Animated Solution for Mathematics - Sequence and Series: Let $\frac{1}{16}$, $a$ and $b$ be in G.P. and $\frac{1}{a}$, $\frac{1}{b}$, $6$ be in A.P., where $a,b>0$. Then $72(a+b)$ is equal to

We begin with the G.P. sequence: $\frac{1}{16},a,b$. In the realm of G.P., for any three consecutive terms $x,y,z$, the square of the middle term is the product of its neighbors: $y^2=xz$.

Applying this to our sequence, we obtain:

By isolating $b$, we establish our primary bridge:

Next, we examine the A.P. sequence: $\frac{1}{a},\frac{1}{b},6$. The A.P. world operates on the principle that the middle term is the arithmetic mean of its neighbors, expressed as $2\cdot \text{middle}=\text{first}+\text{last}$.

This yields the following equation:

We now merge these two worlds by substituting $b=16a^2$ into our A.P. equation:

Simplifying the left side, we obtain:

To clear the denominators, we multiply the entire equation by $8a^2$, resulting in:

Rearranging this into the standard quadratic form, we arrive at:

To solve $48a^2+8a-1=0$, we split the middle term using the factors $12$ and $-4$:

Factoring by grouping, we get:

This provides two potential values for $a$: $a=\frac{1}{12}$ or $a=-\frac{1}{4}$. Since the problem constraints specify $a>0$, we reject the negative root and accept $a=\frac{1}{12}$.

With $a=\frac{1}{12}$ determined, we calculate $b$:

Finally, we compute the value of $72(a+b)$:

The final result is $14$.