Here, $$a_i^2 = {\left( {{a_{i - 1}} + 1} \right)^2}$$
$$ \Rightarrow \,\,a_i^2 - a_{i - 1}^2 = 2{a_{i - 1}} + 1.$$
Putting $$i = 2, 3, . . . . . , n + 1$$    and adding,
$$\eqalign{ & a_{n + 1}^2 - a_1^2 = 2\left( {{a_1} + {a_2} + ..... + {a_n}} \right) + n \cr & \therefore \,\,a_{n + 1}^2 - 0 = 2nA + n; \cr & \therefore \,\,A = \left( {a_{n + 1}^2 - n} \right)\frac{1}{{2n}} = \frac{{a_{n + 1}^2}}{{2n}} - \frac{1}{2} \geqslant - \frac{1}{2}. \cr} $$

$$\eqalign{ & {a_1} = {h_1} = 2,{a_{10}} = {h_{10}} = 3 \cr & 3 = {a_{10}} = 2 + 9d \cr & \Rightarrow \,d = \frac{1}{9} \cr & \therefore \,\,{a_4} = 2 + 3d = \frac{7}{3} \cr & \,\,\,\,\,\,\,3 = {h_{10}} \Rightarrow \frac{1}{3} = \frac{1}{{{h_{10}}}} = \frac{1}{2} + 9D \cr & \therefore \,\,D = - \frac{1}{{54}} \cr & \,\,\,\,\,\frac{1}{{{h_7}}} = \frac{1}{2} + 6D = \frac{1}{2} - \frac{1}{9} = \frac{7}{{18}} \cr & \therefore \,\,{a_4}{h_7} = \frac{7}{3} \times \frac{{18}}{7} = 6. \cr} $$

$$\eqalign{ & S = \sum\limits_{i = 1}^{30} {{a_i} = \frac{{30}}{2}\left[ {2{a_1} + 29d} \right]} {\text{ }} \cr & T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}} = \frac{{15}}{2}\left[ {2{a_1} + 28d} \right]} \cr & {\text{Since, }}S - 2T = 75 \cr & \Rightarrow \,\,30\,{a_1} + 435d - 30\,{a_1} - 420d = 75 \cr & \Rightarrow \,\,d = 5 \cr & {\text{Also, }}{a_5} = 27\,\,\,\, \Rightarrow \,\,{a_1} + 4d = 27 \cr & \Rightarrow \,\,{a_1} = 7, \cr & {\text{Hence, }}{a_{10}} = {a_1} + 9d = 7 + 9 \times 5 = 52 \cr} $$

$$\eqalign{ & {S_n} = 1 + 2a + 3{a^2} + ..... + n{a^{n - 1}} \cr & {\text{or, }}{S_n} = a + 2{a^2} + ..... + \left( {n - 1} \right){a^{n - 1}} + n{a^n} \cr & {\text{where, }}a = 1 + \frac{1}{n} \cr & \therefore \left( {1 - a} \right){S_n} = 1 + a + {a^2} + ..... + {a^{n - 1}} - n{a^n} \cr & \left( {1 - a} \right){S_n} = \frac{{{a^n} - 1}}{{a - 1}} - n{a^n} \cr & \Rightarrow - \frac{1}{n}{S_n} = \frac{{{{\left( {1 + \frac{1}{n}} \right)}^n} - 1}}{{\frac{1}{n}}} - n{\left( {1 + \frac{1}{n}} \right)^n} = - n \cr & \Rightarrow {S_n} = {n^2} \cr} $$

$$\eqalign{ & {\text{In the quadratic equation }}a{x^2} + bx + c = 0 \cr & \Delta = {b^2} - 4ac{\text{ and }}\alpha + \beta = - \frac{b}{a},\alpha \beta = \frac{c}{a} \cr & {\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \cr & = \frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a} = \frac{{{b^2} - 2ac}}{{{a^2}}} \cr & {\text{and }}\,\,{\alpha ^3} + {\beta ^3} = - \frac{{{b^3}}}{{{a^3}}} - \frac{{3c}}{a}\left( { - \frac{b}{a}} \right) \cr & = - \left( {\frac{{{b^3} - 3abc}}{{{a^3}}}} \right) \cr & {\text{Given }}\alpha + \beta ,{\alpha ^2} + {\beta ^2},{\alpha ^3} + {\beta ^3}{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\, - \frac{b}{a}, - \frac{{{b^2} - 2ac}}{{{a^2}}}, - \frac{{\left( {{b^3} - 3\,abc} \right)}}{{{a^3}}}{\text{are in G}}{\text{.P}}{\text{.}} \cr & \Rightarrow {\left( {\frac{{{b^2} - 2\,ac}}{{{a^2}}}} \right)^2} = \frac{b}{a}\left( {\frac{{{b^3} - 3\,abc}}{{{a^3}}}} \right) \cr & \Rightarrow \,{b^4} + 4{a^2}{c^2} - 4a{b^2}c = {b^4} - 3\,a{b^2}c \cr & \Rightarrow \,4{a^2}{c^2} - a{b^2}c = 0 \cr & \Rightarrow \,ac\,\Delta = 0 \cr & \Rightarrow \,\,c\,\Delta = 0\,\,\left( {\because \,{\text{In quadractic }}a \ne 0} \right) \cr} $$

