Let $$a_1,a_2,a_3,...$$ be a G.P. of increasing positive terms. If $$a_1a_5 = 28$$ and $$a_2+a_4 = 29$$, then $$a_6$$ is equal to:

The terms $$a_1,a_2,a_3,...$$ are given to be in GP.

Let the first term of the series $$a_1$$ be $$a$$, and the common ratio be $$r$$.

The $$n^{th}$$ term of the series $$a_n$$ can be calculated as,

$$a_n\ =\ a\left(r\right)^{n-1}$$

$$a_5\ =\ a\left(r\right)^{5-1}\ =\ ar^4$$

$$a_2\ =\ a\left(r\right)^{2-1}\ =\ ar$$

$$a_4\ =\ a\left(r\right)^{4-1}\ =\ ar^3$$

$$a_1\times\ a_5\ =\ a\times\ ar^4\ =\ a^2r^4\ =\ 28$$  $$-(1)$$

Taking the square root on both sides, we get,

$$ar^2\ =\ \sqrt{\ 28}=\ 2\sqrt{\ 7}$$  $$-(2)$$

$$a_2\ +\ a_4\ =\ ar\ +\ ar^3\ =\ ar\left(1\ +\ r^2\right)\ =\ 29$$  $$-(3)$$

dividing $$(2)$$ by $$(3)$$, we get,

$$\dfrac{ar^2}{ar\left(1\ +\ r^2\right)}\ =\ \dfrac{2\sqrt{\ 7}}{29}$$

$$\dfrac{r}{1\ +\ r^2}\ =\ \dfrac{2\sqrt{\ 7}}{29}$$

$$2\sqrt{\ 7}r^2\ -\ 29r\ +\ 2\sqrt{\ 7}\ =\ 0$$

$$\left(r\ -\ 2\sqrt{\ 7}\right)\left(2\sqrt{\ 7}r\ -\ 1\right)\ =\ 0$$

$$r\ =\ 2\sqrt{\ 7}$$ or $$r\ =\dfrac{1}{2\sqrt{\ 7}}$$

We are given that the series is increasing, so the value of $$r$$ has to be greater than 1 and is equal to $$2\sqrt{\ 7}$$

Substituting $$r$$ in (2), we get,

$$a\left(2\sqrt{\ 7}\right)^2\ =\ 2\sqrt{\ 7}$$

$$a\ =\ \dfrac{1}{2\sqrt{\ 7}}$$

$$a_6\ =\ ar^5\ =\ \left(\dfrac{1}{2\sqrt{\ 7}}\right)\left(2\sqrt{\ 7}\right)^5\ =\ \left(2\sqrt{\ 7}\right)^4\ =\ \left(28\right)^2\ =\ 784$$

Hence, the correct answer is option D.