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Question

$If three real numbers x, y, z are in G.P. such that x + y + z = 42 . If x, \frac{5 y}{4}, z are in A.P.,$
$then the largest possible value of x is$

Moderate

Solution

$\begin{array}{l} Let x = a, y = a r, z = a r^{2} are in G.P \\ Since x + y + z = 42 \\ \Rightarrow a + a r + a r^{2} = 42 \dots \dots \dots \dots \dots . . (1) \end{array}$
$\begin{array}{l} Since x, \frac{5 y}{4}, z are in A.P \\ 2 (\frac{5}{4} y) = x + z \\ \frac{5}{2} \cdot a r = a + a r^{2} \dots \dots \dots \dots . . . (2) \end{array}$
Solving , $a r = 12$
Now, from (1)
$\begin{array}{l} a + 12 + a (\frac{144}{a^{2}}) = 42 \\ \Rightarrow a + \frac{144}{a} = 30 \\ \Rightarrow a^{2} - 30 a + 144 = 0 \\ \Rightarrow a^{2} - 24 a - 6 a + 144 = 0 \\ \Rightarrow a (a - 24) - 6 (a - 24) = 0 \\ \Rightarrow a = 6 or 24 \\ Maximum of a is 24 \end{array}$

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