}^{2} = 4 m^{2} n l$
Hence, the value of ${G_{1}}^{4} + 2 {G_{2}}^{4} + {G_{3}}^{4}$ is $4 m^{2} n l .$

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