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:

<p>First, let us denote the first term of the G.P. by $a_1 = A$ and the common ratio (which is $> 1$, since the G.P. is increasing) by $r$. Then the terms are:</p> <p>$ a_1 = A,\quad a_2 = Ar,\quad a_3 = Ar^2,\quad a_4 = Ar^3,\quad a_5 = Ar^4,\quad a_6 = Ar^5,\;\dots $</p> <p>We are given:</p> <p><p>$a_1 \cdot a_5 = 28$, i.e.</p> <p>$ A \cdot (A r^4) \;=\; A^2 r^4 \;=\; 28. \quad (1) $</p></p> <p><p>$a_2 + a_4 = 29$, i.e.</p> <p>$ Ar \;+\; A r^3 \;=\; A(r + r^3) \;=\; 29. \quad (2) $</p></p> <p>We want to find $a_6 = A r^5$.</p> <hr /> <h2>1. Solve for $r$</h2> <p>From $\text{(1)}$: </p> <p>$ A^2 r^4 = 28 \quad\Longrightarrow\quad A^2 = \frac{28}{r^4}. $</p> <p>From $\text{(2)}$: </p> <p>$ A\,\bigl(r + r^3\bigr) = 29 \quad\Longrightarrow\quad A = \frac{29}{r + r^3}. $</p> <p>Plug $A$ from $\text{(2)}$ into $\text{(1)}$. After some algebra (or by a systematic approach), one finds that </p> <p>$ r^2 = 28 \quad\Longrightarrow\quad r = \sqrt{28} \;=\; 2\sqrt{7} $</p> <p>(since $r>1$, we take the positive root).</p> <hr /> <h2>2. Solve for $A$</h2> <p>From (2), using $r = 2\sqrt{7}$:</p> <p>$ r + r^3 = 2\sqrt{7} + (2\sqrt{7})^3 = 2\sqrt{7} + 56\sqrt{7} = 58\sqrt{7}. $</p> <p>Hence,</p> <p>$ A = \frac{29}{\,58\sqrt{7}\,} = \frac{1}{2\sqrt{7}}. $</p> <hr /> <h2>3. Find $a_6$</h2> <p>$ a_6 = A\,r^5 = \frac{1}{2\sqrt{7}}\;\times\;(2\sqrt{7})^5. $</p> <p>Compute $(2\sqrt{7})^5$. First, $(2\sqrt{7})^2 = 28$; thus</p> <p>$ (2\sqrt{7})^5 = (2\sqrt{7})^4 \cdot (2\sqrt{7}) = 784 \times (2\sqrt{7}) = 1568\sqrt{7}. $</p> <p>Therefore,</p> <p>$ a_6 = \frac{1}{2\sqrt{7}} \cdot 1568\sqrt{7} = \frac{1568}{2}\;\;(\text{since }\sqrt{7} \text{ cancels}) = 784. $</p> <hr /> <h3><strong>Answer: $a_6 = 784$.</strong></h3> <p>Hence, the correct option is <strong>Option B</strong>.</p>