$${\text{AM}} \geqslant {\text{GM}}$$   for positive numbers. So, $$\frac{{{4^x} + \frac{4}{{{4^x}}}}}{2} \geqslant \sqrt {{4^x} \cdot \frac{4}{{{4^x}}}} = 2.$$

$$\eqalign{ & a,b,c{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \,\,\,\,\,\,\,\,\,\,\,{b^2} = ac\,\,\,\,\,\,\,.....\left( 1 \right) \cr & \,\,\,\,\,\,\,\,\,\,a{x^2} + 2bx + c = 0 \cr & {\text{and }}d{x^2} + 2ex + f = 0{\text{ have a common root }} \cr & {\text{Let it be }}\alpha {\text{, then }}a{\alpha ^2} + 2b\alpha + c = 0 \cr & \,\,\,\,\,\,\,d{\alpha ^2} + 2e\alpha + f = 0 \cr & \Rightarrow \,\,\frac{{{\alpha ^2}}}{{2\left( {bf - ec} \right)}} = \frac{\alpha }{{cd - af}} = \frac{1}{{2\left( {ae - bd} \right)}} \cr & \Rightarrow \,\,{\alpha ^2} = \frac{{bf - ce}}{{ae - bd}};\alpha = \frac{{cd - af}}{{2\left( {ae - bd} \right)}} \cr & {\text{Substituting the value of }}\alpha {\text{, we get}} \cr & \,\,\,\,\,\,\,\frac{{{{\left( {cd - af} \right)}^2}}}{{4{{\left( {ae - bd} \right)}^2}}} = \frac{{bf - ce}}{{ae - bd}} \cr & \Rightarrow {\left( {cd - af} \right)^2} = 4\left( {ae - bd} \right)\left( {bf - ce} \right) \cr & {\text{Dividing both sides by }}{a^2}{c^2}{\text{ we get}} \cr & {\left( {\frac{d}{a} - \frac{f}{c}} \right)^2} = 4\left( {\frac{e}{c} - \frac{{bd}}{{ac}}} \right)\left( {\frac{{bf}}{{ac}} - \frac{e}{a}} \right) \cr & {\left( {\frac{d}{a} - \frac{f}{c}} \right)^2} = 4\left( {\frac{e}{c} - \frac{d}{b}} \right)\left( {\frac{f}{c} - \frac{e}{a}} \right)\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right] \cr & \Rightarrow \,\frac{{{d^2}}}{{{a^2}}} + \frac{{{f^2}}}{{{c^2}}} - \frac{{2df}}{{ac}} = \frac{{4ef}}{{cb}} - \frac{{4{e^2}}}{{ac}} - \frac{{4df}}{{{b^2}}} + \frac{{4de}}{{ab}} \cr & \Rightarrow \,\frac{{{d^2}}}{{{a^2}}} + \frac{{{f^2}}}{{{c^2}}} + \frac{{4{e^2}}}{{{b^2}}} + 2\frac{d}{a}.\frac{f}{c} - 4\frac{e}{b}.\frac{f}{c} - 4\frac{d}{a}.\frac{e}{b} = 0\,\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right] \cr & \Rightarrow {\left( {\frac{d}{a} + \frac{f}{c} - 2\frac{e}{b}} \right)^2} = 0 \cr & \Rightarrow \frac{d}{a} + \frac{f}{c} = \frac{{2e}}{b} \cr & \Rightarrow \frac{d}{a},\frac{e}{b},\frac{f}{c}{\text{ are in A}}{\text{.P}}{\text{.}} \cr} $$

$$\eqalign{ & \because \,a,b,c{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,{b^2} = ac \cr & {\text{Since, }}a + b + c = xb \cr & \Rightarrow \,a + c = \left( {x - 1} \right)b \cr & {\text{Take square on both sides, we get}} \cr & {a^2} + {c^2} + 2ac = {\left( {x - 1} \right)^2}{b^2} \cr & \Rightarrow \,\,{a^2} + {c^2} = {\left( {x - 1} \right)^2}ac - 2ac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,{b^2} = ac} \right] \cr & \Rightarrow \,\,{a^2} + {c^2} = ac\left[ {{{\left( {x - 1} \right)}^2} - 2} \right] \cr & \Rightarrow \,\,{a^2} + {c^2} = ac\left[ {{x^2} - 2x - 1} \right] \cr & \because \,\,{a^2} + {c^2}{\text{ are positive and }}{b^2} = ac{\text{ which is also positive}}{\text{.}} \cr & {\text{Then, }}{x^2} - 2x - 1{\text{ would be positive but for }}x = 2,{x^2} - 2x - 1{\text{ is negative}}{\text{.}} \cr & {\text{Hence, }}x{\text{ cannot be taken as 2}}{\text{.}} \cr} $$

Clearly the required series is $$1 + a + ca + a\left( {ca} \right) + c\left( {aca} \right) + a\left( {caca} \right) + .....{\text{ to }}2n{\text{ terms}}$$
$$\eqalign{ & = 1 + a + ca + c{a^2} + {c^2}{a^2} + {c^2}{a^3} + .....{\text{ to }}2n{\text{ terms}} \cr & = \left( {1 + ca + {c^2}{a^2} + .....{\text{ to }}n{\text{ terms}}} \right) + \left( {a + c{a^2} + {c^2}{a^3} + .....{\text{ to }}n{\text{ terms}}} \right) \cr & = \frac{{1\left\{ {1 - {{\left( {ca} \right)}^n}} \right\}}}{{1 - ca}} + \frac{{a\left\{ {1 - {{\left( {ca} \right)}^n}} \right\}}}{{1 - ca}} \cr & = \frac{{\left( {1 + a} \right)\left( {1 - {c^n}{a^n}} \right)}}{{1 - ac}} \cr} $$

