$$\eqalign{ & f\left( 1 \right) = \left[ {\frac{1}{2} + \frac{1}{{100}}} \right] = 0,....., \cr & f\left( {49} \right) = \left[ {\frac{1}{2} + \frac{{49}}{{100}}} \right] = \left[ {\frac{{99}}{{100}}} \right] = 0, \cr & f\left( {50} \right) = \left[ {\frac{1}{2} + \frac{{50}}{{100}}} \right] = 1, \cr & f\left( {51} \right) = \left[ {\frac{1}{2} + \frac{{51}}{{100}}} \right] = 1,....., \cr & f\left( {100} \right) = \left[ {\frac{1}{2} + \frac{{100}}{{100}}} \right] = 1. \cr} $$

The given series is arithmetic whose first term $$= 20,$$  and common difference $$ = - \frac{2}{3}$$
As the common difference is negative the terms will become negative after some stage. So the sum is maximum when all positive terms are added
Now, for the positive terms
$$\eqalign{ & {x_n} \geqslant 0 \cr & \Rightarrow 20 + \left( {n - 1} \right) \times - \frac{2}{3} \geqslant 0 \cr & \Rightarrow 60 - 2\left( {n - 1} \right) \geqslant 0 \cr & \Rightarrow n \leqslant 31. \cr} $$
∴ The first 31 terms are non - negative
∴ Maximum sum
$$ = {S_{31}} = \frac{{31}}{2}\left[ {2 \times 20 + \left( {31 - 1} \right) \times - \frac{2}{3}} \right] = 310$$

Use $$A \geqslant G$$  for the numbers $$a{b^2},a{c^2},b{c^2},b{a^2},c{a^2},c{b^2}.$$

Lengths of line segments on one side of the diagonal are
$$\sqrt 2 ,2\sqrt 2 ,3\sqrt 2 ,.....,\left( {n - 1} \right)\sqrt 2 .$$
So, the required sum $$ = 2\left\{ {\sqrt 2 + 2\sqrt 2 + 3\sqrt 2 + ..... + \left( {n - 1} \right)\sqrt 2 } \right\} + n\sqrt 2 $$
$$ = 2\sqrt 2 \left\{ {1 + 2 + 3 + ..... + \left( {n - 1} \right)} \right\} + n\sqrt 2 .$$

$$\eqalign{ & \frac{1}{H} = \frac{1}{2}\left( {\frac{1}{{\rm{\alpha }}} + \frac{1}{{\rm{\beta }}}} \right) \cr & = \frac{{{\rm{\alpha }} + {\rm{\beta }}}}{{2{\rm{\alpha \beta }}}} \cr & = \frac{{\left( {4 + \sqrt 5 } \right)}}{{\left( {5 + \sqrt 2 } \right)}}.\frac{{\left( {5 + \sqrt 2 } \right)}}{{\left( {8 + 2\sqrt 5 } \right)}}.\frac{1}{2} \cr & = \frac{1}{4} \cr & \therefore \,\,H = 4. \cr} $$

$$\eqalign{ & \because \,\,{S_n} = \left( {50 - \frac{{7A}}{2}} \right)n + {n^2} \times \frac{A}{2} \cr & \Rightarrow \,\,{a_1} = 50 - 3A \cr & \therefore \,\,d = {a_2} - {a_1} = \left( {{S_2} - {S_1}} \right) - {S_1} \cr & \Rightarrow \,\,d = \frac{A}{2} \times 2 = A \cr & {\text{Now, }}{a_{50}} = {a_1} + 49 \times d \cr & = \left( {50 - 3A} \right) + 49A = 50 + 46A \cr & {\text{So,}}\,\left( {d,{a_{50}}} \right) = \left( {A,50 + 46A} \right) \cr} $$

$$\eqalign{ & A \geqslant G \cr & \Rightarrow \,\,\frac{{{{\log }_{{a_2}}}{a_1} + {{\log }_{{a_3}}}{a_2} + ..... + {{\log }_{{a_1}}}{a_n}}}{n} \geqslant \root n \of {{{\log }_{{a_2}}}{a_1} \cdot {{\log }_{{a_3}}}{a_2} \cdot .....{{\log }_{{a_1}}}{a_n}} = 1. \cr} $$

Let,
\[\frac{\begin{gathered} S = 3 + 7 + 13 + 21 + 31 + \,\,\,.....\,\,\, + {a_n} \hfill \\ \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,- S = \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,3 + 7 + 13 + 21 + 31 + .... + {a_{n - 1}} + {a_n} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, - \,\,\,\,\, - \,\,\,\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\, - \hfill \\ \end{gathered} }{{0 = 3 + 4 + 6 + 8 + 10 + 12 + ..... - {a_n}\,\,\,\,\,\,\,\,\,\,\,}}\]
$$\eqalign{ & \Rightarrow {a_n} = 3 + \left[ {4 + 6 + 8 + 10 + 12 + .....\,\left( {n - 1} \right){\text{terms}}} \right] \cr & = 3 + \frac{{\left( {n - 1} \right)}}{2}\left[ {8 + \left\{ {\left( {n - 1} \right) - 1} \right\} \times 2} \right] \cr & = 3 + \frac{{\left( {n - 1} \right)}}{2}\left[ {8 + 2n - 4} \right] \cr & = 3 + \frac{{\left( {n - 1} \right)}}{2}\left( {2n + 4} \right) \cr & = 3 + \left( {n - 1} \right)\left( {n + 2} \right) \cr & \therefore {15^{th}}{\text{ term }} = {a_{15}} = 3 + \left( {15 - 1} \right)\left( {15 + 2} \right) \cr & = 3 + 14 \times 17 = 241 \cr} $$

$$\eqalign{ & x + \frac{1}{x} \geqslant 2,.....,{x^n} + \frac{1}{{{x^n}}} \geqslant 2 \cr & {\text{on adding}} \cr & \left( {x + \frac{1}{x}} \right) + \left( {{x^2} + \frac{1}{{{x^2}}}} \right) + ..... + \left( {{x^n} + \frac{1}{{{x^n}}}} \right) \geqslant 2n, \cr & \left( {\frac{1}{{{x^n}}} + \frac{1}{{{x^{n - 1}}}} + ..... + \frac{1}{x}} \right) + 1 + \left( {x + {x^2} + ..... + {x^n}} \right) \geqslant 1 + 2n \cr & \frac{{\left( {1 + x + ..... + {x^{n - 1}} + {x^n}} \right) + {x^{n + 1}} + {x^{n + 2}} + ..... + {x^{2n}}}}{{{x^n}}} \geqslant 1 + 2n \cr & \frac{{{x^n}}}{{1 + x + ..... + {2^{2n}}}} \leqslant \frac{1}{{1 + 2n}} \cr} $$

$$\eqalign{ & {\left( {2a} \right)^2} + {\left( {3b} \right)^2} + {\left( {4c} \right)^2} - \left( {2a} \right)\left( {3b} \right) - \left( {3b} \right)\left( {4c} \right) - \left( {4c} \right)\left( {2a} \right) = 0 \cr & \Rightarrow \,\,{\left( {2a - 3b} \right)^2} + {\left( {3b - 4c} \right)^2} + {\left( {4c - 2a} \right)^2} = 0 \cr & \Rightarrow \,\,2a = 3b = 4c \cr & \Rightarrow \,\,\frac{a}{{\frac{1}{2}}} = \frac{b}{{\frac{1}{3}}} = \frac{c}{{\frac{1}{4}}}. \cr} $$