$$\eqalign{ & 1,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log_3} \left( {{{4.3}^x} - 1} \right){\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,{\text{2lo}}{{\text{g}}_9}\left( {{3^{1 - x}} + 2} \right) = 1 + {\log _3}\left( {{{4.3}^x} - 1} \right) \cr & \Rightarrow \,\,{\log _3}\left( {{3^{1 - x}} + 2} \right) = {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right) \cr & \Rightarrow \,\,{\log _3}\left( {{3^{1 - x}} + 2} \right) = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right] \cr & \Rightarrow \,\,{3^{1 - x}} + 2 = 3\left( {{{4.3}^x} - 1} \right) \cr & \Rightarrow \,\,{3.3^{ - x}} + 2 = {12.3^x} - 3. \cr & {\text{Put }}{{\text{3}}^x} = t \cr & \Rightarrow \,\,\frac{3}{t} + 2 = 12t - 3{\text{ or 12}}{t^2} - 5t - 3 = 0; \cr & {\text{Hence }}t = - \frac{1}{3},\frac{3}{4} \cr & \Rightarrow \,\,{3^x} = \frac{3}{4}\left( {{\text{as }}{{\text{3}}^x} \ne - ve} \right) \cr & \Rightarrow \,\,x = {\log _3}\left( {\frac{3}{4}} \right){\text{ or }}x = {\log _3}3 - {\log _3}4 \cr & \Rightarrow \,\,x = 1 - {\log _3}4 \cr} $$

Given expression is $$\frac{1}{2}\sum {\frac{{2a}}{{b + c - a}}} $$
$$\eqalign{ & = \frac{1}{2}\sum {\left( {\frac{{2a}}{{b + c - a}} + 1} \right) - \frac{3}{2}} \cr & = \frac{1}{2}\left( {a + b + c} \right)\sum {\frac{1}{{b + c - a}} - \frac{3}{2}} \cr & {\text{Now, as }}\left( {a + b + c} \right) = \sum {\left( {b + c - a} \right)} \cr} $$
Applying A.M. $$ \geqslant $$ H.M.
Minimum value of the expression $$ = \frac{1}{2} \times 9 - \frac{3}{2} = 3.$$

$$\eqalign{ & A + C = 2B{\text{ and }}A + B + C = {180^ \circ }{\text{ so, }}B = {60^ \circ } \cr & \therefore \cos {60^ \circ } = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} \cr & \Rightarrow {b^2} = {a^2} + {c^2} - ac \cr} $$

$$\eqalign{ & {\left( {\frac{8}{5}} \right)^2} + {\left( {\frac{{12}}{5}} \right)^2} + {\left( {\frac{{16}}{5}} \right)^2} + {\left( {\frac{{20}}{5}} \right)^2}..... + {\left( {\frac{{44}}{5}} \right)^2} \cr & S = \frac{{16}}{{25}}\left( {{2^2} + {3^2} + {4^2} + ..... + {{11}^2}} \right) \cr & = \frac{{16}}{{25}}\left( {\frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1} \right) \cr & = \frac{{16}}{{25}} \times 505 = \frac{{16}}{5} \times 101 \cr & \Rightarrow \frac{{16}}{5}m = \frac{{16}}{5} \times 101 \cr & \Rightarrow m = 101. \cr} $$

Let $$a,\,b$$  be the first terms of an A.P.
Then first two terms of H.P. is
$$\eqalign{ & \frac{1}{a} = \frac{2}{5} \Rightarrow a = \frac{5}{2}{\text{ and }}\frac{1}{b} = \frac{{12}}{{23}} \Rightarrow b = \frac{{23}}{{12}} \cr & \therefore \,{\text{common difference }}d = b - a \cr & = \frac{{23}}{{12}} - \frac{5}{2} \cr & = \frac{{23 - 30}}{{12}} \cr & = \frac{{ - 7}}{{12}} \cr} $$
Let $${n^{th}}$$ term be the smallest positive term of the A.P.
$$\eqalign{ & \therefore \,{a_n} \geqslant 0 \cr & \Rightarrow a + \left( {n - 1} \right)d \geqslant 0 \cr & \Rightarrow \frac{5}{2} + \left( {n - 1} \right)\left( {\frac{{ - 7}}{{12}}} \right) \geqslant 0 \cr & \Rightarrow 30 - 7\left( {n - 1} \right) \geqslant 0 \cr & \Rightarrow 30 - 7n + 7 \geqslant 0 \cr & \Rightarrow - 7n + 37 \geqslant 0 \cr & \Rightarrow - 7n \geqslant - 37 \cr & \Rightarrow 7n \leqslant 37 \cr & \Rightarrow n \leqslant \frac{{37}}{7} \cr & \Rightarrow n \leqslant 5.3\,\,\left( {{\text{approx}}{\text{.}}} \right) \cr} $$
$$\therefore \,n = 5$$   since $$n$$ is a natural number
$$\eqalign{ & \therefore \,{a_n} = {a_5} \cr & = a + \left( {5 - 1} \right)d \cr & = \frac{5}{2} + 4 \cdot \left( {\frac{{ - 7}}{{12}}} \right) \cr & = \frac{5}{2} - \frac{7}{3} \cr & = \frac{{15 - 14}}{6} \cr & = \frac{1}{6} \cr} $$
$$\therefore $$  Value of largest term of H.P.
$$\eqalign{ & {\text{ = }}\frac{1}{{{a_5}}} \cr & {\text{ = }}\frac{1}{{\frac{1}{6}}} \cr & = 6 \cr} $$

