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.
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