$$\eqalign{ & \because a,b,c{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \therefore \frac{b}{a} = \frac{c}{b} = r \cr & \Rightarrow \frac{{{b^2}}}{{{a^2}}} = \frac{{{c^2}}}{{{b^2}}} = {r^2} \cr & \Rightarrow {a^2},{b^2},{c^2}{\text{ are in G}}{\text{.P}}{\text{.}} \cr} $$

Since an A.P. is either increasing or decreasing, if possible let $$– 4$$ be the first term of an A.P., whose $$m^{th}$$ and $$n^{th}$$ terms are respectively 8 and 13. Then
$$\eqalign{ & 8 = - 4 + \left( {m - 1} \right)d\,\,{\text{and }}13 = - 4 + \left( {n - 1} \right)d \cr & \Rightarrow \frac{{12}}{{m - 1}} = \frac{{17}}{{n - 1}} = d \cr & {\text{Let }}\frac{{m - 1}}{{12}} = \frac{{n - 1}}{{17}} = k,\,{\text{then }}m = 12k + 1,\,{\text{and }}n = 17\,k + 1 \cr} $$
∴ for $$k = 1, 2, 3, .....$$   we get different pairs of values of $$m$$ and $$n,$$ which shows that infinite number of A.P.’s can be obtained

$$\eqalign{ & x = \frac{1}{{{1^4}}} + \frac{1}{{{3^4}}} + \frac{1}{{{5^4}}} + .....\,{\text{to }}\infty \cr & x = \left( {\frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + .....\,{\text{to }}\infty } \right) - \left( {\frac{1}{{{2^4}}} + \frac{1}{{{4^4}}} + .....\,{\text{to }}\infty } \right) \cr & x = \frac{{{\pi ^4}}}{{90}} - \frac{1}{{16}}\left( {\frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + .....\,{\text{to }}\infty } \right) = \frac{{{\pi ^4}}}{{90}} - \frac{1}{{16}} \cdot \frac{{{x^4}}}{{90}}. \cr} $$

We know sum of all interior angles of a polygon having $$n$$ sides $$ = \left( {n - 2} \right){180^ \circ }$$
Given smallest angle $$ = {120^ \circ }$$
$$\eqalign{ & a = 120 \cr & d = 5 \cr & {\text{Sum, }}S = \left( {\frac{n}{2}} \right)\left( {2a + \left( {n - 1} \right)d} \right) \cr & \left( {\frac{n}{2}} \right)\left( {240 + \left( {n - 1} \right)5} \right) = \left( {n - 2} \right){180^ \circ } \cr & n\left( {240 + 5n - 5} \right) = \left( {n - 2} \right){360^ \circ } \cr & 5{n^2} + 235n - 360n + 720 = 0 \cr & 5{n^2} - 125n + 720 = 0 \cr & {n^2} - 25n + 144 = 0 \cr & \left( {n - 9} \right)\left( {n - 16} \right) = 0 \cr & \therefore \,n = 9{\text{ or }}16 \cr} $$
$$n=16$$  cannot be possible since interior angle cannot be greater than $${180^ \circ }$$
Hence option A is the answer.

$$\eqalign{ & {\text{As per question,}} \cr & \,\,\,\,\,\,\,{\text{a}} + {\text{ar}} = 12\,\,\,\,\,.....\left( 1 \right) \cr & \,\,\,\,\,\,{\text{a}}{{\text{r}}^2} + {\text{a}}{{\text{r}}^3} = 48\,\,\,.....\left( 2 \right) \cr & \Rightarrow \,\,\frac{{{\text{a}}{{\text{r}}^2}\left( {1 + {\text{r}}} \right)}}{{{\text{a}}\left( {1 + {\text{r}}} \right)}} = \frac{{48}}{{12}} \cr & \Rightarrow \,\,{{\text{r}}^2} = 4 \cr & \Rightarrow \,\,{\text{r}} = - 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because {\text{ terms are }} = + ve{\text{ and }} - ve{\text{ alternately}}} \right) \cr & \Rightarrow \,\,{\text{a}} = - 12 \cr} $$

$$2x + 1, 4x + 1, 8x +1$$    are in G.P.
$$ \Rightarrow \,\,{\left( {4x + 1} \right)^2} = \left( {2x + 1} \right)\left( {8x + 1} \right)$$
⇒ $$x = 0$$  and for this value $$f\left( x \right),f\left( {2x} \right),f\left( {4x} \right)$$    are equal.

