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The sequence $\sqrt{7},\sqrt{7+\sqrt{7}},{\sqrt{7+\sqrt{7+\sqrt{7}}}},....$ converges to.

1. $\frac{1+\sqrt{33}}{2}$
2. $\frac{1+\sqrt{32}}{2}$
3. $\frac{1+\sqrt{30}}{2}$
4. $\frac{1+\sqrt{29}}{2}$

Let $$x = \sqrt{7 + \sqrt{7 + \sqrt{7 \ldots}}}$$

Then, we can write $x$ recursively as:

$$x = \sqrt{7+x}$$

Solving this equation, we get:

\begin{align}x^2 &= 7+x\\[2em]x&=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot(-7)}}{2}\\[2em]x &= \frac{1\pm\sqrt{29}}{2}\end{align}

Hence, option D is the correct answer.