Since $$\left| r \right| > 1,\frac{1}{{\left| r \right|}} < 1$$
$$\therefore x = \frac{a}{{1 - \frac{1}{r}}} = \frac{{ar}}{{r - 1}}$$
Similarly, $$y = \frac{b}{{1 - \left( { - \frac{1}{r}} \right)}} = \frac{{br}}{{r + 1}}\,{\text{and}}$$
$$\eqalign{ & z = \frac{c}{{1 - \frac{1}{{{r^2}}}}} = \frac{{c{r^2}}}{{{r^2} - 1}}\,\,\,\,\,....\left( 1 \right) \cr & \therefore xy = \frac{{ar}}{{r - 1}} \times \frac{{br}}{{r + 1}} = \frac{{ab{r^2}}}{{{r^2} - 1}}\,\,\,....\left( 2 \right) \cr} $$
Dividing (2) by (1), we get
$$\frac{{xy}}{z} = \frac{{ab{r^2}}}{{{r^2} - 1}} \times \frac{{{r^2} - 1}}{{c{r^2}}} = \frac{{ab}}{c}$$

Let $$b = ar$$  and $$c = ar^2 ,$$  so that $$a, b, c$$  are in G.P.
$$ \therefore a + b + c = xb$$
$$\eqalign{ & \Rightarrow a + ar + a{r^2} = x.ar \cr & \Rightarrow {r^2} + \left( {1 - x} \right)r + 1 = 0\,\,\,\,\,.....\left( 1 \right) \cr} $$
If $$r$$ is real, then discriminant of $$\left( 1 \right) \geqslant 0$$
$$\eqalign{ & \Rightarrow {\left( {1 - x} \right)^2} - 4.1.1 \geqslant 0 \cr & \Rightarrow {x^2} - 2x - 3 \geqslant 0 \cr & \Rightarrow \left( {x + 1} \right)\left( {x - 3} \right) \geqslant 0 \cr & \Rightarrow x \leqslant - 1{\text{ or }}x \geqslant 3. \cr} $$
Now for $$x = 3$$  we get $$r = 1,$$  which will make $$a = b = c$$   Also for $$x = – 1,$$  we get $$r = – 1,$$  for which $$a = c,$$  thus $$x < – 1$$  or $$x > 3$$

$$\eqalign{ & \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = {1^4} + {3^4} + {5^4} + ..... + {{\left( {2n - 1} \right)}^4}} \cr & \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = {1^4} + {2^4} + {3^4} + ..... + } {\left( {2n} \right)^4} - \left\{ {{2^4} + {4^4} + ..... + {{\left( {2n} \right)}^4}} \right\} \cr & \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = f\left( {2n} \right) - 16\left\{ {{1^4} + {2^4} + ..... + {n^4}} \right\} = f\left( {2n} \right) - 16f\left( n \right).} \cr & {\text{If }}n = 2m,\,{\text{then }}\sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = {1^4} + {3^4} + {5^4} + ..... + {{\left( {4m - 1} \right)}^4}} \cr & = {1^4} + {2^4} + {3^4} + ..... + {\left( {2m} \right)^4} + {\left( {2m + 1} \right)^4} + ..... + {\left( {4m - 1} \right)^4} + {\left( {4m} \right)^4} - \left\{ {{2^4} + {4^4} + ..... + {{\left( {4m} \right)}^4}} \right\} \cr & = f\left( {4m} \right) - 16\left\{ {{1^4} + {2^4} + ..... + {{\left( {2m} \right)}^4}} \right\} \cr & = f\left( {2n} \right) - 16f\left( n \right),{\text{e}}{\text{.t}}{\text{.c}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Let, }}{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right){\text{and }}{\log _{10}}\left( {{2^x} + 3} \right){\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore 2{\log _{10}}\left( {{2^x} - 1} \right) = {\log _{10}}2 + {\log _{10}}\left( {{2^x} + 3} \right) \cr & \Rightarrow {\log _{10}}{\left( {{2^x} - 1} \right)^2} = {\log _{10}}2\left( {{2^x} + 3} \right) \cr & \Rightarrow {2^{2x}} + 1 - {2^{x + 1}} = {2.2^x} + 6 \cr & \Rightarrow {a^2} + 1 - 2a = 2a + 6\,\,\,\,{\text{where }}a = {2^x}. \cr & \Rightarrow {a^2} - 4a - 5 = 0 \cr & \Rightarrow a = 5\,\,{\text{or }}a = - 1 \cr & {2^x} = 5 \cr & \Rightarrow \log 2 = \log 5 \cr & \Rightarrow x = \frac{{\log 5}}{{\log 2}} \cr & \Rightarrow x = {\log _2}5 \cr} $$

$$\eqalign{ & {\text{Here, }}a + b = {a_1} + {a_{2n}} = {a_2} + {a_{2n - 1}} = ..... = {a_n} + {a_{n + 1}} \cr & {\text{and }}ab = {g_1}.{g_{2n}} = {g_2}.{g_{2n - 1}} = ..... = {g_n}.{g_{n + 1}}\,\,{\text{and }}h = \frac{{2ab}}{{a + b}}. \cr} $$

$$\eqalign{ & 2{\log _{100}}a - {\log _a}{\left( {\frac{1}{{100}}} \right)^2} = 2\left\{ {{{\log }_{100}}a + {{\log }_a}100} \right\}. \cr & {\text{Also, }}\frac{{{{\log }_{100}}a + {{\log }_a}100}}{2} \geqslant \sqrt {{{\log }_{100}}a \times {{\log }_a}100} = 1. \cr} $$