Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms. If $a_1 a_5 = 28$ and $a_2 + a_4 = 29$, then $a_6$ is equal to:

Step 1: Express terms in terms of $a_1$ and $r$

In a geometric progression (G.P.), each term can be expressed using the first term $a_1$ and the common ratio $r$. The $n$ -th term is given by the formula $a_n = a_1 r^{n-1}$. Since the terms are increasing and positive, the common ratio $r$ must be greater than 1.

Step 2: Formulate equations from given conditions

Using the general formula for the $n$ -th term, we can rewrite the given conditions. The first condition $a_1 a_5 = 28$ becomes $a_1 (a_1 r^4) = 28$, which simplifies to $a_1^2 r^4 = 28$. The second condition $a_2 + a_4 = 29$ becomes $a_1 r + a_1 r^3 = 29$, which can be factored as $a_1 r (1 + r^2) = 29$.

Step 3: Solve for $a_1$ and $r$

From equation (1), $a_1^2 r^4 = 28$, taking the square root gives $a_1 r^2 = \sqrt{28} = 2\sqrt{7}$. We can express $a_1$ as $a_1 = \frac{2\sqrt{7}}{r^2}$. Substitute this into equation (2): $\frac{2\sqrt{7}}{r^2} r (1 + r^2) = 29$. This simplifies to $2\sqrt{7} \frac{1+r^2}{r} = 29$, which leads to the quadratic equation $2\sqrt{7} r^2 - 29r + 2\sqrt{7} = 0$. Solving this quadratic equation for $r$ using the quadratic formula yields two possible values for $r$: $2\sqrt{7}$ and $\frac{1}{2\sqrt{7}}$. Since the G.P. has increasing terms, $r$ must be greater than 1. Therefore, $r = 2\sqrt{7}$. Substituting this value back into the expression for $a_1$ gives $a_1 = \frac{\sqrt{7}}{14}$.

Step 4: Calculate $a_6$

Now that we have $a_1 = \frac{\sqrt{7}}{14}$ and $r = 2\sqrt{7}$, we can find the 6th term, $a_6$. Using the formula $a_6 = a_1 r^5$, we substitute the values: $a_6 = \left(\frac{\sqrt{7}}{14}\right) (2\sqrt{7})^5$. This simplifies to $a_6 = \frac{\sqrt{7}}{14} \cdot 32 \cdot (\sqrt{7})^5 = \frac{\sqrt{7}}{14} \cdot 32 \cdot 49\sqrt{7}$. Multiplying the terms, we get $a_6 = \frac{7}{14} \cdot 32 \cdot 49 = \frac{1}{2} \cdot 32 \cdot 49 = 16 \cdot 49 = 784$.

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