$$\eqalign{ & \ln \left( {a + c} \right),\ln \left( {a - c} \right),\ln \left( {a - 2b + c} \right){\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,a + c,a - c,a - 2b + c{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,{\left( {c - a} \right)^2} = \left( {a + c} \right)\left( {a - 2b + c} \right) \cr & \Rightarrow {\left( {c - a} \right)^2} = {\left( {a + c} \right)^2} - 2b\left( {a + c} \right) \cr & \Rightarrow \,2b\left( {a + c} \right) = {\left( {a + c} \right)^2} - {\left( {c - a} \right)^2} \cr & \Rightarrow 2b\left( {a + c} \right) = 4ac \cr & \Rightarrow \,b = \frac{{2ac}}{{a + c}} \cr & \Rightarrow \,a,b,c{\text{ are in H}}{\text{.P}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{The product is }}P = {2^{\frac{1}{4}}}{.2^{\frac{2}{8}}}{.2^{\frac{3}{{16}}}}...... \cr & = {2^{\frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + ......\,\,\infty }} \cr & {\text{Now let }}S = \frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + ......\,\infty \,\,\,\,\,\,......\left( 1 \right) \cr & \frac{1}{2}S = \frac{1}{8} + \frac{2}{{16}} + ......\,\infty \,\,\,\,\,\,\,\,......\left( 2 \right) \cr & {\text{Subtracting }}\left( 2 \right){\text{ from }}\left( 1 \right) \cr & \Rightarrow \,\,\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + ......\,\infty \cr & {\text{or }}\frac{1}{2}S = \frac{{\frac{1}{4}}}{{1 - \frac{1}{2}}} = \frac{1}{2} \cr & \Rightarrow \,\,S = 1 \cr & \therefore \,\,P = {2^S} = 2 \cr} $$

$$n$$ = index of the highest power of $$x = 1 + 2 + 4 + . . . . . + 128.$$

$$\eqalign{ & \frac{1}{{{a_2}}} - \frac{1}{{{a_1}}} = \frac{1}{{{a_3}}} - \frac{1}{{{a_2}}} = ...... = \frac{1}{{{a_n}}} - \frac{1}{{{a_{n - 1}}}} = d\,\,\,\left( {{\text{say}}} \right) \cr & {\text{Then }}{a_1}{a_2} = \frac{{{a_1} - {a_2}}}{d},\,{a_2}{a_3} = \frac{{{a_2} - {a_3}}}{d},......,{a_{n - 1}}{a_n} = \frac{{{a_{n - 1}} - {a_n}}}{d} \cr & \therefore \,\,{a_1}{a_2} + {a_2}{a_3} + ...... + {a_{n - 1}}{a_n} \cr & = \frac{{{a_1} - {a_2}}}{d} + \frac{{{a_2} - {a_3}}}{d} + ...... + \frac{{{a_{n - 1}}{a_n}}}{d} \cr & = \frac{1}{d}\left[ {{a_1} - {a_2} + {a_2} - {a_3} + ...... + {a_{n - 1}}{a_n}} \right] = \frac{{{a_1} - {a_n}}}{d} \cr & {\text{Also, }}\frac{1}{{{a_n}}} = \frac{1}{{{a_1}}} + \left( {n - 1} \right)d \cr & \Rightarrow \,\,\frac{{{a_1} - {a_n}}}{{{a_1}{a_n}}} = \left( {n - 1} \right)d \cr & \Rightarrow \,\,\frac{{{a_1} - {a_n}}}{d} = \left( {n - 1} \right){a_1}{a_n} \cr & {\text{Which is the required result}}{\text{.}} \cr} $$

Three numbers $$a, b$$  and $$c$$ will be in G.P. if $$b^2 = ac.$$  Only option $$(D)$$ i.e. $${\tan ^2}{30^ \circ },{\tan^2}{45^ \circ }{\text{and}}\,{\tan^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
$$\eqalign{ & \because {\text{ta}}{{\text{n}}^2}{30^ \circ } = \frac{1}{3} \cr & {\tan ^2}{45^ \circ } = 1 \cr & {\text{and }}{\tan ^2}{60^ \circ } = 3 \cr & \therefore {\tan ^2}{30^ \circ },{\tan^2}{45^ \circ }{\text{and}}\,{\tan^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}} \cr} $$