Let the arithmetic mean of frac{1}{a} and fra | vedclass

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(A) Given $a, 4, \alpha, b$ are in $A$.$P$. Let the common difference be $d$. Then $4 = a + d$, $\alpha = 4 + d$, and $b = 4 + 2d$. Since $a = 4 - d$,

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One root in $(1, 4)$ and another in $(-2, 0)$

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(A) Given $a, 4, \alpha, b$ are in $A$.$P$. Let the common difference be $d$. Then $4 = a + d$, $\alpha = 4 + d$, and $b = 4 + 2d$. Since $a = 4 - d$, we have $a, 4, \alpha, b$ as $4-d, 4, 4+d, 4+2d$. The arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$, so $\frac{1}{2}(\frac{1}{4-d} + \frac{1}{4+2d}) = \frac{5}{16}$. $\frac{4+2d+4-d}{(4-d)(4+2d)} = \frac{5}{8} \Rightarrow \frac{8+d}{16+4d-2d^2} = \frac{5}{8}$. $64 + 8d = 80 + 20d - 10d^2$ $\Rightarrow 10d^2 - 12d - 16 = 0$ $\Rightarrow 5d^2 - 6d - 8 = 0$. $(5d+4)(d-2) = 0$. Since $a > 2$, $4-d > 2 \Rightarrow d 2$. Re-evaluating: if $d=2$, $a=2$. If $d=-4/5$, $a=24/5=4.8 > 2$. Using $d = -4/5$: $a = 4.8, \alpha = 3.2, b = 2.4$. Equation: $3.2x^2 - 4.8x + 2(3.2 - 4.8) = 0$ $\Rightarrow 3.2x^2 - 4.8x - 3.2 = 0$ $\Rightarrow x^2 - 1.5x - 1 = 0$. Roots are $x = \frac{1.5 \pm \sqrt{2.25 + 4}}{2} = \frac{1.5 \pm 2.5}{2} = 2, -0.5$. One root is $2 \in (0, 4)$ and another is $-0.5 \in (-2, 0)$.

Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}$, where $a > 2$. If $\alpha$ is such that $a, 4, \alpha, b$ are in $A$.$P$.,then the equation $\alpha x^2 - ax + 2(\alpha - 2b) = 0$ has :

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