When $$n$$ is even, let $$n = 2m.$$  From the question,
$${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + .....\,{\text{to }}2m{\text{ terms}} = \frac{{2m{{\left( {2m + 1} \right)}^2}}}{2}.$$
When $$n$$ is odd, let $$n = 2m + 1.$$
$$\eqalign{ & \therefore \,\,\left( {{1^2} + {{2.2}^2} + {3^2} + {{2.4}^2} + {5^2} + {{2.6}^2} + ..... + \,{\text{to }}2m{\text{ terms}}} \right) + {\left( {2m + 1} \right)^2} \cr & = \frac{{2m{{\left( {2m + 1} \right)}^2}}}{2} + {\left( {2m + 1} \right)^2} = \frac{{\left( {n - 1} \right){n^2}}}{2} + {n^2}. \cr} $$

We know that in an A.P.
$${a_1} + {a_n} = {a_2} + {a_{n - 1}} = {a_3} + {a_{n - 2}}\,\,\,\,.....\left( {\text{i}} \right)$$
[see the properties of A.P.]
$$\eqalign{ & \therefore \frac{1}{{{a_1}{a_n}}} + \frac{1}{{{a_2}{a_{n - 1}}}} + \frac{1}{{{a_3}{a_{n - 2}}}} + ..... + \frac{1}{{{a_n}{a_1}}} \cr & = \frac{1}{{{a_1} + {a_n}}}\left[ {\frac{{{a_1} + {a_n}}}{{{a_1}{a_n}}} + \frac{{{a_1} + {a_n}}}{{{a_2}{a_{n - 2}}}} + \frac{{{a_1} + {a_n}}}{{{a_3}{a_{n - 2}}}} + ..... + \frac{{{a_1} + {a_n}}}{{{a_n}{a_1}}}} \right] \cr & = \frac{2}{{{a_1} + {a_n}}}\left[ {\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + ..... + \frac{1}{{{a_n}}}} \right] \cr} $$

The G.P. is $$a,ar,a{r^2},.....,a{r^{2n}}$$
So, $$P = a \cdot ar \cdot a{r^2} \cdot ..... \cdot a{r^{2n}}$$
$$\eqalign{ & = {a^{2n + 1}} \cdot {r^{1 + 2 + ..... + 2n}} \cr & = {a^{\left( {2n + 1} \right)}}{r^{\frac{{2n\left( {2n + 1} \right)}}{2}}} = {a^{2n + 1}}{r^{n\left( {2n + 1} \right)}} = {\left( {a{r^n}} \right)^{\left( {2n + 1} \right)}} \cr} $$
$$ = {\left( {2n + 1} \right)^{th}}$$   power of the $${\left( {n + 1} \right)^{th}}$$   term of G.P.

Let the required three numbers of G.P. be $$\frac{a}{r},a{\text{ and }}ar.$$
Then, their sum $$ = \frac{a}{r} + a + ar = 38$$
$$\eqalign{ & \Rightarrow a\left( {\frac{{1 + r + {r^2}}}{r}} \right) = 38\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{product}} = \frac{a}{r} \times a \times ar = 1728 \cr & \Rightarrow {a^3} = {\left( {12} \right)^3} \cr & \therefore a = 12\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Substitute the value of $$a,$$ in equation (i), we get
$$\eqalign{ & \therefore 12 \times \left( {\frac{{1 + r + {r^2}}}{r}} \right) = 38 \cr & \Rightarrow 6 + 6r + 6{r^2} = 19r \cr & \Rightarrow 6{r^2} - 13r + 6 = 0 \cr & \Rightarrow \left( {3r - 2} \right)\left( {2r - 3} \right) = 0 \cr & \therefore r = \frac{2}{3}{\text{ or }}\frac{3}{2} \cr} $$
Hence, the required numbers are 18, 12, 8 or 8, 12, 18
∴ Greatest number = 18

$$y = 2\sqrt x ,$$   being in the first quadrant
The sequence of $$x$$-coordinates: 1, 2, 4, 8, . . . . .
∴ the sequence of $$y$$-coordinates $$2,2\sqrt 2 ,2\sqrt 4 ,2\sqrt 8 ,.....,$$     where the common ratio is $$\sqrt 2 .$$
$$\therefore \,\,{y_n} = 2 \cdot {\left( {\sqrt 2 } \right)^{n - 1}}.$$

$$\eqalign{ & 2 = {2^n}\sin \alpha + \frac{1}{{{2^n}\sin \alpha }} \cr & 2 \cdot {2^n}\sin \alpha = {\left( {{2^n}\sin \alpha } \right)^2} + 1 \cr & {\left( {{2^n}\sin \alpha - 1} \right)^2} = 0 \cr & \sin \alpha = \frac{1}{{{2^n}}} \cr & {\text{for, }}n = 1,\sin {\alpha _1} = \frac{1}{2} \cr & {\text{for, }}n = 2,\sin{\alpha _2} = \frac{1}{4} \cr & {\text{for, }}n = 3,\sin {\alpha _3} = \frac{1}{8} \cr & S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .....\,{\text{upto }}\infty \cr & = \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = \frac{{\frac{1}{2}}}{{\frac{1}{2}}} = 1 \cr} $$

Let the $${150^{th}}$$  term = $$n.$$ Then
$$\eqalign{ & 1 + 2 + 3 + ..... + \left( {n - 1} \right) < 150 < 1 + 2 + 3 + ..... + n \cr & \therefore \,\,n\left( {n - 1} \right) < 300 < n\left( {n + 1} \right) \cr & \Rightarrow \,\,n = 17. \cr} $$

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

For G.P., $${t_n} = {2^{n - 1}};$$   for A.P., $${T_m} = 1 + \left( {m - 1} \right)3.$$    They are common if $${2^{n - 1}} = 3m - 2\,\,{\text{or, }}{{\text{2}}^{n - 2}} + 1 = \frac{{3m}}{2} \leqslant 150$$
$$ \Rightarrow \,\,n \leqslant 9,m \leqslant 100.$$
By trial, $$n = 1, m = 1; n = 3, m = 2; n = 5, m = 6; n = 7, m = 22; n = 9, m = 86.$$

Given $${a^2},{b^2},{c^2}$$  are in A.P.
$$\eqalign{ & \Rightarrow {a^2} + \left( {ab + bc + ca} \right),{b^2} + \left( {ab + bc + ca} \right),{c^2} + \left( {ab + bc + ca} \right)\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \left( {a + b} \right)\left( {a + c} \right),\left( {b + c} \right)\left( {b + a} \right),\left( {c + a} \right)\left( {c + b} \right)\,{\text{are in A}}{\text{.P}}. \cr & \Rightarrow \frac{1}{{b + c}},\frac{1}{{c + a}},\frac{1}{{a + b}}\,{\text{are in A}}{\text{.P}}. \cr & \left[ {{\text{Divide by}}\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)} \right] \cr & {\text{Again, }}{a^2},{b^2},{c^2}\,{\text{are in A}}{\text{.P}}. \cr & \Rightarrow \frac{1}{{b + c}},\frac{1}{{c + a}},\frac{1}{{a + b}}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \frac{{a + b + c}}{{b + c}},\frac{{a + b + c}}{{c + a}},\frac{{a + b + c}}{{a + b}}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \frac{a}{{b + c}} + 1,\frac{b}{{c + a}} + 1,\frac{c}{{a + b}} + 1\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}}\,{\text{are in A}}{\text{.P}}. \cr} $$