$$\eqalign{ & A = \frac{{{a_1} + {a_2} + ..... + {a_n}}}{n},G = \root n \of {{a_1}{a_2}.....{a_n}} = 1; \cr & A \geqslant G \cr & \Rightarrow \,\,{a_1} + {a_2} + ..... + {a_n} \geqslant n. \cr} $$

$${n^{th}}$$ term is the middle term in each case. So $$a, b, c$$  are the AM, GM, HM respectively of the same two numbers. For any two numbers AM, GM, HM are in GP.

$$\eqalign{ & 2.\overline {357} = 2 + .357 + 0.000357 + ....\infty \cr & = 2 + \frac{{357}}{{{{10}^3}}} + \frac{{357}}{{{{10}^6}}} + ....\infty = 2 + \frac{{\frac{{357}}{{{{10}^3}}}}}{{1 - \frac{1}{{{{10}^3}}}}} \cr & = 2 + \frac{{357}}{{999}} = \frac{{2355}}{{999}} \cr} $$

$$\eqalign{ & {\log _e}5 + {\log _e}\left( {{5^x} - \frac{{11}}{5}} \right) = 2{\log _e}\left( {{5^x} - 1} \right) \cr & \Rightarrow {5^{x + 1}} - 11 = {5^{2x}} + 1 - 2 \times {5^x} \cr & \Rightarrow {5^{2x}} - {7.5^x} + 12 = 0 \cr & {\text{Let, }}{5^x} = t,{t^2} - 7t + 12 = 0 \cr & \Rightarrow t = 4,3 \cr & {5^x} = 4 \cr & {\log _5}{5^x} = {\log _5}4 \cr & x = {\log _5}4 \cr & {5^x} = 3 \cr & {\log _5}{5^x} = {\log _5}3 \cr & x = {\log _5}3 \cr} $$

$$\eqalign{ & {\text{Given that for an A}}{\text{.P}}{\text{., }}{S_n} = c{n^2} \cr & {\text{Then }}\,{T_n} = {S_n} - {S_{n - 1}} = c{n^2} - c{\left( {n - 1} \right)^2} \cr & \,\,\,\,\,\, = \left( {2n - 1} \right)c \cr & \therefore \,\,{\text{Sum of squares of }}n{\text{ terms of this A}}{\text{.P}}{\text{.}} \cr & {\text{ = }}\sum {T_n^2 = \sum {{{\left( {2n - 1} \right)}^2}.{c^2}} } \cr & = \,\,{c^2}\left[ {4\sum {{n^2} - 4} \sum {n + n} } \right] \cr & = \,\,{c^2}\left[ {\frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{4n\left( {n + 1} \right)}}{2} + n} \right] \cr & = \,\,{c^2}n\left[ {\frac{{2\left( {2{n^2} + 3n + 1} \right) - 6\left( {n + 1} \right) + 3}}{3}} \right] \cr & = \,\,{c^2}n\left[ {\frac{{4{n^2} - 1}}{3}} \right] = \frac{{n\left( {4{n^2} - 1} \right){c^2}}}{3} \cr} $$

$$\eqalign{ & AM \geqslant HM \cr & \Rightarrow \frac{{a + b + c}}{3} \geqslant \frac{3}{{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}} \cr & \Rightarrow \left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \geqslant 9 \cr} $$

$$\eqalign{ & {\text{Given that }}a,b,c{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,{\text{2}}b = a + c \cr & \Rightarrow \,\,{\text{but given }}a + b + c = \frac{3}{2} \cr & \Rightarrow \,\,3b = \frac{3}{2} \cr & \Rightarrow \,\,b = \frac{1}{2}{\text{ and then }}a + c = 1 \cr & {\text{Again }}{a^2},{b^2},{c^2},{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,{b^4} = {a^2}{c^2} \cr & \Rightarrow \,\,{b^2} = \pm ac \cr & \Rightarrow \,\,ac = \frac{1}{4}{\text{ or }} - \frac{1}{4}\,\,{\text{and }}a + c = 1\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Considering }}a + c = 1{\text{ and }}ac = \frac{1}{4} \cr & \Rightarrow \,\,{\left( {a - c} \right)^2} = 1 - 1 = 0 \cr & \Rightarrow \,\,a = c{\text{ but }}a \ne c{\text{ as given that }}a < b < c \cr & \therefore \,\,{\text{ We consider }}a + c = 1{\text{ and }}ac = - \frac{1}{4} \cr & \Rightarrow \,\,{\left( {a - c} \right)^2} = 1 + 1 = 2 \cr & \Rightarrow \,\,a - c = \pm \sqrt 2 \,\,{\text{but }}a < c \cr & \Rightarrow \,\,a - c = - \sqrt 2 \,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{Solving }}\left( 1 \right){\text{ and }}\left( 2 \right){\text{ we get }}a = \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr} $$

