Sequences and Series MCQ Questions & Answers in Algebra | Maths

$${\text{G}}{\text{.P}}{\text{.}} = x$$
$$\frac{a}{{1 - r}} = x$$   ( where, $$a = {1^{st}}$$  term and $$r$$ = common ratio )
$$\eqalign{ & \Rightarrow \frac{2}{{1 - r}}x\,\,\,.....\left( {\text{i}} \right)\,\,\left( {\because {\text{Given }}a = 2{\text{ and }}\left| r \right| < 1} \right) \cr & \Rightarrow - 1 < r < 1 \cr & \Rightarrow 1 > - r < - 1 \cr & \Rightarrow 1 + 1 > 1 - r > 1 - 1 \cr & \Rightarrow 0 < 1 - r < 2 \cr & \Rightarrow \frac{1}{{1 - r}} > \frac{1}{2},\frac{2}{{1 - r}} > 1 \cr} $$
from equation (i) $$x > 1$$
Hence, $$1 < x < \infty .$$

$$\eqalign{ & {\text{Let }}S = \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ......n{\text{ terms}} \cr & {\text{ = }}\left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{8}} \right) + \left( {1 - \frac{1}{{16}}} \right).....n\,{\text{terms}} \cr & {\text{ = }}\left( {1 + 1 + 1 + .....n{\text{ terms}}} \right) - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + .... + \frac{1}{{{2^n}}}} \right) \cr & = n - \left[ {\frac{{\frac{1}{2}\left( {1 - \frac{1}{{{2^n}}}} \right)}}{{1 - \frac{1}{2}}}} \right] \cr & = n - 1 + {2^{ - n}} \cr} $$

$$\eqalign{ & A = \frac{{a + ar + a{r^2} + ..... + a{r^{n - 1}}}}{n} = \frac{{a\left( {1 - {r^n}} \right)}}{{n\left( {1 - r} \right)}}; \cr & H = \frac{1}{{\frac{{\frac{1}{a} + \frac{1}{{ar}} + ..... + \frac{1}{{a{r^{n - 1}}}}}}{n}}} = \frac{{n\left( {r - 1} \right){r^n}}}{{\frac{r}{a}\left( {{r^n} - 1} \right)}}. \cr} $$

Given that $$1 + n\left[ {\frac{x}{{1 - x}}} \right] + \left[ {\frac{{n\left( {n + 1} \right)}}{{2\,!}}} \right]{\left[ {\frac{x}{{1 - x}}} \right]^2} + .....\,\infty\,\, $$         is expansion of $${\left[ {1 - \frac{x}{{1 - x}}} \right]^{ - n}}.$$
So, it is $$ = {\left[ {1 - \frac{x}{{1 - x}}} \right]^{ - n}}$$
$$ = {\left[ {\frac{{1 - x - x}}{{1 - x}}} \right]^{ - n}} = {\left[ {\frac{{1 - x}}{{1 - 2x}}} \right]^n}$$

Let required number of months = $$n$$
$$\eqalign{ & \therefore \,\,200 \times 3 + \left( {240 + 280 + 320 + ..... + {{\left( {n - 3} \right)}^{th}}{\text{term}}} \right) = 11040 \cr & \Rightarrow \,\,\frac{{n - 3}}{2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right] = 11040 - 600 \cr & \Rightarrow \,\,\left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440 \cr & \Rightarrow \,\,\left( {n - 3} \right)\left( {20n + 160} \right) = 10440 \cr & \Rightarrow \,\,\left( {n - 3} \right)\left( {n + 8} \right) = 522 \cr & \Rightarrow \,\,{n^2} + 5n - 546 = 0 \cr & \Rightarrow \,\,\left( {n + 26} \right)\left( {n - 21} \right) = 0 \cr & \therefore \,\,n = 21 \cr} $$

