Sequences and Series Test

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Sequences and Series Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The sum to the first n terms of the series $$\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + {\rm{ }} \ldots $$ is $$9 + {2^{ - 10}}$$. The value of n is

A
$$10$$

B
$$9$$

C
$$8$$

D
none of these

Solution

$$\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{7}{8}+\dfrac{15}{16}+........$$

$$=\left ( 1-\dfrac{1}{2} \right )+\left ( 1-\dfrac{1}{4} \right )+\left ( 1-\dfrac{1}{8} \right)+....$$

$$=(1+1+1+.......upto\; n\; terms)-\left ( \dfrac{1}{2} +\dfrac{1}{4}+\dfrac{1}{8}+.....n\; terms\right )$$

$$=n-\left [ \dfrac{\dfrac{1}{2}\left \{ 1-(\dfrac{1}{2})^{n} \right \}}{1-\dfrac{1}{2}} \right ]$$

$$=n-1+2^{-n}$$

Now,

$$n-1+2^{-n}=9+2^{-10}$$

$$\Rightarrow n=10$$
Question 2
1 / -0

$$A = (2 + 1)({2^2} + 1)({2^4} + 1).....\left( {{2^{2016}} + 1} \right)$$. The value of $${\left( {A + 1} \right)^{\frac{1}{{2016}}}}$$ is

A
$$4$$

B
$$2016$$

C
$$^{{2^{4032}}}$$

D
$$2$$

Solution

$$A=(2+1)(2^2+1)(2^4+1)(2^8+1).....(2^{2016}+1)$$
$$A=(2-1) (2+1) (2^2+1)(2^4+1)(2^8+1)....(2^{2016}+1)$$
$$A=(2^2-1)(2^2+1)(2^4+1)......(2^{2016}+1)$$
Further simplification gives
$$A=2^{4032}-1$$
$$\Rightarrow \ A+1=2^{4032}$$
$$\Rightarrow \ (A+1)^{1/2016} =(2^{4032})^{1/2016}$$
$$=2^2$$
$$=4$$
Question 3
1 / -0

The value of $$S=\sum _{ k=1 }^{ 6 }{ \left( \sin { \dfrac { 2\pi k }{ 7 } } -i\cos { \dfrac { 2\pi k }{ 7 } } \right) }$$ ?

A
$$-1$$

B
$$0$$

C
$$-i$$

D
$$i$$

Solution

$$ sin \dfrac{2 \pi k}{7} - i cos\dfrac{2 \pi k}{7}$$
$$ = -i \left ( cos\dfrac{2\pi k}{7}+i sin \dfrac{2 \pi k}{7} \right )$$
$$ = -ie^{2\pi k/7i}$$
$$ \displaystyle \Rightarrow S = \sum_{k=1}^{6} -i\left ( e^{2\pi k/i} \right )$$
$$\displaystyle = -i \left ( \sum_{k=1}^{6} e^{i2\pi k/7} \right )$$
$$ = -i $$ [sum of roots of the equation
$$ x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+1 = 0].$$
$$ = -i \left ( -1 \right )$$
$$ = i $$
Question 4
1 / -0

If $$P,\,Q$$ be two arithmetic means between $$\dfrac{1}{3}$$ and $$\dfrac{1}{24}$$, then their value are

A
$$\dfrac{7}{72},\dfrac{5}{36}$$

B
$$\dfrac{17}{72},\dfrac{5}{36}$$

C
$$\dfrac{7}{36},\dfrac{5}{72}$$

D
$$\dfrac{5}{72},\dfrac{17}{72}$$

Solution

P and Q are the arithmetic means between $$\dfrac{1}{3} $$ and $$\dfrac{1}{24}$$

So the resulting sequence formed an AP
$$\dfrac{1}{3},P,Q, \dfrac{1}{24}$$ are in AP

$$\Rightarrow a_4=a+(4-1)d$$
$$\Rightarrow\dfrac{1}{24}=\dfrac{1}{3}+(4-1)d$$

$$\Rightarrow \dfrac{-7}{24}=3d$$

$$\Rightarrow d=\dfrac{-7}{72}$$

$$\therefore P=a+d=\dfrac{1}{3}+\dfrac{-7}{72}$$

=$$\dfrac{24-7}{72}$$

=$$\dfrac{17}{72}$$

and $$Q=a+2d=\dfrac{1}{3}+\dfrac{-14}{72}$$

=$$\dfrac{24-14}{72}$$

=$$\dfrac{5}{36}$$
Question 5
1 / -0

If $$\dfrac{1}{x (x + 1) (x + 2)....(x + n)} = \dfrac{A_0}{x} + \dfrac{A_1}{x + 1} + \dfrac{A_2}{x + 2} + .... + \dfrac{A_n}{x + n} $$, then $$ A_r$$ $$=$$

