CMAT Progressions and Series Questions

The series is an infinite Geometric progression with first term as $$2\sqrt{\ 2}$$ and common ratio $$\sqrt{\ \frac{2}{3}}$$
Now we know that sum of an infinite geometric progression = $$\frac{a}{1-r}$$ where r is the common ratio and a is first term
so we get sum =$$\frac{2\sqrt{\ 2}}{1-\sqrt{\ \frac{2}{3}}}$$
                      =$$=\ \frac{2\sqrt{\ 6}}{\sqrt{\ 3}-\sqrt{\ 2}}$$
Rationalizing the denominator
we get :$$=\frac{2\sqrt{\ 6}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)}{3-2}$$
= $$2\sqrt{\ 6}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)$$

The geometric mean proportion between two numbers 'a' and 'b' is given by $$\sqrt{a\times b}$$

The given two numbers are 30 + $$\sqrt200$$ and 54 - $$\sqrt648$$

which are also equal to 10(3+$$\sqrt2$$) and 18(3-$$\sqrt2$$).

Hence the geometric mean proportion = $$\sqrt {10(3+\sqrt2)\times 18(3-\sqrt2)}$$ = $$\sqrt {180\times (3^2 - (\sqrt2)^2)}$$ = $$\sqrt {1260}$$ = $$6\times \sqrt35$$.