Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \frac{{A{L_1}}}{{AB}} = \frac{{{L_1}{M_1}}}{{BC}} \cr & \therefore \frac{1}{{n + 1}} = \frac{{{L_1}{M_1}}}{a} \cr & \therefore {L_1}{M_1} = \frac{a}{{n + 1}}; \cr} $$

$$\eqalign{ & \frac{{A{L_2}}}{{AB}} = \frac{{{L_2}{M_2}}}{{BC}} \cr & \therefore \frac{2}{{n + 1}} = \frac{{{L_2}{M_2}}}{a} \cr & \therefore {L_2}{M_2} = \frac{{2a}}{{n + 1}},{\text{etc}}{\text{.}} \cr & \therefore \,{\text{The required sum }} = \frac{a}{{n + 1}} + \frac{{2a}}{{n + 1}} + \frac{{3a}}{{n + 1}} + .... + \frac{{na}}{{n + 1}} \cr & = \frac{a}{{n + 1}}\left( {1 + 2 + .... + n} \right) = \frac{a}{{n + 1}} \cdot \frac{{n\left( {n + 1} \right)}}{2} = \frac{{an}}{2} \cr} $$

$$\eqalign{ & {a_1} + \left( {p - 1} \right)d,{a_1} + \left( {q - 1} \right)d,{a_1} + \left( {r - 1} \right)d\,\,{\text{are}}\,{\text{in}}\,{\text{G}}{\text{.P}}. \cr & \therefore \,\,\frac{{{a_q}}}{{{a_p}}} = \frac{{{a_1} + \left( {q - 1} \right)d}}{{{a_1} + \left( {p - 1} \right)d}} = \frac{{{a_1} + \left( {r - 1} \right)d}}{{{a_1} + \left( {q - 1} \right)d}} = \frac{{r - p}}{{q - p}}. \cr} $$

The roots of equation are 2 and 3
$$\eqalign{ & \therefore g = \sqrt {xy} = 2 \cr & \Rightarrow xy = 4 \cr & G = \sqrt {\left( {x + 1} \right)\left( {y + 1} \right)} = 3 \cr & \Rightarrow \left( {x + 1} \right)\left( {y + 1} \right) = 9 \cr & \therefore x = y = 2 \cr} $$

$$\eqalign{ & S = \frac{1}{{1 - \frac{1}{2}}} = 2 \cr & {S_n} = \frac{{1\left( {1 - \frac{1}{{{2^n}}}} \right)}}{{1 - \frac{1}{2}}} = 2 - \frac{1}{{{2^{n - 1}}}} \cr & {\text{Given, }}S - {S_n} < {10^{ - 6}} \cr & \therefore \frac{1}{{{2^{n - 1}}}} < {10^{ - 6}} \cr & \Rightarrow {2^{n - 1}} > {10^6} \cr & \therefore n - 1 > 6\,{\log _2}10 = \frac{6}{{0.3010}}n > 20\left[ {\because \frac{6}{{.3018}} < 20} \right] \cr & \therefore n = 21 \cr} $$

As given, $$n^{th}$$ term is : $${T_n} = 3n + 7$$
Sum of $$n$$ term, $${S_n} = \sum {{T_n}} $$
$$\eqalign{ & = \sum {\left( {3n + 7} \right) = 3\sum {n + 7} \sum 1 } \cr & = \frac{{3n\left( {n + 1} \right)}}{2} + 7n = n\left[ {\frac{{3n + 3 + 14}}{2}} \right] \cr & = n\left[ {\frac{{3n + 17}}{2}} \right] \cr} $$
Sum of 50 terms $$ = {S_{50}} = 50\left[ {\frac{{3 \times 50 + 17}}{2}} \right]$$
$$ = 50\left[ {\frac{{167}}{2}} \right] = 25 \times 167 = 4175$$

$$\eqalign{ & 8 + \left( {n - 1} \right)\left( { - 2} \right) = 2\left( {2 + \left( {n - 1} \right)\left( { - 1} \right)} \right) \cr & \Rightarrow 10 = 6\,{\text{which is absurd}} \cr} $$

