Let a$_1$, a$_2$, a$_3$, .... be a G.P. of increasing positive numbers. Let the sum of its 6^th and 8^th terms be 2 and the product of its 3^rd and 5^th terms be $\frac{1}{9}$. Then $6(a_2+a_4)(a_4+a_6)$ is equal to

<p>Given the conditions :</p> <ol> <li>$a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$</li> <li>$a_3 \cdot a_5 = \frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \frac{1}{9} \Rightarrow a r^3 = \frac{1}{3}$</li> </ol> <p>From this, we can form the equation $\frac{r^2}{3} + \frac{r^4}{3} = 2$, which simplifies to $r^4 + r^2 = 6$.</p> <p>This can be factored to give $\left(r^2 - 2\right)\left(r^2 + 3\right) = 0$, yielding $r^2 = 2$ (since $r^2$ cannot be $-3$ for real $r$).</p> <p>So, we have $r = \sqrt{2}$.</p> <p>Substituting $r = \sqrt{2}$ into the equation $a r = \frac{1}{6}$, we get $a = \frac{1}{6\sqrt{2}}$.</p> <p>Now, we find the value of $6(a_2+a_4)(a_4+a_6)$:</p> <p>$6(a_2+a_4)(a_4+a_6) = 6\left(a r + a r^3\right)\left(a r^3 + a r^5\right)$</p> <p>$= 6\left(\frac{1}{6\sqrt{2}} + \frac{1}{3\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}} + \frac{2}{3\sqrt{2}}\right)$</p> <p>$= 6 \cdot \frac{1}{2} \cdot 1 = 3$.</p>