Key Concepts and Formulas

• Geometric Progression (G.P.): 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 is given by $a_n = ar^{n-1}$, where $a$ is the first term.
• Properties of G.P. Terms: For a G.P., the product of terms equidistant from the beginning and end is constant. Specifically, $a_k \cdot a_{n-k+1} = a_1 \cdot a_n$. Also, $a_i \cdot a_j = a_k \cdot a_l$ if $i+j = k+l$. This implies $a_1 a_5 = a_2 a_4 = a_3^2$.
• Increasing Positive Terms: For a G.P. with increasing positive terms, the first term $a$ must be positive ($a > 0$) and the common ratio $r$ must be greater than 1 ($r > 1$).

Step-by-Step Solution

Step 1: Express the given conditions using the G.P. formula. Let the first term be $a$ and the common ratio be $r$. The terms of the G.P. are $a, ar, ar^2, ar^3, \ldots$. We are given:

1. $a_1 a_5 = 28$
2. $a_2 + a_4 = 29$

Using $a_n = ar^{n-1}$: $a_1 = a$ $a_2 = ar$ $a_4 = ar^3$ $a_5 = ar^4$

Substituting these into the given conditions:

1. $a \cdot (ar^4) = 28 \implies a^2 r^4 = 28$
2. $ar + ar^3 = 29 \implies ar(1+r^2) = 29$

Step 2: Simplify the first condition using G.P. properties. The condition $a^2 r^4 = 28$ can be rewritten as $(ar^2)^2 = 28$. We know that $a_3 = ar^2$. Therefore, $a_3^2 = 28$. Since the terms are positive, $a_3 = \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$. So, we have $ar^2 = 2\sqrt{7}$. This is a more convenient form than $a^2 r^4 = 28$.

Step 3: Solve the system of equations for $a$ and $r$. We have the following two equations: (i) $ar^2 = 2\sqrt{7}$ (ii) $ar(1+r^2) = 29$

From equation (i), we can express $ar$ as $ar = \frac{2\sqrt{7}}{r}$. Substitute this into equation (ii): $\left(\frac{2\sqrt{7}}{r}\right)(1+r^2) = 29$ Multiply both sides by $r$: $2\sqrt{7}(1+r^2) = 29r$ Expand and rearrange into a quadratic equation in $r$: $2\sqrt{7} + 2\sqrt{7}r^2 = 29r$ $2\sqrt{7}r^2 - 29r + 2\sqrt{7} = 0$

Solve this quadratic equation for $r$ using the quadratic formula $r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, where $A = 2\sqrt{7}$, $B = -29$, $C = 2\sqrt{7}$. $r = \frac{-(-29) \pm \sqrt{(-29)^2 - 4(2\sqrt{7})(2\sqrt{7})}}{2(2\sqrt{7})}$ $r = \frac{29 \pm \sqrt{841 - 4(4 \times 7)}}{4\sqrt{7}}$ $r = \frac{29 \pm \sqrt{841 - 112}}{4\sqrt{7}}$ $r = \frac{29 \pm \sqrt{729}}{4\sqrt{7}}$ Since $\sqrt{729} = 27$: $r = \frac{29 \pm 27}{4\sqrt{7}}$

This gives two possible values for $r$: $r_1 = \frac{29 + 27}{4\sqrt{7}} = \frac{56}{4\sqrt{7}} = \frac{14}{\sqrt{7}} = \frac{14\sqrt{7}}{7} = 2\sqrt{7}$ $r_2 = \frac{29 - 27}{4\sqrt{7}} = \frac{2}{4\sqrt{7}} = \frac{1}{2\sqrt{7}}$

Step 4: Apply the condition for increasing positive terms to select the correct common ratio. The problem states that the G.P. has increasing positive terms. This means $r > 1$. For $r_1 = 2\sqrt{7}$: Since $\sqrt{7} > \sqrt{1}=1$, $2\sqrt{7} > 2$, which is greater than 1. This is a valid common ratio. For $r_2 = \frac{1}{2\sqrt{7}}$: Since $2\sqrt{7} > 1$, $\frac{1}{2\sqrt{7}} < 1$. This would result in a decreasing sequence, so it is not valid. Therefore, the common ratio is $r = 2\sqrt{7}$.

Step 5: Calculate the first term $a$. Using equation (i): $ar^2 = 2\sqrt{7}$ Substitute $r = 2\sqrt{7}$: $a(2\sqrt{7})^2 = 2\sqrt{7}$ $a(4 \times 7) = 2\sqrt{7}$ $28a = 2\sqrt{7}$ $a = \frac{2\sqrt{7}}{28} = \frac{\sqrt{7}}{14}$ Since $a = \frac{\sqrt{7}}{14} > 0$, this is consistent with positive terms.

Step 6: Calculate the sixth term, $a_6$. The formula for the $n$-th term is $a_n = ar^{n-1}$. For $a_6$: $a_6 = ar^{6-1} = ar^5$ Substitute $a = \frac{\sqrt{7}}{14}$ and $r = 2\sqrt{7}$: $a_6 = \left(\frac{\sqrt{7}}{14}\right) (2\sqrt{7})^5$ Calculate $(2\sqrt{7})^5$: $(2\sqrt{7})^5 = 2^5 \cdot (\sqrt{7})^5 = 32 \cdot (\sqrt{7}^2)^2 \cdot \sqrt{7} = 32 \cdot 7^2 \cdot \sqrt{7} = 32 \cdot 49 \cdot \sqrt{7}$ Now substitute this back into the expression for $a_6$: $a_6 = \frac{\sqrt{7}}{14} \cdot (32 \cdot 49 \cdot \sqrt{7})$ $a_6 = \frac{\sqrt{7} \cdot \sqrt{7}}{14} \cdot (32 \cdot 49)$ $a_6 = \frac{7}{14} \cdot (32 \cdot 49)$ $a_6 = \frac{1}{2} \cdot (32 \cdot 49)$ $a_6 = 16 \cdot 49$ $a_6 = 16 \cdot (50 - 1) = 800 - 16 = 784$

Common Mistakes & Tips

• Forgetting the "increasing positive terms" condition: This condition is crucial for selecting the correct value of the common ratio $r$. Always verify that $a>0$ and $r>1$.
• Algebraic Errors with Radicals: Be careful when simplifying expressions involving square roots, especially when raising them to powers.
• Using the Property $a_1 a_5 = a_3^2$: Recognizing this property can significantly simplify the initial equations and make solving for $a$ and $r$ more straightforward.

Summary We leveraged the definition of a Geometric Progression and the given conditions to set up a system of equations. By using the property that $a_1 a_5 = a_3^2$, we simplified the problem. Solving the resulting quadratic equation for the common ratio $r$, we selected the valid ratio that satisfied the "increasing positive terms" condition ($r > 1$). Subsequently, we found the first term $a$ and then calculated the sixth term $a_6$ using the formula $a_n = ar^{n-1}$.

The final answer is $\boxed{784}$.