$$\eqalign{ & {\text{Given}}{{\text{ }}^m}{C_{r - 1}}{,^m}{C_r}{,^m}{C_{r + 1}}{\text{ are in A}}{\text{.P}}{\text{.}} \cr & {{\text{2}}^m}{C_r}{ = ^m}{C_{r - 1}}{ + ^m}{C_{r + 1}} \cr & \Rightarrow \,\,2 = \frac{{^m{C_{r - 1}}}}{{^m{C_r}}} + \frac{{^m{C_{r + 1}}}}{{^m{C_r}}} \cr & = \frac{r}{{m - r + 1}} + \frac{{m - r}}{{r + 1}} \cr & \Rightarrow \,\,{m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0 \cr} $$

Let the G.P. be $$a,ar,a{r^2},.....\left( {0 < r < 1} \right).$$     From the question, $$\frac{a}{{1 - r}} = {3^3} + 3.3 - 9$$
$$\eqalign{ & \left\{ {\because \,\,f'\left( x \right) = 3{x^2} + 3 > 0;\,{\text{so, }}f\left( x \right)\,{\text{is monotonically increasing;}}} \right. \cr & \left. {\therefore \,\,f\left( 3 \right)\,{\text{is the greatest value in }}\left[ { - 2,3} \right].} \right\} \cr & {\text{Also, }}f'\left( 0 \right) = 3.\,\,{\text{So, }}a - ar = 3. \cr & {\text{Solving, }}a = 27\left( {1 - r} \right)\,{\text{and }}a\left( {1 - r} \right) = 3 \cr & {\text{we get }}r = \frac{2}{3},\frac{4}{3}. \cr & {\text{But }}r < 1. \cr} $$

$$\eqalign{ & {A_n} = 704 + \frac{{704}}{2} + \frac{{704}}{4} + ....\,{\text{to }}n{\text{ terms}} \cr & = \frac{{704\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 - \frac{1}{2}}} \cr & = 704 \times 2\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right) \cr & {B_n} = 1984 - \frac{{1984}}{2} + \frac{{1984}}{4} + ....\,{\text{to }}n{\text{ terms}} \cr & = \frac{{1984\left( {1 - {{\left( {\frac{{ - 1}}{2}} \right)}^n}} \right)}}{{1 - \left( {\frac{{ - 1}}{2}} \right)}} \cr & = 1984 \times \frac{2}{3}\left( {1 - {{\left( {\frac{{ - 1}}{2}} \right)}^n}} \right) \cr & {\text{Now}},\,{A_n} = {B_n} \cr & \Rightarrow 704 \times 2\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right) \cr & = 1984 \times \frac{2}{3} \times \left( {1 - {{\left( {\frac{{ - 1}}{2}} \right)}^n}} \right) \cr & \Rightarrow 33 - 31 = 33{\left( {\frac{1}{2}} \right)^n} - 31{\left( {\frac{{ - 1}}{2}} \right)^n} \cr & \Rightarrow {2^{n + 1}} = 33 - 31{\left( { - 1} \right)^n} \cr & \Rightarrow n = 5 \cr} $$

The value of $$0.0\overline {37} $$  stands for the number $$0.0373737\, . . . . . = 0.037 + 0.00037 +\, . . . . .$$
$$\eqalign{ & = \frac{{37}}{{{{10}^3}}} + \frac{{37}}{{{{10}^5}}} + ..... = \frac{{37}}{{{{10}^3}}}\left[ {1 + \frac{1}{{100}} + .....} \right] \cr & = \frac{{37}}{{{{10}^3}}}\left[ {\frac{1}{{1 - \frac{1}{{100}}}}} \right] = \frac{{37}}{{990}} \cr} $$

$$\eqalign{ & S = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},P = {a^n} \cdot {r^{\frac{{n\left( {n - 1} \right)}}{2}}} \cr & R = \frac{1}{a} + \frac{1}{{ar}} + \frac{1}{{a{r^2}}} + .....\,n{\text{ terms}} = \frac{{1 - {r^n}}}{{a\left( {1 - r} \right){r^{n - 1}}}} \cr & {S^n} = {R^n}{P^k} \cr & \Rightarrow {\left( {\frac{S}{R}} \right)^n} = {P^k} \cr & \Rightarrow {\left( {{a^2}{r^{n - 1}}} \right)^n} = {P^k} \cr & \Rightarrow {P^2} = {P^k} \cr & \Rightarrow k = 2 \cr} $$

