Animated Solution for Mathematics - Sequence and Series: Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16},a>2$. 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:

Imagine you are standing on a number line. We are given four points: $a$, $4$, $\alpha$, and $b$. They are locked in an Arithmetic Progression (A.P.), meaning the gap between any two consecutive terms is a constant common difference, $d$.

If we start at $a$, the next term is $a+d=4$. This gives us the relation $d=4-a$.

We can now express the remaining terms in terms of $a$:

- The third term, $\alpha$, is $a+2d=a+2(4-a)=8-a$. - The fourth term, $b$, is $a+3d=a+3(4-a)=12-2a$.

By doing this, we have reduced the complexity of the problem by expressing all variables in terms of $a$.

The problem provides the condition that the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$. Mathematically, this is expressed as:

Simplifying the left side, we obtain:

Substituting $b=12-2a$ into the equation, we get:

This simplifies to:

Cross-multiplying the terms leads to a quadratic equation in $a$:

Dividing the entire equation by $2$ yields:

Factoring the quadratic, we find:

This provides two potential values: $a=4.8$ or $a=2$. Given the constraint $a>2$, we must discard $a=2$ and accept $a=4.8$.

With $a=4.8$, we calculate the remaining values:

- $\alpha =8-4.8=3.2$ - $b=12-2(4.8)=2.4$

We now substitute these into the final quadratic equation $\alpha x^2-ax+2(\alpha -2b)=0$:

Dividing by $1.6$ simplifies the equation to:

Factoring this quadratic gives $(2x+1)(x-2)=0$. The roots are $x=2$ and $x=-0.5$.