$$\eqalign{ & x = \sum\limits_{n = 0}^\infty {{a^n}} = \frac{1}{{1 - a}};\,\,\,\,a = 1 - \frac{1}{x} \cr & y = \sum\limits_{n = 0}^\infty {{b^n}} = \frac{1}{{1 - b}};\,\,\,b = 1 - \frac{1}{y} \cr & z = \sum\limits_{n = 0}^\infty {{c^n}} = \frac{1}{{1 - c}};\,\,\,\,c = 1 - \frac{1}{z} \cr & a,b,c{\text{ are in A}}{\text{.P}}{\text{. OR }}2b = a + c \cr & 2\left( {1 - \frac{1}{y}} \right) = 1 - \frac{1}{x} + 1 - \frac{1}{z} \cr & \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \cr & \Rightarrow x,y,z{\text{ are in H}}{\text{.P}}{\text{.}} \cr} $$

$$\eqalign{ & a{r^2} = 4\,\,.....\left( 1 \right) \cr & a.ar.a{r^2}.a{r^3}.a{r^4} = {a^5}{r^{10}} \cr & = {\left( {a{r^2}} \right)^5} = {4^5} \cr} $$

Since $$n^{th}$$ term of A.P. $$ = a + \left( {n - 1} \right)d$$
$$\therefore p = 1 + \left( {n - 1} \right)2$$
( ∵ First term $$= a = 1$$  and common difference $$= d = 2$$  )
$$\eqalign{ & \Rightarrow n = \frac{{p + 1}}{2} \cr & \therefore \left( {1 + 3 + 5 + ..... + p} \right) + \left( {1 + 3 + 5 + ..... + q} \right) = \left( {1 + 3 + 5 + ..... + r} \right) \cr & \Rightarrow \frac{{\frac{{p + 1}}{2}}}{2}\left[ {2 \times 1 + \left( {\frac{{p + 1}}{2} - 1} \right)2} \right] + \frac{{\left( {\frac{{q + 1}}{2}} \right)}}{2}\left[ {2 \times 1 + \left( {\frac{{q + 1}}{2} - 1} \right)2} \right] \cr & = \frac{{r + 1}}{4}\left[ {2 \times 1 + \left( {\frac{{r + 1}}{2} - 1} \right)2} \right] \cr & \Rightarrow \frac{{p + 1}}{4}\left[ {2 + \left( {p - 1} \right)} \right] + \frac{{q + 1}}{4}\left[ {2 + \left( {q - 1} \right)} \right] \cr & = \frac{{r + 1}}{4}\left[ {2 + r - 1} \right] \cr & \Rightarrow {\left( {p + 1} \right)^2} + {\left( {q + 1} \right)^2} = {\left( {r + 1} \right)^2} \cr} $$
This is the possible only when $$p = 7, q = 5, r = 9$$
$$ \therefore p + q + r = 7 + 5 + 9 = 21$$

$$9900 = {3^2} \times {2^2} \times {5^2} \times 11$$
∴ sum of all divisors
$$ = \left( {{3^0} + {3^1} + {3^2}} \right)\left( {{2^0} + {2^1} + {2^2}} \right)\left( {{5^0} + {5^1} + {5^2}} \right)\left( {{{11}^0} + {{11}^1}} \right).$$
∴ sum of all proper divisors
$$ = 13 \times 7 \times 31 \times 12 - 1 - 9900$$
$$= 23951.$$

$$\frac{1}{{24}} - \frac{1}{{30}} = \frac{1}{{20}} - \frac{1}{{24}}$$     and so in H.P.

