Let $$S_{1}, S_{2},...$$ be the squares such that for each n ≥ 1, the length of the diagonal of $$S_{n}$$ is equal to the length of the side of $$S_{n+1}$$. If the length of the side of $$S_{3}$$ is 4 cm, what is the length of the side of $$S_{n}$$ ?

Length of side of $$S_{n + 1}$$ = Length of diagonal of $$S_n$$

=> Length of side of $$S_{n + 1}$$ = $$\sqrt{2}$$ (Length of side of $$S_{n}$$)

=> $$\frac{\textrm{Length of side of }S_{n + 1}}{\textrm{Length of side of }S_n} = \sqrt{2}$$

=> Sides of $$S_1 , S_2 , S_3 , S_4,........, S_n$$ form a G.P. with common ratio, $$r = \sqrt{2}$$

It is given that, $$S_3 = ar^2 = 4$$

=> $$a (\sqrt{2})^2 = 4$$

=> $$a = \frac{4}{2} = 2$$

$$\therefore$$ $$n^{th}$$ term of G.P. = $$a (r^{n - 1})$$

= $$2 (\sqrt{2})^{n - 1}$$

=$$2^[{\frac{n+1}{2}}]$$