Let the $$1025^{th}$$  term = $$2^n.$$ Then
$$\eqalign{ & 1 + 2 + 4 + 8 + ..... + {2^{n - 1}} < 1025 \leqslant 1 + 2 + 4 + 8 + ..... + {2^n} \cr & \therefore \,\,{2^n} - 1 < 1025 < {2^{n + 1}} - 1\,\,{\text{or, }}{2^n} < 1026 < {2^{n + 1}} \cr & \Rightarrow \,\,n = 10. \cr} $$

$$\eqalign{ & a,b,c,d\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\,\,\,d,c,b,a{\text{ are also in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,\,\frac{d}{{abcd}},\frac{c}{{abcd}},\frac{b}{{abcd}},\frac{a}{{abcd}}{\text{ are also in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,\,\frac{1}{{abc}},\frac{1}{{abd}},\frac{1}{{acd}},\frac{1}{{bcd}}{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,abc,abd,acd,bcd{\text{ are in H}}{\text{.P}}{\text{.}} \cr} $$

$${\text{HM}}\,{\text{ = }}\,\frac{{2ab}}{{a + b}} = \frac{2}{{{a^{ - 1}} + {b^{ - 1}}}} = \frac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}},\,{\text{where}}\,\,n = 0.$$

$$\eqalign{ & \because \,\,{a_1},{a_2},{a_3},.....{\text{ are in H}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\frac{1}{{{a_1}}},\frac{1}{{{a_2}}},\frac{1}{{{a_3}}}.....{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\,\,\frac{1}{{{a_1}}} = \frac{1}{5}{\text{ and }}\frac{1}{{{a_{20}}}} = \frac{1}{{25}} \cr & \frac{1}{{{a_1}}} + 19d = \frac{1}{{{a_{20}}}} \cr & \Rightarrow \,\,\frac{1}{5} + 19d = \frac{1}{{25}} \cr & \Rightarrow \,\,d = \frac{{ - 4}}{{475}} \cr & {\text{Now }}\frac{1}{{{a_n}}} = \frac{1}{5} + \left( {n - 1} \right)\left( {\frac{{ - 4}}{{475}}} \right) \cr & {\text{Clearly }}\,{a_n} < 0{\text{ }}\,{\text{if }}\,\frac{1}{{{a_n}}} < 0 \cr & \Rightarrow \,\,\frac{1}{5} - \frac{{4n}}{{475}} + \frac{4}{{475}} < 0 \cr & \Rightarrow \,\, - 4n < - 99{\text{ or }}n > \frac{{99}}{4} = 24\frac{3}{4} \cr & \therefore \,\,n \geqslant 25 \cr & {\text{Hence least value of }}n{\text{ is 25}}{\text{.}} \cr} $$

Let $$d$$ be the common difference of the A.P.
Then $${a_{2r}} = {a_{2r - 1}} + d.$$
$$\eqalign{ & \therefore \sum\limits_{r = 1}^{100} {{a_{2r}}} = \sum\limits_{r = 1}^{100} {\left( {{a_{2r - 1}} + d} \right)} = \sum\limits_{r = 1}^{100} {{a_{2r - 1}} + 100d} \cr & \Rightarrow \alpha = \beta + 100d \cr & \Rightarrow d = \frac{{\alpha - \beta }}{{100}} \cr} $$

$$\eqalign{ & f\left( x \right) = a{x^2} + bx + c \cr & f\left( 1 \right) = a + b + c = 3 \cr & \Rightarrow f\left( 1 \right) = 3 \cr & {\text{Now }}f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Put }}x = y = 1{\text{ in eqn}}{\text{.}}\left( 1 \right) \cr & f\left( 2 \right) = f\left( 1 \right) + f\left( 1 \right) + 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 2f\left( 1 \right) + 1 \cr & f\left( 2 \right) = 7 \cr & \Rightarrow \,f\left( 3 \right) = 12 \cr & {\text{Now, }}{S_n} = 3 + 7 + 12 + ...... + {t_{n\,\,\,\,}}\,\,\,\,\,\,.....\left( 1 \right) \cr & {S_n} = 3 + 7 + ...... + {t_{n - 1}} + {t_n}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{Subtract}}\left( 2 \right){\text{from}}\left( 1 \right) \cr & {t_n} = 3 + 4 + 5 + ......{\text{ upto }}n{\text{ terms}} \cr & {t_n} = \frac{{\left( {{n^2} + 5n} \right)}}{2} \cr & {S_n} = \sum {{t_n}\, = \sum {\frac{{\left( {{n^2} + 5n} \right)}}{2}} } \cr & {S_n} = \frac{1}{2}\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{5n\left( {n + 1} \right)}}{2}} \right] \cr & \,\,\,\,\,\, = \frac{{n\left( {n + 1} \right)\left( {n + 8} \right)}}{6} \cr & {S_{10}} = \frac{{10 \times 11 \times 18}}{6} = 330 \cr} $$

$$\eqalign{ & a = 1 + 10 + {10^2} + ..... + {10^{54}} \cr & = \frac{{{{10}^{55}} - 1}}{{10 - 1}} = \frac{{{{10}^{55}} - 1}}{{{{10}^5} - 1}} \cdot \frac{{{{10}^5} - 1}}{{10 - 1}} = bc \cr} $$