Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\log _{{e}}}{{{b}}_1},{\log _{{e}}}{{{b}}_2}, - - - ,{\log _{{e}}}{{{b}}_{{{101}}}}\,{\text{are}}\,{\text{in}}\,{\text{A}}.{\text{P}}. \cr & \Rightarrow \,\,{b_1},{{{b}}_2}, - - - {{{b}}_{{{101}}}}\,{\text{are}}\,{\text{in}}\,{\text{G}}{\text{.P}}{\text{.}} \cr & {\text{Also}}\,{{{a}}_1},\,{{{a}}_2}, - - - {{{a}}_{{{101}}}}\,{\text{are}}\,{\text{in}}\,{\text{A}}{\text{.P}}{\text{.}} \cr & {\text{where}}\,{{{a}}_1} = {{{b}}_{{1}}}\,{\text{are}}\,{{{a}}_{{{51}}}}{{ = }}{{{b}}_{{{51}}}}{{.}} \cr & \therefore \,\,\,{{{b}}_2},{{{b}}_3}, - - - ,{{{b}}_{50}}{\text{ and }}{\text{GM's}}\,\,{{{a}}_2},{{{a}}_3}, - - - ,{{{a}}_{50}}{\text{ are in AM's}} \cr & {\text{between }}{{{b}}_1}{\text{ and }}{{{b}}_{51}}. \cr & \because {\text{ GM}} < {\text{AM}} \cr & \Rightarrow \,\,{{{b}}_2} < {{{a}}_2},{{{b}}_3} < {{{a}}_3}, - - - ,{{{b}}_{50}} < {{{a}}_{50}} \cr & \because \,\,\,\,{{{b}}_1} + {{{b}}_2} + - - - + {{{b}}_{51}} < {{{a}}_1} + {{{a}}_2} + - - - + {{{a}}_{51}} \cr & \Rightarrow \,\,\,{{t}}\, < {{s}} \cr & {\text{Also }}{{{a}}_1},{{{a}}_{51}},{{{a}}_{101}}{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \,\,\,\,\,\,\,\,\,\,\,{{{b}}_1},{{{b}}_{51}},{{{b}}_{101}}{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \because \,\,\,\,\,\,\,{{{a}}_1} = {{{b}}_1}{\text{ and }}{{{a}}_{51}} = {{{b}}_{51}} \cr & \therefore \,\,\,\,\,\,\,{{{b}}_{101}} > {{{a}}_{101}} \cr} $$

If in A.P., $$\sqrt y = \sqrt x + \left( {n - 1} \right)d$$     and $$\sqrt z = \sqrt x + \left( {m - 1} \right)d$$
$$\therefore \,\,\frac{{\sqrt y - \sqrt x }}{{\sqrt z - \sqrt x }} = \frac{{n - 1}}{{m - 1}},$$     a rational number. As $$x, y, z$$  are prime, $$\frac{{\sqrt y - \sqrt x }}{{\sqrt z - \sqrt x }}$$   is irrational.
∴ irrational = rational (absurd). So, $$\sqrt x ,\sqrt y ,\sqrt z $$   are not in A.P.
Similarly, they are not in G.P. or H.P.

Let the $$3n$$ terms of G.P. be
$$\eqalign{ & a,ar,a{r^2},.....\,a{r^{n - 1}},a{r^n},a{r^{n + 1}},.....\,a{r^{2n - 1}},a{r^{2n}},{a^{2n + 1}},.....,a{r^{3n - 1}}.\,\,{\text{Then}} \cr & {S_1} = a + ar + a{r^2} + ..... + a{r^{n - 1}} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} \cr & {S_2} = a{r^n} + a{r^{n + 1}} + ..... + a{r^{2n - 1}} = \frac{{a{r^n}\left( {1 - {r^n}} \right)}}{{1 - r}} \cr & {S_3} = a{r^{2n}} + a{r^{2n + 1}} + ..... + a{r^{3n - 1}} = \frac{{a{r^{2n}}\left( {1 - {r^n}} \right)}}{{1 - r}} \cr & {\text{Clearly, }}\frac{{{S_2}}}{{{S_1}}} = \frac{{{S_3}}}{{{S_2}}} = {r^n} \cr} $$

$$\eqalign{ & {\text{Let the G}}{\text{.P}}{\text{. be }}a,ar{\text{ and }}a{r^2}{\text{ then }}a = A + d;ar = A + 4d; \cr & a{r^2} = A + 8d \cr & \Rightarrow \,\,\frac{{a{r^2} - ar}}{{ar - a}} = \frac{{\left( {A + 8d} \right) - \left( {A + 4d} \right)}}{{\left( {A + 4d} \right) - \left( {A + d} \right)}} \cr & r = \frac{4}{3} \cr} $$

Let $$a$$ = first term of G.P. and $$r$$ = common ratio of G.P.;
Then G.P. is $$a,ar,a{r^2}$$
$$\eqalign{ & {\text{Given }}{S_\infty } = 20 \Rightarrow \frac{a}{{1 - r}} = 20 \cr & \Rightarrow a = 20\left( {1 - r} \right)\,\,\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{Also }}{a^2} + {a^2}{r^2} + {a^2}{r^4} + ..... + {\text{to}}\,\infty = 100 \cr & \Rightarrow \,\,\frac{{{a^2}}}{{1 - {r^2}}} = 100 \cr & \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{From }}\left( {\text{i}} \right),{a^2} = 400{\left( {1 - r} \right)^2}; \cr & {\text{From }}\left( {{\text{ii}}} \right),{\text{ we get 100}}\left( {1 - r} \right)\left( {1 + r} \right) = 400{\left( {1 - r} \right)^2} \cr & \Rightarrow \,\,1 + r = 4 - 4r \cr & \Rightarrow \,\,5r = 3 \cr & \Rightarrow \,\,r = \frac{3}{5}. \cr} $$