Let $$n^{th}$$ term of series is $$T_n$$ then
$$\eqalign{ & {S_n} = 12 + 16 + 24 + 40 + .... + {T_n} \cr & {\text{Again }}\,{S_n} = 12 + 16 + 24 + .... + {T_n} \cr} $$
On subtraction
$$\eqalign{ & 0 = \left( {12 + 4 + 8 + 16 + .... + \,{\text{upto }}n{\text{ terms}}} \right) - {T_n} \cr & {\text{or }}{T_n} = 12 + \left[ {4 + 8 + 16 + .... + \,{\text{upto}}\left( {n - 1} \right){\text{terms}}} \right] \cr & = 12 + \frac{{4\left( {{2^{n + 1}} - 1} \right)}}{{2 - 1}} = {2^{n - 1}} + 8 \cr} $$
On putting $$n = 1, 2, 3 ....$$
$$\eqalign{ & {T_1} = {2^2} + 8,{T_2} = {2^3} + 8,{T_3} = {2^4} + 8.....\,{\text{etc}}{\text{.}} \cr & {S_n} = {T_1} + {T_2} + {T_3} + .... + {T_n} \cr & = \left( {{2^2} + {2^3} + {2^4} + ....\,{\text{upto }}n{\text{ terms}}} \right) + \left( {8 + 8 + 8 + ....\,{\text{upto }}n{\text{ terms}}} \right) \cr & = \frac{{{2^2}\left( {{2^n} - 1} \right)}}{{2 - 1}} + 8n = 4\left( {{2^n} - 1} \right) + 8n. \cr} $$

As $$x, y, z$$   are A.M. of $$a$$ and $$b$$
$$\eqalign{ & \therefore x + y + z = 3\left( {\frac{{a + b}}{2}} \right) \cr & \therefore 15 = \frac{3}{2}\left( {a + b} \right) \cr & \Rightarrow a + b = 10\,\,\,.....\left( 1 \right) \cr} $$
Again $$\frac{1}{x} , \frac{1}{y} , \frac{1}{z}$$   are A.M. of $$\frac{1}{a}$$ and $$\frac{1}{b}$$
$$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2}\left( {\frac{1}{a} + \frac{1}{b}} \right) \cr & \therefore \frac{5}{3} = \frac{3}{2} \cdot \frac{{a + b}}{{ab}} \cr & \Rightarrow \frac{{10}}{9} = \frac{{10}}{{ab}} \cr & \Rightarrow ab = 9\,\,\,\,\,.....\left( 2 \right) \cr} $$
Solving (1) and (2), we get
$$a = 9, 1, b = 1, 9$$

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

$$\eqalign{ & {\text{We have}} \cr & S = 1 + \frac{2}{3} + \frac{6}{{{3^2}}} + \frac{{10}}{{{3^3}}} + \frac{{14}}{{{3^4}}} + .....\infty \,\,\,\,\,\,......\left( 1 \right) \cr & {\text{Multiplying both sides by }}\frac{1}{3}{\text{ we get}} \cr & \frac{1}{3}S = \frac{1}{3} + \frac{2}{{{3^2}}} + \frac{6}{{{3^3}}} + \frac{{10}}{{{3^4}}} + .....\infty \,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \cr & {\text{Subtracting eqn}}{\text{.}}\left( 2 \right){\text{from eqn}}{\text{.}}\left( 1 \right){\text{we get}} \cr & \frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{{{3^2}}} + \frac{4}{{{3^3}}} + \frac{4}{{{3^4}}} + .....\infty \cr & \Rightarrow \,\,\frac{2}{3}S = \frac{4}{3} + \frac{4}{{{3^2}}} + \frac{4}{{{3^3}}} + \frac{4}{{{3^4}}} + .....\infty \cr & \Rightarrow \,\,\frac{2}{3}S = \frac{{\frac{4}{3}}}{{1 - \frac{1}{3}}} = \frac{4}{3} \times \frac{3}{2} \cr & \Rightarrow \,\,S = 3 \cr} $$

If the three terms of the G.P. be $$\frac{a}{r},a\,{\text{and }}ar$$   then
$$\eqalign{ & S = \frac{a}{r} + a + ar = \frac{a}{r}\left( {1 + r + {r^2}} \right) \cr & P = {a^3}\,{\text{and }}R = \frac{r}{a} + \frac{1}{a} + \frac{1}{{ar}} = \frac{1}{{ar}}\left( {{r^2} + r + 1} \right) \cr & {\text{Now,}}\frac{{{P^2}{R^3}}}{{{S^3}}} = \frac{{{a^6}\frac{1}{{{a^3}{r^3}}}{{\left( {{r^2} + 2 + 1} \right)}^3}}}{{\frac{{{a^3}}}{{{r^3}}}{{\left( {{r^2} + r + 1} \right)}^3}}} = 1 \cr} $$
So, the required ratio is $$1 : 1 .$$