We have, $${a_2} - {a_1} = {a_3} - {a_2} = .....\,{a_{2k}} - {a_{2k - 1}} = d$$
Hence,
$$\eqalign{ & a_1^2 - a_2^2 = \left( {{a_1} - {a_2}} \right)\left( {{a_1} + {a_2}} \right) = - d\left( {{a_1} + {a_2}} \right) \cr & a_3^2 - a_4^2 = \left( {{a_3} - {a_4}} \right)\left( {{a_3} + {a_4}} \right) = - d\left( {{a_3} + {a_4}} \right) \cr & ......................... \cr & ......................... \cr & a_{2k - 1}^2 - a_{2k}^2 = \left( {{a_{2k - 1}} - {a_{2k}}} \right)\left( {{a_{2k - 1}} + {a_{2k}}} \right) = - d\left( {{a_{2k - 1}} + {a_{2k}}} \right) \cr & {\text{Adding, we get}} \cr & a_1^2 - a_2^2 + a_3^2 - a_4^2 + ..... + a_{2k - 1}^2 - a_{2k}^2 \cr & = - d\left( {{a_1} + {a_2} + {a_3} + {a_4} + .....\,{a_{2k - 1}} + {a_{2k}}} \right) \cr & = - d \cdot \frac{{2k}}{2}\left( {{a_1} + {a_{2k}}} \right) = - dk\left( {{a_1} + {a_{2k}}} \right) \cr & {\text{But, }}{a_{2k}} = {a_1} + \left( {2k - 1} \right)d \cr & \Rightarrow - d = \frac{{{a_1} - {a_{2k}}}}{{2k - 1}} \cr & \therefore {\text{The required sum}} = \frac{k}{{2k - 1}}\left( {a_1^2 - a_{2k}^2} \right) \cr & \Rightarrow M = \frac{k}{{2k - 1}} \cr} $$

$$\frac{{{x^{\log y - \log z}} + {y^{\log z - \log x}} + {z^{\log x - \log y}}}}{3} \geqslant \root 3 \of {{x^{\log y - \log z}} \cdot {y^{\log z - \log x}} \cdot {z^{\log x - \log y}}} = 1$$
because $$\log \left\{ {{x^{\log y - \log z}} \cdot {y^{\log z - \log x}} \cdot {z^{\log x - \log y}}} \right\} = 0\left( {{\text{on simplification}}} \right).$$

$$\eqalign{ & \frac{x}{{1 - r}} = 5 \cr & \Rightarrow \,r = 1 - \frac{x}{5} \cr & {\text{Since G}}{\text{.P}}{\text{. contains infinite terms}} \cr & \therefore \,\,\,\, - 1 < r < 1 \cr & \Rightarrow \,\, - 1 < 1 - \frac{x}{5} < 1 \cr & \Rightarrow \,\, - 2 < - \frac{x}{5} < 0 \cr & \Rightarrow \,\, - 10 < x < 0. \cr & \Rightarrow \,\,0 < \frac{x}{5} < 2 \cr & \Rightarrow \,\,0 < x < 10. \cr} $$

$$\eqalign{ & \frac{{\frac{p}{2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]}}{{\frac{q}{2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = \frac{{{p^2}}}{{{q^2}}} \cr & \Rightarrow \,\,\frac{{2{a_1} + \left( {p - 1} \right)d}}{{2{a_1} + \left( {q - 1} \right)d}} = \frac{p}{q} \cr & \frac{{{a_1} + \left( {\frac{{p - 1}}{2}} \right)d}}{{{a_1} + \left( {\frac{{q - 1}}{2}} \right)d}} = \frac{p}{q} \cr & {\text{For }}\frac{{{a_6}}}{{{a_{21}}}},\,\,p = 11,\,\,q = 41 \cr & \Rightarrow \,\,\frac{{{a_6}}}{{{a_{21}}}} = \frac{{11}}{{41}} \cr} $$

$$\eqalign{ & {T_m} = a + \left( {m - 1} \right)d = \frac{1}{n}\,\,\,......\left( 1 \right) \cr & {T_n} = a + \left( {n - 1} \right)d = \frac{1}{m}\,\,\,......\left( 2 \right) \cr & \left( 1 \right) - \left( 2 \right)\,\, \Rightarrow \left( {m - n} \right)d = \frac{1}{n} - \frac{1}{m} \cr & \Rightarrow \,d = \frac{1}{{mn}} \cr & {\text{From }}\left( 1 \right)\,\,a = \frac{1}{{mn}} \cr & \Rightarrow \,a - d = 0 \cr} $$