$$\eqalign{ & S = 1 + 4x + 7{x^2} + 10{x^3} + ..... \cr & x.S = x + 4{x^2} + 7{x^3} + ..... \cr} $$
Subtract
$$\eqalign{ & S\left( {1 - x} \right) = 1 + 3x + 3{x^2} + 3{x^3} + ..... \cr & S\left( {1 - x} \right) = 1 + 3x\left( {\frac{1}{{1 - x}}} \right) \cr & \because \left| x \right| < 1 \cr & S = \frac{{1 + 2x}}{{{{\left( {1 - x} \right)}^2}}} \cr & {\text{Given : }}\frac{{1 + 2x}}{{{{\left( {1 - x} \right)}^2}}} = \frac{{35}}{{16}} \cr & \Rightarrow 16 + 32x = 35 + 35{x^2} - 70x \cr & \Rightarrow x = \frac{1}{5},\frac{{19}}{7} \cr & {\text{But, }}\left| x \right| < 1 \cr & \therefore x = \frac{1}{5} \cr} $$

$$\eqalign{ & {S_n} = \frac{1}{9}\left( 9 \right) + \frac{2}{9}\left( {99} \right) + \frac{3}{9}\left( {999} \right) + .... \cr & = \frac{1}{9}\left[ {10 + {{2.10}^2} + {{3.10}^3} + ....} \right] - \frac{1}{9}\left[ {1 + 2 + 3 + ....} \right] \cr & = \frac{1}{9}S - \frac{1}{9}\frac{{n\left( {n + 1} \right)}}{2} \cr & S = 10 + {2.10^2} + {3.10^3} + .... + n{.10^n} \cr & \frac{{10S = {{10}^2} + {{2.10}^3} + .... + \left( {n - 1} \right){{10}^n} + n{{.10}^{n + 1}}}}{{ - 9S = \left( {10 + {{10}^2} + .... + {{10}^n}} \right) - n{{.10}^{n + 1}}}} \cr & S = \frac{n}{9}{10^{n + 1}} - \frac{{{{10}^{n + 1}} - 10}}{{81}} \cr & \therefore {S_n} = \frac{n}{{81}}{10^{n + 1}} - \frac{{{{10}^{n + 1}} - 10}}{{9.81}} - \frac{1}{9}\frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore 9{S_n} = \frac{{\left( {9n - 1} \right){{10}^{n + 1}}}}{{81}} + \frac{{10}}{{81}} - \frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore 9\left( {{S_n} - {S_{n - 1}}} \right) = \frac{{{{10}^n}}}{{81}}\left\{ {10\left( {9n - 1} \right) - \left( {9n - 10} \right)} \right\} - n = n\left( {{{10}^n} - 1} \right) \cr} $$

$$\eqalign{ & {\text{We know that }}{e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ......\infty \cr & {\text{Put }}x = - 1 \cr & \therefore \,\,{e^{ - 1}} = 1 - 1 + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}}......\infty \cr & \therefore \,\,{e^{ - 1}} = \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \frac{1}{{5!}}......\infty \cr} $$

The general term is
$$\eqalign{ & {T_n} = \frac{{\frac{n}{2} \cdot \frac{{n + 1}}{2}}}{{{1^3} + {2^3} + {3^3} + ..... + {n^3}}} = \frac{1}{{n\left( {n + 1} \right)}} \cr & = \frac{1}{n} - \frac{1}{{n + 1}} \cr & \therefore {S_n} = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}} \cr} $$

Since, $$0 < x < \pi , - 1 < \cos x < 1$$
$$ \Rightarrow 0 \leqslant \left| {\cos x} \right| < 1.$$
We can write the given expression as $${11^{\frac{1}{{\left( {1 - \left| {\cos x} \right|} \right)}}}} = 121$$
$$\eqalign{ & \Rightarrow \frac{1}{{1 - \left| {\cos x} \right|}} = 2 \cr & \Rightarrow 1 - \left| {\cos x} \right| = \frac{1}{2} \cr & \Rightarrow \left| {\cos x} \right| = \frac{1}{2} \cr & \Rightarrow \cos x = \pm \frac{1}{2} \cr & \Rightarrow x = \pm \frac{\pi }{3}, \pm \frac{{2\pi }}{3} \cr} $$