Key Concepts and Formulas

• Geometric Progression (G.P.): For three terms $x, y, z$ in G.P., the middle term is the geometric mean of the other two, meaning $y^2 = xz$.
• Arithmetic Progression (A.P.): For three terms $x, y, z$ in A.P., the middle term is the arithmetic mean of the other two, meaning $2y = x + z$.

Step-by-Step Solution

Step 1: Formulate equations from the given information. We are given that ${1 \over {16}}$, a, and b are in G.P. Using the G.P. property, we have: $a^2 = \left( {{1 \over {16}}} \right) \cdot b$ $16a^2 = b \quad \text{(Equation 1)}$ This equation expresses 'b' in terms of 'a' based on the G.P. condition.

We are also given that ${1 \over a}$, ${1 \over b}$, and 6 are in A.P. Using the A.P. property, we have: $2 \cdot \left( {{1 \over b}} \right) = {{1 \over a}} + 6$ ${2 \over b} = {1 \over a} + 6 \quad \text{(Equation 2)}$ This equation relates 'a' and 'b' based on the A.P. condition.

Step 2: Substitute Equation 1 into Equation 2 to solve for 'a'. Substitute $b = 16a^2$ from Equation 1 into Equation 2: ${2 \over {16a^2}} = {1 \over a} + 6$ Simplify the left side: ${1 \over {8a^2}} = {1 \over a} + 6$ To eliminate the denominators, multiply the entire equation by $8a^2$. Since $a > 0$, $a \neq 0$. $8a^2 \left( {{1 \over {8a^2}}} \right) = 8a^2 \left( {{1 \over a}} \right) + 8a^2 (6)$ $1 = 8a + 48a^2$ Rearrange into a standard quadratic equation: $48a^2 + 8a - 1 = 0$

Step 3: Solve the quadratic equation for 'a'. We can solve the quadratic equation $48a^2 + 8a - 1 = 0$ by factoring. We look for two numbers that multiply to $(48)(-1) = -48$ and add up to 8. These numbers are 12 and -4. $48a^2 + 12a - 4a - 1 = 0$ Factor by grouping: $12a(4a + 1) - 1(4a + 1) = 0$ $(12a - 1)(4a + 1) = 0$ This gives two possible values for 'a': $12a - 1 = 0 \implies a = {1 \over {12}}$ $4a + 1 = 0 \implies a = -{1 \over 4}$

Step 4: Apply the condition $a > 0$ to select the correct value of 'a'. The problem states that $a > 0$. Therefore, we must reject the negative solution. The correct value for 'a' is $a = {1 \over {12}}$.

Step 5: Calculate the value of 'b' using Equation 1. Substitute the valid value of $a = {1 \over {12}}$ into Equation 1 ($b = 16a^2$): $b = 16 \left( {{1 \over {12}}} \right)^2$ $b = 16 \left( {{1 \over {144}}} \right)$ $b = {{16} \over {144}}$ Simplify the fraction: $b = {1 \over 9}$ Since $b = {1 \over 9} > 0$, this value is valid.

Step 6: Calculate the value of the expression $72(a + b)$. Substitute the values $a = {1 \over {12}}$ and $b = {1 \over 9}$ into the expression: $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right)$ Find a common denominator for ${1 \over {12}}$ and ${1 \over 9}$. The least common multiple (L.C.M.) of 12 and 9 is 36. ${1 \over {12}} = {3 \over {36}}$ ${1 \over 9} = {4 \over {36}}$ So, ${1 \over {12}} + {1 \over 9} = {3 \over {36}} + {4 \over {36}} = {7 \over {36}}$ Now multiply by 72: $72 \left( {7 \over {36}} \right) = {{72 \cdot 7} \over {36}}$ Since $72 = 2 \times 36$, we can simplify: $= 2 \times 7$ $= 14$

Common Mistakes & Tips

• Algebraic Manipulation: Be meticulous when simplifying fractions and solving quadratic equations. Errors in these steps are common.
• Constraint Application: Always use the given constraints ($a, b > 0$ in this case) to filter out extraneous solutions from equations.
• Fraction Arithmetic: Ensure accurate addition and multiplication of fractions by using the correct common denominator.

Summary The problem required us to use the definitions of Geometric Progression and Arithmetic Progression to set up a system of two equations involving variables 'a' and 'b'. By substituting one equation into the other, we derived a quadratic equation for 'a'. After solving the quadratic equation and applying the condition $a > 0$, we found the unique value of 'a'. This value was then used to find 'b'. Finally, we substituted these values of 'a' and 'b' into the expression $72(a + b)$ to obtain the numerical result.

The final answer is $\boxed{14}$.