Sequences and Series MCQ Questions & Answers in Algebra | Maths

Sum = 4 and second term = $$\frac{3}{4}$$ , it is given that
first term is $$a$$ and common ratio $$r$$
$$\eqalign{ & \Rightarrow \,\,\frac{a}{{1 - r}} = 4{\text{ and }}ar = \frac{3}{4} \cr & \Rightarrow r = \frac{3}{{4a}} \cr & {\text{Therefore, }}\frac{a}{{1 - \frac{3}{{4a}}}} = 4 \cr & \Rightarrow \,\frac{{4{a^2}}}{{4a - 3}} = 4 \cr & {\text{or }}{a^2} - 4a + 3 = 0 \cr & \Rightarrow \,\,\left( {a - 1} \right)\left( {a - 3} \right) = 0 \cr & \Rightarrow \,\,a = 1{\text{ or 3}} \cr & {\text{When }}a = 1,r = \frac{3}{4}{\text{ and when }}a = 3,r = \frac{1}{4} \cr} $$

Sum $$ = \frac{2}{{10}} + \frac{4}{{{{10}^3}}} + \frac{6}{{{{10}^5}}} + \frac{8}{{{{10}^7}}} + .....\,{\text{to }}\infty $$        which is an arithmetico-geometric series

$$\eqalign{ & \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ..... \cr & = \left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{8}} \right) + \left( {1 - \frac{1}{{16}}} \right) + ..... \cr & = n - \frac{{\frac{1}{2}\left\{ {1 - \frac{1}{{{2^n}}}} \right\}}}{{1 - \frac{1}{2}}} = n - 1 + {2^{ - n}} \cr} $$

$$\eqalign{ & {S_n} = \frac{1}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{1}{{^n{C_2}}} + ..... + \frac{1}{{^n{C_n}}} \cr & {t_n} = \frac{0}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{2}{{^n{C_2}}} + ..... + \frac{n}{{^n{C_n}}} \cr & {t_n} = \frac{n}{{^n{C_n}}} + \frac{{n - 1}}{{^n{C_{n - 1}}}} + \frac{{n - 2}}{{^{^n}{C_{n - 2}}}} + ..... + \frac{0}{{^n{C_0}}} \cr & {\text{Add,}}\,\,{{2}}{{{t}}_n} = \left( n \right)\left[ {\frac{1}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + .....\frac{1}{{^n{C_n}}}} \right] = n{S_n} \cr & \therefore \,\,\frac{{{t_n}}}{{{S_n}}} = \frac{n}{2} \cr} $$

\[\begin{array}{l} l = A{R^{p - 1}} \Rightarrow \log l = \log A + \left( {p - 1} \right)\log R\\ m = A{R^{q - 1}} \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R\\ n = A{R^{r - 1}} \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R\\ {\rm{Now,}}\\ \left| \begin{array}{l} \log l\,\,\,\,\,\,p\,\,\,\,\,\,1\\ \log m\,\,\,\,q\,\,\,\,\,\,1\\ \log n\,\,\,\,\,\,r\,\,\,\,\,\,1 \end{array} \right| = \left| \begin{array}{l} \log A + \left( {p - 1} \right)\log R\,\,\,\,\,\,\,p\,\,\,\,\,\,\,\,1\\ \log A + \left( {q - 1} \right)\log R\,\,\,\,\,\,\,\,q\,\,\,\,\,\,\,\,1\\ \log A + \left( {r - 1} \right)\log R\,\,\,\,\,\,\,\,\,r\,\,\,\,\,\,\,\,1 \end{array} \right|\\ {\rm{Operating }}\,\,{{\rm{C}}_1} - \left( {\log R} \right){{\rm{C}}_2} + \left( {\log R - \log A} \right){{\rm{C}}_3}\\ = \left| \begin{array}{l} 0\,\,\,\,\,\,p\,\,\,\,\,\,1\\ 0\,\,\,\,\,\,q\,\,\,\,\,\,1\\ 0\,\,\,\,\,\,r\,\,\,\,\,\,\,1 \end{array} \right| = 0 \end{array}\]

$$\eqalign{ & {\text{Let the series }}a,ar,a{r^2},.....{\text{ are in geometric progression}}{\text{.}} \cr & {\text{given, }}a = ar + a{r^2} \cr & \Rightarrow \,\,1 = r + {r^2} \cr & \Rightarrow \,\,{r^2} + r - 1 = 0 \cr & \Rightarrow \,\,r = \frac{{ - 1 \pm \sqrt {1 - 4 \times - 1} }}{2} \cr & \Rightarrow \,\,r = \frac{{ - 1 \pm \sqrt 5 }}{2} \cr & \Rightarrow \,\,r = \frac{{\sqrt 5 - 1}}{2}\left[ {\because {\text{ terms of G}}{\text{.P}}{\text{. are positive}}} \right. \cr & \therefore r {\text{ should be }}\left. {{\text{positive}}} \right] \cr} $$

Product $$ = \left( { - 1} \right)\left( { - 2} \right)\left( { - {2^2}} \right).....\left( { - {2^{15}}} \right)\left( {x - 1} \right)\left( {x - \frac{1}{2}} \right)\left( {x - \frac{1}{{{2^2}}}} \right)\left( {x - \frac{1}{{{2^3}}}} \right).....\left( {x - \frac{1}{{{2^{15}}}}} \right)$$
$$ = {2^{1 + 2 + 3 + ..... + 15}}\left( {x - 1} \right)\left( {x - \frac{1}{2}} \right)\left( {x - \frac{1}{{{2^2}}}} \right).....\left( {x - \frac{1}{{{2^{15}}}}} \right)$$
∴ co-efficient of $${x^{15}} = {2^{1 + 2 + 3 + ..... + 15}}.\left( { - 1 - \frac{1}{2} - \frac{1}{{{2^2}}} - ..... - \frac{1}{{{2^{15}}}}} \right).$$

$$a = \frac{1}{{1 - \left( {\sqrt 3 - 1} \right)}} = \frac{1}{{2 - \sqrt 3 }} = 2 + \sqrt 3 $$
and $$ab = 1$$
$$ \Rightarrow b = 2 - \sqrt 3 $$
so, $$a$$ and $$b$$ are roots of $${x^2} - 4x + 1 = 0$$

Clearly $${a_n},{b_n},{c_n}$$  are the middle terms of the given A.P., G.P., H.P. respectively. So, $${a_n}$$ is the AM of $$a,b;{b_n}$$  is the GM of $$a, b$$  and $${c_n}$$ is the HM of $$a, b.$$  Also $${a_n},{b_n},{c_n}$$  are positive because $$a, b$$  are positive.
∴ $${a_n},{b_n},{c_n}$$  are in G.P.; so discriminant $$ = b_n^2 - 4{a_n}{c_n} = - 3{a_n}{c_n} < 0.$$

Till $${10^{th}}$$ minute number of counted notes = 1500
$$\eqalign{ & 3000 = \frac{n}{2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\, = n\left[ {148 - n + 1} \right] \cr & {n^2} - 149n + 3000 = 0 \cr & \Rightarrow \,\,n = 125,24 \cr & {\text{But }}n = 125{\text{ is not possible}} \cr & \therefore \,\,{\text{total time = }}24 + 10 = 34{\text{ minutes}}{\text{.}} \cr} $$