\[\frac{\begin{gathered} {\text{Let, }}{S_1} = 2 + 3 + 5 + 9 + 16 + ..... + {x_n} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{S_1} = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 + 3 + 5 + 9 + .....{x_{n - 1}} + {x_n} \hfill \\ \end{gathered} }{{O = 2 + \left[ {1 + 2 + 4 + 7 + ..... + {\text{to }}\left( {n - 1} \right){\text{terms}}} \right] - {x_n}}}\]
$$\therefore {x_n} = 2 + \left[ {1 + 2 + 4 + 7 + ..... + {\text{to }}\left( {n - 1} \right){\text{terms}}} \right]$$
Again let
\[\frac{\begin{gathered} {S_2} = 1 + 2 + 4 + 7 + ..... + {t_{n - 1}} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{S_2} =\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,1 + 2 + 4 + 7 + ..... + {t_{n - 2}} + {t_{n - 1}} \hfill \\ \end{gathered} }{{O = 1 + \left[ {1 + 2 + 3 + ..... + \left( {n - 2} \right){\text{terms}}} \right] - {t_{n - 1}}}}\]
$$\eqalign{ & {t_{n - 1}} = 1 + \frac{{\left( {n - 2} \right)\left( {n - 1} \right)}}{2} = \frac{{{n^2} - 3n + 4}}{2} \cr & \therefore {S_2} = \sum\limits_{n = 1}^{n - 1} {{t_{n - 1}}} = \frac{1}{2}\sum {{n^2} - \frac{3}{2}\sum {n + 2\sum 1 } } \cr & = \frac{1}{2}\frac{{\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6} - \frac{3}{2}\frac{{n\left( {n - 1} \right)}}{2} + 2\left( {n - 1} \right) \cr & = \left( {n - 1} \right)\left[ {\frac{{2{n^2} - n}}{{12}} - \frac{{3n}}{4} + 2} \right] \cr & = \frac{{n - 1}}{{12}}\left[ {2{n^2} - n - 9n + 24} \right] \cr & = \frac{{n - 1}}{6}\left[ {{n^2} - 5n + 12} \right] \cr & = \frac{{{n^3} - 6{n^2} + 17n - 12}}{6} \cr & \therefore {x_n} = 2 + {S_2} \cr & = 2 + \frac{{{n^3} - 6{n^2} + 17n - 12}}{6} \cr & = \frac{{{n^3} - 6{n^2} + 17n}}{6} \cr & {\text{So, }}{x_{20}} = 990 \cr} $$

Roots of Given equation $${x^3} - 12{x^2} + 39x - 28 = 0$$      are in A.P.
Let $$\alpha - \beta ,\alpha ,\alpha + \beta $$    be the roots of the equation.
Sum of the roots $$ = \alpha - \beta + \alpha + \alpha + \beta = \frac{{ - \left( { - 12} \right)}}{1} = 12$$
$$\eqalign{ & 3\alpha = 12 \cr & \Rightarrow \alpha = 4{\text{ and }}\left( {\alpha - \beta } \right)\alpha + \alpha \left( {\alpha + \beta } \right) + \left( {\alpha + \beta } \right)\left( {\alpha - \beta } \right) = 39 \cr & \Rightarrow {\alpha ^2} - \alpha \beta + {\alpha ^2} + \alpha \beta + {\alpha ^2} - {\beta ^2} = 39 \cr & \Rightarrow 3{\alpha ^2} - {\beta ^2} = 39 \cr & \Rightarrow 3{\left( 4 \right)^2} - {\beta ^2} = 39 \cr & \Rightarrow 48 - {\beta ^2} = 39 \cr & \Rightarrow - {\beta ^2} = 39 - 48 \cr & \Rightarrow - {\beta ^2} = - 9 \cr & \Rightarrow \beta = \pm 3 \cr} $$

We know that
$$\frac{{{e^x} + {e^{ - x}}}}{2} = 1 + \frac{{{x^2}}}{{2\,!}} + \frac{{{x^4}}}{{4\,!}} + \frac{{{x^6}}}{{6\,!}} + .....$$
keeping $$x = 2,$$  we get
Expression $$ = \frac{1}{2}\left[ {\frac{{{e^2} + {e^{ - 2}}}}{2}} \right] - 1 = \frac{{{{\left( {{e^2} - 1} \right)}^2}}}{{2{e^2}}}$$

We have
$$\eqalign{ & {8^{1 + \left| {\cos x} \right| + {{\cos }^2}x + \left| {{{\cos }^3}x} \right| + .....\,{\text{to }}\infty }} = {4^3} \cr & \left[ {\because {{\cos }^2}x = \left| {{{\cos }^2}x} \right|{\text{ also }}\left| {{{\cos }^n}x} \right| = {{\left| {\cos x} \right|}^n}} \right] \cr & \Rightarrow {8^{\frac{1}{{1 - \left| {\cos x} \right|}}}} = {4^3} \cr & \Rightarrow {2^{^{\frac{3}{{1 - \left| {\cos x} \right|}}}}} = {2^6} \cr & \therefore \frac{3}{{1 - \left| {\cos x} \right|}} = 6 \cr & \Rightarrow 1 - \left| {\cos x} \right| = \frac{1}{2} \cr & \therefore \left| {\cos x} \right| = \frac{1}{2} \cr & \Rightarrow \cos x = \pm \frac{1}{2} \cr & \therefore x = \frac{\pi }{3},\frac{{2\pi }}{3} \cr} $$