Let A 1 and A 2 be two arithmetic means and G 1 , G 2 , G 3 be three geometric means of two distinct positive numbers. Then G 1 4 + G 2 4 + G 3 4 + G 1 2 G 3 2 is equal to [2023]

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Solution

Q.
Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}, G_{3}$ be three geometric means of two distinct positive numbers. Then $G_{1}^{4} + G_{2}^{4} + G_{3}^{4} + G_{1}^{2} G_{3}^{2}$ is equal to [2023]

1 $2 (A_{1} + A_{2}) G_{1}^{2} G_{3}^{2}$

2 ${(A_{1} + A_{2})}^{2} G_{1} G_{3}$

3 $(A_{1} + A_{2}) G_{1}^{2} G_{3}^{2}$

4 $2 (A_{1} + A_{2}) G_{1} G_{3}$

Ans.
(2)

$Given: a, A_{1}, A_{2}, b be in an A.P.$

$∴$    $d = \frac{b - a}{3}$

$\Rightarrow$    $A_{1} = a + d = a + (\frac{b - a}{3}) = \frac{2 a + b}{3}$

and   $A_{2} = a + 2 d = a + 2 \frac{(b - a)}{3} = \frac{a + 2 b}{3}$

$∴$ $A_{1} + A_{2} = \frac{2 a + b}{3} + \frac{a + 2 b}{3} = a + b ...(i)$

$Also, let a, G_{1}, G_{2}, G_{3}, b be in a G.P.$

$Now, a r^{4} = b \Rightarrow r = {(\frac{b}{a})}^{1 / 4}$

$∴$    $G_{1} = a^{3 / 4} b^{1 / 4}, G_{2} = \sqrt{a b}, G_{3} = a^{1 / 4} b^{3 / 4}$

$Now, G_{1}^{4} + G_{2}^{4} + G_{3}^{4} + G_{1}^{2} G_{3}^{2}$

   $= a^{3} b + a^{2} b^{2} + a b^{3} + a^{3 / 2} b^{1 / 2} a^{1 / 2} b^{3 / 2}$

   $= a^{3} b + a b^{3} + a^{2} b^{2} + a^{2} b^{2}$

   $= a b (a^{2} + b^{2}) + 2 a^{2} b^{2}$

   $= a b (a^{2} + b^{2} + 2 a b) = {(A_{1} + A_{2})}^{2} \cdot G_{1} G_{3} [Using (i)]$

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