Given, $$0.5 + 0.55 + 0.555 + .....{\text{to}} \,n$$
$$\eqalign{ & = 5\left[ {0.1 + 0.11 + 0.111 + .....{\text{ to }}n{\text{ terms}}} \right] \cr & = \frac{5}{9}\left[ {0.9 + 0.99 + 0.999 + .....{\text{ to }}n{\text{ terms}}} \right] \cr & = \frac{5}{9}\left[ {\frac{9}{{10}} + \frac{{99}}{{100}} + \frac{{999}}{{1000}} + .....{\text{ to }}n{\text{ terms}}} \right] \cr & = \frac{5}{9}\left[ {\left( {1 - \frac{1}{{10}}} \right) + \left( {1 - \frac{1}{{100}}} \right) + \left( {1 - \frac{1}{{1000}}} \right) + .....{\text{ to }}n{\text{ terms}}} \right] \cr & = \frac{5}{9}\left[ {\left( {1 - \frac{1}{{10}}} \right) + \left( {1 - \frac{1}{{{{10}^2}}}} \right) + \left( {1 - \frac{1}{{{{10}^3}}}} \right) + .....\,\left( {1 - \frac{1}{{{{10}^n}}}} \right)} \right] \cr & = \frac{5}{9}\left[ {n - \left( {\frac{1}{{10}} + \frac{1}{{{{10}^2}}} + .....\,\frac{1}{{{{10}^n}}}} \right)} \right] \cr & = \frac{5}{9}\left[ {n - \frac{1}{{10}}\frac{{\left\{ {1 - {{\left( {\frac{1}{{10}}} \right)}^n}} \right\}}}{{\left( {1 - \frac{1}{{10}}} \right)}}} \right] \cr & = \frac{5}{9}\left[ {n - \frac{1}{9}\left( {1 - \frac{1}{{{{10}^n}}}} \right)} \right] \cr} $$

$$\eqalign{ & {S_k} = 1 + \frac{1}{{k + 1}} + \frac{1}{{{{\left( {k + 1} \right)}^2}}} + .....\,{\text{to }}\infty = \frac{1}{{1 - \frac{1}{{k + 1}}}} = \frac{{k + 1}}{k} \cr & \therefore \,\,k{S_k} = k + 1 \cr & \therefore \,\,\sum\limits_{k = 1}^n {k{S_k} = \sum\limits_{k = 1}^n {\left( {k + 1} \right) = 2 + 3 + ..... + \left( {n + 1} \right).} } \cr} $$

As $$n$$ is odd, the value of the given expression
$$\eqalign{ & = {1^3} - {2^3} + {3^3} - ..... + {n^3} \cr & = \left( {{1^3} + {2^3} + {3^3} + ..... + {n^3}} \right) - 2\left\{ {{2^3} + {4^3} + .....{{\left( {n - 1} \right)}^3}} \right\} \cr & = {\left\{ {\frac{{n\left( {n + 1} \right)}}{2}} \right\}^2} - 16\left\{ {{1^3} + {2^3} + ..... + {{\left( {\frac{{n - 1}}{2}} \right)}^3}} \right\} \cr & = \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{2} - 16 \cdot {\left\{ {\frac{{\frac{{n - 1}}{2} \cdot \frac{{n + 1}}{2}}}{2}} \right\}^2}. \cr} $$

$$\eqalign{ & {\text{ATQ }}2 + 5 + 8 + .....2n{\text{ terms}} = 57 + 59 + 61 + .....n{\text{ terms}} \cr & \Rightarrow \,\frac{{2n}}{2}\left[ {4 + \left( {2n - 1} \right)3} \right] \cr & = \frac{n}{2}\left[ {114 + \left( {n - 1} \right)2} \right] \cr & \Rightarrow \,6n + 1 = n + 56 \cr & \Rightarrow \,5n = 55 \cr & \Rightarrow n = 11 \cr} $$