$$\eqalign{ & AL_1^2 + {L_1}M_1^2 = \left( {{a^2} + {1^2}} \right) + \left\{ {{{\left( {a - 1} \right)}^2} + {1^2}} \right\} \cr & AL_2^2 + {L_2}M_2^2 = \left( {{a^2} + {2^2}} \right) + \left\{ {{{\left( {a - 2} \right)}^2} + {2^2}} \right\} \cr & ................................................ \cr & AL_{a - 1}^2 + {L_{a - 1}}M_{a - 1}^2 = {a^2} + {\left( {a - 1} \right)^2} + \left\{ {{1^2} + {{\left( {a - 1} \right)}^2}} \right\} \cr} $$
∴ The required sum

$$\eqalign{ & = \left( {a - 1} \right){a^2} + \left\{ {{1^2} + {2^2} + ..... + {{\left( {a - 1} \right)}^2}} \right\} + 2\left\{ {{1^2} + {2^2} + ..... + {{\left( {a - 1} \right)}^2}} \right\} \cr & = \left( {a - 1} \right){a^2} + 3 \cdot \frac{{\left( {a - 1} \right)a\left( {2a - 1} \right)}}{6} \cr & = a\left( {a - 1} \right)\left\{ {a + \frac{{2a - 1}}{2}} \right\} = \frac{{a\left( {a - 1} \right)\left( {4a - 1} \right)}}{2} \cr} $$

$${t_n} = 2 + 1 + 3 + 5 + 7 + .....\,{\text{to }}n{\text{ terms}} = 2 + {\left( {n - 1} \right)^2}.$$

$$\eqalign{ & {\text{Let }}a,ar,a{r^2}{\text{ are in G}}{\text{.P}}{\text{.}} \cr & {\text{According to the question}} \cr & a,2ar,a{r^2}{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,{\text{2}} \times {\text{2}}ar = a + a{r^2} \cr & \Rightarrow \,4r = 1 + {r^2} \cr & \Rightarrow \,{r^2} - 4r + 1 = 0 \cr & r = \frac{{4 \pm \sqrt {16 - 4} }}{2} = 2 \pm \sqrt 3 \cr & {\text{Since }}r > 1 \cr & \therefore \,\,\,r = 2 - \sqrt 3 {\text{ is rejected}} \cr & {\text{Hence, }}r = 2 + \sqrt 3 \cr} $$

Let the progression be $$a, a + d, a + 2d$$
Then $${x_4} = 3{x_1}$$
$$\eqalign{ & \Rightarrow a + 3d = 3a \cr & \Rightarrow 3d = 2a\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{Again, }}{x_7} = 2{x_3} + 1 \cr & \Rightarrow a + 6d = 2\left( {a + 2d} \right) + 1 \cr & \Rightarrow 2d = a + 1\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Solving (i) and (ii) we get
$$a = 3, d = 2$$

$$\eqalign{ & \alpha \,,\beta \,{\text{ are the roots of }}{x^2} - x + p = 0 \cr & \therefore \,\,\,\,\alpha + \beta = 1\,\,\,\,\,\,.....\left( 1 \right) \cr & \,\,\,\,\,\,\,\alpha \beta = p\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & \gamma \,,\delta \,\,{\text{are the roots of }}{x^2} - 4x + q = 0 \cr & \therefore \,\,\,\,\,\gamma + \delta = 4\,\,\,\,\,.....\left( 3 \right) \cr & \,\,\,\,\,\,\,\,\,\gamma \delta = q\,\,\,\,\,\,\,\,.....\left( 4 \right) \cr & \alpha ,\,\,\beta ,\,\,\gamma ,\,\,\delta \,\,{\text{are in G}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\,{\text{Let }}\alpha = a;\beta = ar,\gamma = a{r^2},\delta = a{r^3}. \cr & {\text{Substituting these values in equations }}\left( {\text{1}} \right){\text{,}}\left( {\text{2}} \right){\text{, }}\left( 3 \right){\text{ and}}\,\left( 4 \right),\,{\text{we get }} \cr & \,\,\,\,a + ar = 1\,\,\,\,\,\,\,\,.....\left( 5 \right) \cr & \,\,\,\,{a^2}r = p\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 6 \right) \cr & \,\,\,\,a{r^2} + a{r^3} = 4\,\,\,.....\left( 7 \right) \cr & \,\,\,\,{a^2}{r^5} = q\,\,\,\,\,\,\,\,\,\,\,.....\left( 8 \right) \cr & {\text{Dividing (7) by (5) we get}} \cr & \frac{{a{r^2}\left( {1 + r} \right)}}{{a\left( {1 + r} \right)}} = \frac{4}{1} \cr & \Rightarrow {r^2} = 4 \cr & \Rightarrow r = 2, - 2 \cr & \left( 5 \right)\,\,\,\,\,\,\,\,\, \Rightarrow a = \frac{1}{{1 + r}} = \frac{1}{{1 + 2}}{\text{ or }}\frac{1}{{1 - 2}} = \frac{1}{3}{\text{ or }} - 1 \cr & {\text{As }}p{\text{ is an integer (given), }}r{\text{ is also an integer}}\left( {{\text{2 or}} - 2} \right) \cr & \therefore \,\left( 6 \right)\,\,\,\,\, \Rightarrow a \ne \frac{1}{3}\,{\text{ Hence }}a = - 1{\text{ and }}r = - 2. \cr & \therefore \,\,\,\,\,p = {\left( { - 1} \right)^2} \times \left( { - 2} \right) = - 2 \cr & q = {\left( { - 1} \right)^2} \times {\left( { - 2} \right)^5} = - 32 \cr} $$