$$\eqalign{ & \frac{{{S_{nx}}}}{{{S_x}}} = \frac{{\frac{{nx}}{2}\left[ {2a + \left( {nx - 1} \right)d} \right]}}{{\frac{x}{2}\left[ {2a + \left( {x - 1} \right)d} \right]}} \cr & = \frac{{n\left[ {\left( {2a - d} \right) + nxd} \right]}}{{\left( {2a - d} \right) + xd}} \cr} $$
For $$\frac{{{S_{nx}}}}{{{S_x}}}$$ to be independent of $$x$$     $$2a – d = 0$$
∴ $$2a=d$$
now, $${S_p} = \frac{P}{2}\left[ {2a + \left( {P - 1} \right)d} \right] = {P^2}a$$

As given the series is
$$S = 1 + \frac{1}{8} + \frac{{1.3}}{{8.16}} + \frac{{1.3.5}}{{8.16.24}} + .....\,\infty $$
On comparing this series with
$$S = {\left( {1 + x} \right)^n} = 1 + nx + \frac{{n\left( {n - 1} \right)}}{{2\,!}}{x^2} + .....\,\infty ,$$
we get,
$$\eqalign{ & nx = \frac{1}{8}\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}\frac{{n\left( {n - 1} \right)}}{{2\,!}}{x^2} = \frac{{1.3}}{{8.16}}\,\,\,\,\,.....\left( 2 \right) \cr} $$
From Eqs. (1) and (2), we get
$$\eqalign{ & \frac{{\frac{{n\left( {n - 1} \right)}}{{2\,!}}{x^2}}}{{{n^2}{x^2}}} = \frac{{\frac{{1.3}}{{8.16}}}}{{\frac{1}{8} \cdot \frac{1}{8}}} \cr & \Rightarrow \frac{{n - 1}}{{2n}} = \frac{3}{2} \cr & \Rightarrow n - 1 = 3n \cr & \Rightarrow n = - \frac{1}{2} \cr} $$
On putting this value in Eq. (i)
$$\eqalign{ & \Rightarrow \left( { - \frac{1}{2}} \right)x = \frac{1}{8} \cr & \Rightarrow x = - \frac{1}{4}. \cr & {\text{But, }}S = {\left( {1 + x} \right)_n} = {\left( {1 - \frac{1}{4}} \right)^{ - \frac{1}{2}}} \cr & = {\left( {\frac{3}{4}} \right)^{ - \frac{1}{2}}} = \frac{2}{{\sqrt 3 }}. \cr} $$

Let the sides of the triangle be $$a - d, a, a + d$$    then
Perimeter $$ = \left( {a + d} \right) + a + \left( {a - d} \right) = 21$$
∴ $$a = 7$$
Again, $$\left( {a - d} \right)\left( {a + d} \right) = a + 6$$
$$\eqalign{ & \Rightarrow {a^2} - {d^2} = a + 6 \cr & \Rightarrow 49 - {d^2} = 13 \cr} $$
$$\therefore d = \pm 6.$$   Hence, the sides of the triangle are $$1\,cm, 7\,cm, 13\,cm.$$

There are 50 factors. So, the co-efficient of $${x^{49}} = - 1 - 3 - 5 - ..... - 99.$$

If $$a,ar,a{r^2},a{r^3},.....$$    are in G.P., then sum of infinite G.P. $$ = a + ar + ..... + \infty = \frac{a}{{1 - r}}$$
where $$'a'$$ is the first term and $$'r'$$ is the common ratio of G.P.
Given, $$x = 1 + a + {a^2} + .....\,\infty $$
This is a G.P., with common ratio $$'a'.$$
$$\eqalign{ & \Rightarrow x = \frac{1}{{1 - a}} \cr & \Rightarrow x - ax = 1 \cr & \Rightarrow a = \frac{{x - 1}}{x} \cr} $$
Again, $$y = 1 + b + {b^2} + .....\,\infty $$
This is also a G.P., with common ratio $$'b'.$$
$$\eqalign{ & \Rightarrow y = \frac{1}{{1 - b}} \cr & \Rightarrow b = \frac{{y - 1}}{y} \cr} $$
Now, consider $$1 + ab + {a^2}{b^2} + .....\,\infty $$
which is again $$a$$ G.P. with common ratio $$'ab'.$$
∴ Sum $$ = \frac{1}{{1 - ab}} = \frac{1}{{1 - \frac{{x - 1}}{x} \cdot \frac{{y - 1}}{y}}}$$
$$ = \frac{{xy}}{{xy - xy + x + y - 1}} = \frac{{xy}}{{x + y - 1}}$$

