If $$t_n$$ denotes the $$n^{th}$$ term of a G.P. whose common ratio is $$r,$$ then the progression whose $$n^{th}$$ term is $$\frac{1}{{t_n^2 + t_{n + 1}^2}}{\text{is}}$$

Solution :
If $$a$$ be the first term of G.P. then given
$$\eqalign{ & {x_n} = \frac{1}{{t_n^2 + t_{n + 1}^2}} = \frac{1}{{{a^2}{r^{2\left( {n - 1} \right)}} + {a^2}{r^{2n}}}} \cr & = \frac{1}{{{a^2}{r^{2n}}}} \cdot \frac{1}{{{r^{ - 2}} + 1}} = \frac{1}{{{a^2}{r^{2n}}}} \cdot \frac{{{r^2}}}{{1 + {r^2}}} \cr & \therefore {x_{n - 1}} = \frac{1}{{{a^2}{r^{2n - 2}}}} \cdot \frac{{{r^2}}}{{1 + {r^2}}} \cr & \therefore \frac{{{x_n}}}{{{x_{n - 1}}}} = \frac{1}{{{r^2}}} = {\text{constant}} \cr & \therefore {\text{The sequence}}\left\langle {{x_n}} \right\rangle {\text{is a G}}{\text{.P}}{\text{.}} \cr} $$