Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \left( {1 - p} \right)\left( {1 + 3x + 9{x^2} + 27{x^3} + 81{x^4} + 243{x^5}} \right) = 1 - {p^6} \cr & \Rightarrow \left( {1 + 3x + 9{x^2} + 27{x^3} + 81{x^4} + 243{x^5}} \right) = \frac{{1 - {p^6}}}{{1 - p}} \cr & \Rightarrow \left( {1 + 3x + 9{x^2} + 27{x^3} + 81{x^4} + 243{x^5}} \right) = 1 + p + {p^2} + {p^3} + {p^4} + {p^5} \cr} $$
Comparing we get $$p = 3x\,\,{\text{or }}\frac{p}{x} = 3$$

Given sequence can be written as
$$\eqalign{ & \frac{7}{{10}} + \frac{{77}}{{100}} + \frac{{777}}{{{{10}^3}}} + ..... + {\text{ up to 20 terms}} \cr & = {\text{7}}\left[ {\frac{1}{{10}} + \frac{{11}}{{100}} + \frac{{111}}{{{{10}^3}}} + ..... + {\text{up to 20 terms}}} \right] \cr & {\text{Multiply and divide by 9}} \cr & = \frac{7}{9}\left[ {\frac{9}{{10}} + \frac{{99}}{{100}} + \frac{{999}}{{{{10}^3}}} + ..... + {\text{up to 20 terms}}} \right] \cr & = \frac{7}{9}\left[ {\left( {1 - \frac{1}{{10}}} \right) + \left( {1 - \frac{1}{{{{10}^2}}}} \right) + \left( {1 - \frac{1}{{{{10}^3}}}} \right) + ..... + {\text{up to 20 terms}}} \right] \cr & = \frac{7}{9}\left[ {20 - \frac{{\frac{1}{{10}}\left( {1 - {{\left( {\frac{1}{{10}}} \right)}^{20}}} \right)}}{{1 - \frac{1}{{10}}}}} \right] \cr & = \frac{7}{9}\left[ {\frac{{179}}{9} + \frac{1}{9}{{\left( {\frac{1}{{10}}} \right)}^{20}}} \right] \cr & = \frac{7}{{81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right] \cr} $$

$${t_n} = n\left( {n!} \right) = \left\{ {\left( {n + 1} \right) - 1} \right\}n! = \left( {n + 1} \right)!\, - n!.$$

Use $${a_r} = {a_1} + \left( {r - 1} \right)d,$$    where $$d = {a_2} - {a_1}.$$

Let the edges be $$\frac{a}{r},a,ar,$$  where $$r > 1.$$
From the question, $$\frac{a}{r} \cdot a \cdot ar = 216,\,{\text{i}}{\text{.e}}{\text{., }}{a^3} = 216,\,{\text{i}}{\text{.e}}{\text{., }}a = 6\,\,{\text{and }}2\left( {\frac{a}{r} \cdot a + a \cdot ar + \frac{a}{r} \cdot ar} \right) = 252;$$
$$\eqalign{ & \therefore \,\,\frac{1}{r} + r + 1 = \frac{7}{2} \cr & \Rightarrow \,\,r = \frac{1}{2},2. \cr & \therefore \,\,a = 6,r = 2;\,\,{\text{so the longest side}} = ar = 12. \cr} $$

$$\eqalign{ & \frac{{a + b + \frac{c}{2} + \frac{c}{2}}}{4} \geqslant \root 4 \of {a \cdot b \cdot \frac{c}{2} \cdot \frac{c}{2}} \,\,{\text{or }}\frac{{a + b + c}}{4} \geqslant \root 4 \of {\frac{{ab{c^2}}}{4}} \cr & \therefore \,\,\frac{{{{\left( {a + b + c} \right)}^4}}}{{{4^4}}} \geqslant \frac{{ab{c^2}}}{4};\,\,{\text{or }}ab{c^2} \leqslant \frac{1}{{64}}{\left( {a + b + c} \right)^4}. \cr} $$
∴ the greatest value of $$ab{c^2} = \frac{1}{{64}}{\left( {a + b + c} \right)^4}.$$
Also for the greatest value of $$ab{c^2}$$  the numbers have to be equal, i.e., $$a = b = \frac{c}{2}.$$
Also, given the greatest value $$ = \frac{1}{{64}}.$$ 
So, $$a + b + c = 1.$$

The given equation
$$\left( {{a^2} + {b^2}} \right){x^2} - 2b\left( {a + c} \right)x + \left( {{b^2} + {c^2}} \right) = 0$$
has equal roots, so, discriminant = 0
Hence, $$\left\{ {2b{{\left( {a + c} \right)}^2} - 4\left( {{a^2} + {b^2}} \right)\left( {{b^2} + {c^2}} \right) = 0} \right.$$
$$\eqalign{ & \Rightarrow 4{b^2}\left( {{a^2} + {c^2} + 2ca} \right) - 4\left( {{a^2}{b^2} + {a^2}{c^2} + {b^4} + {b^2}{c^2}} \right) = 0 \cr & \Rightarrow {b^2}{a^2} + {b^2}{c^2} + 2{b^2}ca - {a^2}{b^2} - {a^2}{c^2} - {b^4} - {b^2}{c^2} = 0 \cr & \Rightarrow 2{b^2}ca = {b^4} + {a^2}{c^2} \cr & \Rightarrow {b^4} - 2{b^2}ca + {a^2}{c^2} = 0 \cr & \Rightarrow {\left( {{b^2} - ac} \right)^2} = 0 \cr & \Rightarrow {b^2} = ac \cr} $$
⇒ $$a, b, c$$  are in G.P.

$$\eqalign{ & {\text{Given,}}{{\text{ }}^m}{C_{r - 1}},{\,^m}{C_r},{\,^m}{C_{r + 1}}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & {{\text{2}}^m}{C_r} = {\,^m}{C_{r - 1}} + {\,^m}{C_{r + 1}} \cr & \Rightarrow 2 = \frac{{^m{C_{r - 1}}}}{{^m{C_r}}} + \frac{{^m{C_{r + 1}}}}{{^m{C_r}}} = \frac{r}{{m - r + 1}} + \frac{{m - r}}{{r + 1}} \cr & \Rightarrow {m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0. \cr} $$

$$\eqalign{ & \frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c} \cr & \therefore \left( {\frac{1}{a} + \frac{1}{b} - \frac{1}{c}} \right)\left( {\frac{1}{b} + \frac{1}{c} - \frac{1}{a}} \right) \cr & = \left( {\frac{2}{a} - \frac{1}{b}} \right)\left( {\frac{2}{c} - \frac{1}{b}} \right) = \frac{4}{{ac}} - \frac{1}{b}\left( {\frac{2}{a} + \frac{2}{c}} \right) + \frac{1}{{{b^2}}} \cr & = \frac{4}{{ac}} - \frac{2}{b}\left( {\frac{2}{b}} \right) + \frac{1}{{{b^2}}} = \frac{4}{{ac}} - \frac{3}{{{b^2}}} \cr} $$

We have
$$\eqalign{ & \frac{{\left( {{a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n}} \right)}}{n} \geqslant {\left( {{a_1}{a_2}.....{a_{n - 1}}2{a_n}} \right)^{\frac{1}{n}}} \cr & \Rightarrow {a_1} + {a_2} + {a_3} + ..... + {a_{n - 1}} + 2{a_n} \geqslant n{\left( {2c} \right)^{\frac{1}{n}}} \cr} $$