$$\eqalign{ & \because \sum\limits_{k = 0}^{12} {{a_{4k + 1}} = 416} \cr & \Rightarrow \,\,\frac{{13}}{2}\left[ {2{a_1} + 48d} \right] = 416 \cr & \Rightarrow \,\,{a_1} + 24d = 32\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Now, }}{a_9} + {a_{43}} = 66 \cr & \Rightarrow \,\,2{a_1} + 50d = 66\,\,\,\,\,\,\,......\left( 2 \right) \cr & {\text{From eq}}{\text{.}}\left( 1 \right)\& \left( 2 \right){\text{we get; }}d = 1{\text{ and }}{a_1} = 8 \cr & {\text{Also, }}\sum\limits_{r = 1}^{17} {a_r^2} = \sum\limits_{r = 1}^{17} {{{\left[ {8 + \left( {r - 1} \right)1} \right]}^2} = 140m} \cr & \Rightarrow \,\,\sum\limits_{r = 1}^{17} {{{\left( {r + 7} \right)}^2} = 140m} \cr & \Rightarrow \,\,\sum\limits_{r = 1}^{17} {\left( {{r^2} + 14r + 49} \right) = 140m} \cr & \Rightarrow \,\,\left( {\frac{{17 \times 18 \times 35}}{6}} \right) + 14\left( {\frac{{17 \times 18}}{2}} \right) + \left( {49 \times 17} \right) = 140 \cr & \Rightarrow \,\,m = 34 \cr} $$

$$\eqalign{ & \frac{{{a_1} + {a_4}}}{{{a_1}{a_4}}} = \frac{{{a_2} + {a_3}}}{{{a_2}{a_3}}}; \cr & {\text{so,}}\,\frac{1}{{{a_4}}} + \frac{1}{{{a_1}}} = \frac{1}{{{a_3}}} + \frac{1}{{{a_2}}} \cr & {\text{or,}}\,\frac{1}{{{a_4}}} - \frac{1}{{{a_3}}} = \frac{1}{{{a_2}}} - \frac{1}{{{a_1}}}\,\,\,.....\left( 1 \right) \cr & {\text{Also}},\,\frac{{3\left( {{a_2} - {a_3}} \right)}}{{{a_2}{a_3}}} = \frac{{{a_1} - {a_4}}}{{{a_1}{a_4}}}; \cr & {\text{so,}}\,{\text{3}}\left( {\frac{1}{{{a_3}}} - \frac{1}{{{a_2}}}} \right) = \frac{1}{{{a_4}}} - \frac{1}{{{a_1}}}\,\,\,.....\left( 2 \right) \cr} $$
Clearly, (1) and (2)
$$ \Rightarrow \,\,\frac{1}{{{a_2}}} - \frac{1}{{{a_1}}} = \frac{1}{{{a_3}}} - \frac{1}{{{a_2}}} = \frac{1}{{{a_4}}} - \frac{1}{{{a_3}}};$$
$${\text{so, }}\frac{1}{{{a_1}}},\frac{1}{{{a_2}}},\frac{1}{{{a_3}}}$$    are in A.P.