Let $$S = i - 2 - 3i + 4 + 5i.... 100\,{\text{terms,}}$$
$$\eqalign{ & \Rightarrow S = i + 2{i^2} + 3{i^3} + 4{i^4} + 5{i^5}.... + 100{i^{100}} \cr & \Rightarrow iS = {i^2} + 2{i^3} + 3{i^4}.... + 99{i^{100}} + 100{i^{101}} \cr & \Rightarrow S - iS = i + {i^2} + {i^3} + {i^4} + .... + {i^{100}} - 100{i^{101}} \cr & \Rightarrow S\left( {1 - i} \right) = \frac{{i\left( {1 - {i^{100}}} \right)}}{{1 - i}} - 100{i^{101}} \cr & \Rightarrow S\left( {1 - i} \right) = - 100i \cr & \Rightarrow S = \frac{{ - 100i}}{{1 - i}} = - 50i\left( {1 + i} \right) = - 50\left( {i - 1} \right) \cr & = 50\left( {1 - i} \right) \cr} $$

$${\left( {1 - 1 + 2 - 2 + ..... + 5 - 5} \right)^2} = {1^2} + {1^2} + {2^2} + {2^2} + ..... + {5^2} + {5^2} + 2S,$$              where $$S$$ is the required sum
or, $$0 = 2\left( {{1^2} + {2^2} + {3^2} + {4^2} + {5^2}} \right) + 2S.$$

We have,
$$\eqalign{ & \sum\limits_{r{\text{ }} = {\text{ }}1}^n {\left( {x{\text{ }} + {\text{ }}r - 1} \right)\left( {x{\text{ }} + {\text{ }}r} \right) = 10n} \cr & \sum\limits_{r{\text{ }} = {\text{ }}1}^n {\left( {{x^2} + xr{\text{ }} + \left( {r - 1} \right)x{\text{ }} + {\text{ }}{r^2} - r} \right) = 10n} \cr & \Rightarrow \,\,\sum\limits_{r{\text{ }} = {\text{ }}1}^n {\left( {{x^2} + \left( {2r - 1} \right)x{\text{ }} + {\text{ }}r\left( {r - 1} \right)} \right) = 10n} \cr & \Rightarrow \,n{x^2} + \left\{ {1 + 3 + 5 + ..... + \left( {2n - 1} \right)} \right\}x{\text{ }} + \left\{ {1.2 + 2.3 + ..... + \left( {n - 1} \right)n} \right\} = 10n \cr & \Rightarrow \,n{x^2} + {n^2}x + \frac{{\left( {n - 1} \right)n\left( {n + 1} \right)}}{3} = 10n \cr & \Rightarrow \,{x^2} + nx + \frac{{{n^2} - 31}}{3} = 0 \cr} $$
Let $$\alpha $$ and $$\alpha + 1$$  be its two solutions ( $$\because $$ it has two consequtive integral solutions )
$$\eqalign{ & \Rightarrow \,\,\alpha + \left( {\alpha + 1} \right) = - n \cr & \Rightarrow \,\,\alpha = \frac{{ - n - 1}}{2}\,\,\,\,\,\,\,\,......\left( 1 \right) \cr & {\text{Also }}\alpha \left( {\alpha + 1} \right) = \frac{{{n^2} - 31}}{3}\,\,\,\,\,\,\,......\left( 2 \right) \cr} $$
Putting value of (1) in (2), we get
$$\eqalign{ & - \left( {\frac{{n + 1}}{2}} \right)\left( {\frac{{1 - n}}{2}} \right) = \frac{{{n^2} - 31}}{3} \cr & \Rightarrow \,\,{n^2} = 121 \cr & \Rightarrow \,\,n = 11 \cr} $$