Key Concepts and Formulas

• A Geometric Progression (G.P.) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$).
• The $n$-th term of a G.P. is given by $a_n = a \cdot r^{n-1}$, where $a$ is the first term.
• For a G.P. of "increasing positive numbers," the first term $a > 0$ and the common ratio $r > 1$.

Step-by-Step Solution

Step 1: Expressing the given conditions in terms of the first term ($a$) and common ratio ($r$)

We are given that $a_1, a_2, a_3, \ldots$ is a G.P. of increasing positive numbers. Let the first term be $a$ and the common ratio be $r$. The $n$-th term is $a_n = ar^{n-1}$. The condition "increasing positive numbers" implies $a > 0$ and $r > 1$.

We are given two conditions:

1. $a_3 a_5 = 729$
2. $a_2 + a_4 = \frac{111}{4}$

Let's express the terms involved using the formula $a_n = ar^{n-1}$:

• $a_2 = ar$
• $a_3 = ar^2$
• $a_4 = ar^3$
• $a_5 = ar^4$

Substitute these into the first condition: $(ar^2)(ar^4) = 729$ $a^2 r^{2+4} = 729$ $a^2 r^6 = 729$ This can be rewritten as: $(ar^3)^2 = 729$ Since $729 = 27^2$, we have: $(ar^3)^2 = 27^2$ Taking the square root of both sides, we get $ar^3 = \pm 27$. Since the G.P. consists of positive numbers, $a > 0$ and $r > 0$. Therefore, $ar^3$ must be positive. Thus, we have: $ar^3 = 27 \quad (*)$

Now, substitute the terms into the second condition: $ar + ar^3 = \frac{111}{4}$

Step 2: Solving the system of equations to find $r$ and $a$

We have the equations:

1. $ar^3 = 27$
2. $ar + ar^3 = \frac{111}{4}$

Substitute the value of $ar^3$ from equation (1) into equation (2): $ar + 27 = \frac{111}{4}$ Subtract 27 from both sides: $ar = \frac{111}{4} - 27$ To subtract, find a common denominator: $27 = \frac{27 \times 4}{4} = \frac{108}{4}$. $ar = \frac{111}{4} - \frac{108}{4}$ $ar = \frac{3}{4} \quad (**)$

Now we have a system of two equations: (*) $ar^3 = 27$ (**) $ar = \frac{3}{4}$

To find $r$, divide equation (*) by equation (**): $\frac{ar^3}{ar} = \frac{27}{\frac{3}{4}}$ $r^{3-1} = 27 \times \frac{4}{3}$ $r^2 = 9 \times 4$ $r^2 = 36$ Taking the square root, $r = \pm 6$. Since the G.P. is of increasing positive numbers, the common ratio $r$ must be greater than 1. Therefore, we choose $r = 6$.

Now, substitute $r=6$ into equation (**) to find $a$: $a(6) = \frac{3}{4}$ $a = \frac{3}{4 \times 6}$ $a = \frac{3}{24}$ $a = \frac{1}{8}$ So, the first term is $a = \frac{1}{8}$ and the common ratio is $r=6$. These values satisfy the conditions $a>0$ and $r>1$.

Step 3: Calculating the required sum $24(a_1+a_2+a_3)$

We need to find the value of $24(a_1+a_2+a_3)$. The sum $a_1+a_2+a_3$ can be written as: $a_1+a_2+a_3 = a + ar + ar^2$ Factor out $a$: $a_1+a_2+a_3 = a(1+r+r^2)$ Substitute the values $a = \frac{1}{8}$ and $r = 6$: $a(1+r+r^2) = \frac{1}{8}(1 + 6 + 6^2)$ $= \frac{1}{8}(1 + 6 + 36)$ $= \frac{1}{8}(43)$

Now, multiply this sum by 24: $24(a_1+a_2+a_3) = 24 \times \left(\frac{1}{8} \times 43\right)$ $= \frac{24}{8} \times 43$ $= 3 \times 43$ $= 129$

Common Mistakes & Tips

• Ignoring the conditions: Always remember to check if your calculated values of $a$ and $r$ satisfy the given conditions (e.g., increasing positive numbers implies $a>0$ and $r>1$). This is crucial for selecting the correct root when solving equations like $r^2=36$.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and exponents. Double-check substitutions and simplifications.
• Using the geometric mean property: Notice that $a_3 a_5 = (ar^2)(ar^4) = a^2r^6 = (ar^3)^2$, which means $a_3 a_5 = a_4^2$. This is a useful shortcut for problems involving products of terms in a G.P.

Summary

The problem involves a geometric progression of increasing positive numbers. We used the given conditions $a_3 a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$ to set up a system of equations involving the first term ($a$) and the common ratio ($r$). By expressing the terms in the standard form $a_n = ar^{n-1}$ and solving these equations, we found $a = \frac{1}{8}$ and $r = 6$. The constraint that the G.P. consists of increasing positive numbers was essential in selecting the correct value for $r$. Finally, we calculated the required sum $24(a_1+a_2+a_3)$ using the determined values of $a$ and $r$.

The final answer is $\boxed{129}$.