Taking the sequence $$3, 9, 27, 81, . . . . .$$
Its $$n^{th}$$ term $$ = 3{\left( 3 \right)^{n - 1}} = {3^n}$$
Also take the sequence $$2,8,26,80,.....{\text{ or }}\left( {3 - 1} \right),\left( {9 - 1} \right),\left( {27 - 1} \right),\left( {81 - 1} \right),.....$$
Its $$n^{th}$$ term $$= {3^n} - 1$$
Hence, $$n^{th}$$ term of the sequence
$$\frac{2}{3} + \frac{8}{9} + \frac{{26}}{{27}} + \frac{{80}}{{81}} + .....\,\,{\text{is }}\frac{{{3^n} - 1}}{{{3^n}}}{\text{ or }}1 - {3^{ - n}}$$
Now the sum $$\left( {{S_n}} \right) = \sum {\left( {1 - {3^{ - n}}} \right)} $$
$$\eqalign{ & = n - \left( {{3^{ - 1}} + {3^{ - 2}} + ..... + {3^{ - n}}} \right) \cr & = n - \frac{{{3^{ - 1}}\left\{ {1 - {{\left( {{3^{ - 1}}} \right)}^n}} \right\}}}{{1 - {3^{ - 1}}}} = n - \frac{1}{2}\left( {1 - {3^{ - n}}} \right) \cr & = n + \frac{1}{2}\left( {{3^{ - n}} - 1} \right). \cr} $$

$$\eqalign{ & {\text{Let, }}\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \frac{{a + b}}{2} \cr & \Rightarrow 2{a^{n + 1}} + 2{b^{n + 1}} = {a^{n + 1}} + {b^{n + 1}} + a{b^n} + b{a^n} \cr & \Rightarrow {a^{n + 1}} - {a^n}b + {b^{n + 1}} - a{b^n} = 0 \cr & \Rightarrow \left( {a - b} \right)\left( {{a^n} - {b^n}} \right) = 0 \cr} $$
$$ \Rightarrow {a^n} = {b^n}$$   it is possible for unequal numbers $$a$$ and $$b$$ if $$n = 0$$
$$\eqalign{ & {\text{Let, }}\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \sqrt {ab} \cr & \Rightarrow {a^{n + 1}} + {b^{n + 1}} = {a^{n + \frac{1}{2}}}{b^{\frac{1}{2}}} + {a^{\frac{1}{2}}}{b^{n + \frac{1}{2}}} \cr & \Rightarrow \left( {{a^{n + \frac{1}{2}}} - {b^{n + \frac{1}{2}}}} \right)\left( {\sqrt a - \sqrt b } \right) = 0 \cr & \Rightarrow {a^{n + \frac{1}{2}}} - {b^{n + \frac{1}{2}}} = 0 \cr} $$
which holds true if $$n + \frac{1}{2} = 0 \Rightarrow n = - \frac{1}{2}$$
$$\eqalign{ & {\text{Let, }}\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \frac{{2ab}}{{a + b}} \cr & \Rightarrow {a^{n + 2}} + {a^{n + 1}}b + a{b^{n + 1}} + {b^{n + 2}} = 2{a^{n + 1}}b + 2a{b^{n + 1}} \cr & \Rightarrow \left( {a - b} \right)\left( {{a^{n + 1}} - {b^{n + 1}}} \right) = 0 \cr & \Rightarrow {a^{n + 1}} - {b^{n + 1}} = 0 \cr & \Rightarrow n = - 1 \cr} $$

$$\eqalign{ & S = 3.6 + 4.7 + .....\,{\text{upto }}n - 2{\text{ terms}} \cr & = \left( {1.4 + 2.5 + 3.6 + 4.7 + .....\,{\text{upto }}n{\text{ terms}}} \right) - 14 \cr & = \sum n \left( {n + 3} \right) - 14 = \frac{1}{6}\left( {2{n^3} + 12{n^2} + 10n} \right) - 14 \cr & = \left( {\frac{{2{n^3} + 12{n^2} + 10n - 84}}{6}} \right),{\text{ where }}n = 3,4,5..... \cr} $$