We have, $$S = 2 + 4 + 7 + 11 + 16 + ..... + {T_n}$$
Again, $$S = 2 + 4 + 7 + 11 + 16 + ..... + {T_{n - 1}} + {T_n}$$
Subtracting, we get
$$\eqalign{ & 0 = 2 + \left\{ {2 + 3 + 4 + 5 + ..... + \left( {{T_n} - {T_{n - 1}}} \right)} \right\} - {T_n} \cr & {T_n} = 2 + \frac{1}{2}\left( {n - 1} \right)\left( {4 + \left\{ {n - 2} \right\}1} \right) = \frac{1}{2}\left( {{n^2} + n + 2} \right) \cr & {\text{Now,}} \cr & S = \sum {{T_n} = \frac{1}{2}\sum {\left( {{n^2} + n + 2} \right) = \frac{1}{2}\left( {\sum {{n^2} + \sum {n + 2\sum 1 } } } \right)} } \cr & = \frac{1}{2}\left\{ {\frac{1}{6}n\left( {n + 1} \right)\left( {2n + 1} \right) + \frac{1}{2}n\left( {n + 1} \right) + 2n} \right\} \cr & = \frac{n}{{12}}\left\{ {\left( {n + 1} \right)\left( {2n + 1 + 3} \right) + 12} \right\} \cr & = \frac{n}{6}\left\{ {\left( {n + 1} \right)\left( {n + 2} \right) + 6} \right\} = \frac{n}{6}\left( {{n^2} + 3n + 8} \right) \cr} $$

$$2b = a + c,$$   so the given expression is
$$\left( {a + a + c - c} \right)\left( {a + c + c - a} \right)\left( {2b - b} \right) = 4\,abc$$

The given eq. can be written as
$$\eqalign{ & {\left| {x - 1} \right|^2} - 4\left| {x - 1} \right| + 3 = 0 \cr & \Rightarrow \left( {\left| {x - 1} \right| - 3} \right)\left( {\left| {x - 1} \right| - 1} \right) = 0 \cr & {\text{If }}\left| {x - 1} \right| - 3 = 0 \cr & \Rightarrow x - 1 = \pm 3 \cr & \Rightarrow x = - 2\,\,{\text{or 4}} \cr & {\text{If }}\left| {x - 1} \right| - 1 = 0 \cr & \Rightarrow x - 1 = \pm 1 \cr & \Rightarrow x = 0\,\,{\text{or 2}} \cr} $$
The four roots are $$– 2, 0, 2, 4$$  and are in A.P.

$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... + \frac{1}{{{2^{n - 1}}}} < 2 - \frac{1}{{1000}}$$
LHS of given inequality is in G.P.
$$\eqalign{ & \therefore \frac{{1 - \frac{1}{{{2^n}}}}}{{1 - \frac{1}{2}}} < 2 - \frac{1}{{1000}} \cr & \Rightarrow 2 - \frac{1}{{{2^{n - 1}}}} < 2 - \frac{1}{{100}} \cr & \Rightarrow {2^{n - 1}} < 1000 \cr & {\text{Now, }}{\left( 2 \right)^9} = 512\,\,\& \,\,{\left( 2 \right)^{10}} = 1024 \cr & \therefore n - 1 = 9 \cr & \Rightarrow n = 10. \cr} $$

$$\eqalign{ & {\text{Let }}{10^9} + 2 \cdot \left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ..... + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9} \cr & {\text{Let }}x = {10^9} + 2 \cdot \left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ..... + 10{\left( {11} \right)^9} \cr & {\text{Multiplied by }}\frac{{11}}{{10}}{\text{ on both the sides}} \cr & \frac{{11}}{{10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + ..... + 9{\left( {11} \right)^9} + {11^{10}} \cr & x\left( {1 - \frac{{11}}{{10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + {11^2} \times {\left( {10} \right)^7} + ..... + {11^9} - {11^{10}} \cr & \Rightarrow \,\,\, - \frac{x}{{10}} = {10^9}\left[ {\frac{{{{\left( {\frac{{11}}{{10}}} \right)}^{10}} - 1}}{{\frac{{11}}{{10}} - 1}}} \right] - {11^{10}} \cr & \Rightarrow \,\, - {\frac{x}{{10}}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}} \cr & \Rightarrow \,\,x = {10^{11}} = k{.10^9} \cr & {\text{Given }} \Rightarrow \,\,{\text{k = 100}} \cr} $$