A
$$\dfrac{(-1)^r r !}{(n - r) !}$$

B
$$\dfrac{(-1)^r}{r ! (n - r) !}$$

C
$$\dfrac{1}{r ! (n - r) !}$$

D
$$\dfrac{r !}{(n - r) !}$$
Question 6
1 / -0

What is the sum of the series $$\dfrac{1}{{{3^2} - 4}} + \dfrac{1}{{{7^2} - 4}} + \dfrac{1}{{{{11}^2} - 4}} + ..... + \dfrac{1}{{{{39}^2} - 4}}$$

A
$$\dfrac{{10}}{{41}}$$

B
$$\dfrac{{40}}{{41}}$$

C
$$\dfrac{{20}}{{41}}$$

D
$$\dfrac{{30}}{{41}}$$

Solution

$$\begin{array}{l}\dfrac{1}{{{3^2} - 4}} + \dfrac{1}{{{7^2} - 4}} + \dfrac{1}{{{{11}^2} - 4}} + ..... + \dfrac{1}{{{{39}^2} - 4}}\\ = \dfrac{1}{{\left( {3 - 2} \right)\left( {3 + 2} \right)}} + \dfrac{1}{{\left( {7 - 2} \right)\left( {7 + 2} \right)}} + \dfrac{1}{{\left( {11 - 2} \right)\left( {11 + 2} \right)}} + ..... + \dfrac{1}{{\left( {39 - 4} \right)\left( {39 + 4} \right)}}\\ = \dfrac{1}{4}\left[ {\dfrac{{5 - 1}}{{5 \times 1}} + \dfrac{{9 - 5}}{{9 \times 5}} + \dfrac{{13 - 9}}{{13 \times 9}} + .....} \right]\\ = \dfrac{1}{4}\left[ {\dfrac{{5 - 1}}{{5 \times 1}} + \dfrac{{9 - 5}}{{9 \times 5}} + \dfrac{{13 - 9}}{{13 \times 9}} + .....} \right]\\ = \dfrac{1}{4}\left[ {1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{{13}} + .........\dfrac{1}{{37}} - \dfrac{1}{{41}}} \right]\\ = \dfrac{1}{4}\left[ {1 - \dfrac{1}{{41}}} \right] = \dfrac{{10}}{{41}}\end{array}$$
Question 7
1 / -0

Number of identical terms in the sequence $$2, 5, 8, 11,$$___ upto $$100$$ terms and $$3, 5, 7, 9, 11$$____ upto $$100$$ terms are

A
$$17$$

B
$$33$$

C
$$50$$

D
$$147$$

Solution

Fierst series: $$2, 5, 8, 11.........$$
$$a_1 =2;\ d=3;\ n=100$$
$$l=20(100-1)^*5=2+99^*3=2+297=299$$
so, series $$=2, 5, 8, 11,.....,197, 197,200,203,....,299$$
second series; $$3, 5, 9, 11,.....$$
$$a_1=3;\ d=2;\ n=100$$
$$l=3+(100-1)^*2=3+99^*2=3+198=201$$
so series $$=3, 5, 7, 9, 11, ...., 201$$
series having similar terms; $$5, 11, 17, ....., 197$$
$$a_1=5;\ d=6;\ l=197$$
$$l=a_1+(n-1)d$$
$$197=5+(n-1)6$$
$$192/6=n-1$$
$$32+1=n$$
$$n=33$$
Question 8
1 / -0

Find the value of ?.

A
$$125$$

B
$$25$$

C
$$625$$

D
$$156$$

Solution

$${\left( {3 + 1} \right)^2} = {\left( 4 \right)^2} = 16$$

$${\left( {15 + 6} \right)^2} = {\left( {21} \right)^2} = 441$$

$${\left( {10 + 5} \right)^2} = {\left( {15} \right)^2} = 225$$

$$so,\,{\left( {12 + 13} \right)^2} = {\left( {25} \right)^2} = 625$$

therefore the answer is $$625$$
Question 9
1 / -0

The value of the expression $$\left(\dfrac{1}{2^2 -1}\right) + \left(\dfrac{1}{4^2 -1}\right) + \left(\dfrac{1}{6^2 -1}\right) + ... + \left(\dfrac{1}{20^2 -1}\right)$$ is

A
$$\dfrac{9}{19}$$

B
$$\dfrac{10}{19}$$

C
$$\dfrac{10}{21}$$

D
$$\dfrac{11}{21}$$

Solution

$$n$$ th term in the given series can be written as

$$\dfrac{1}{(2n)^{2}-1}$$

$$=\dfrac{1}{(2n+1)(2n-1)}$$

$$=\dfrac{1}{2}\left \{ \dfrac{1}{2n-1} -\dfrac{1}{2n+1}\right \}$$

So, the given series can be written as

$$\dfrac{1}{2}\left ( 1-\dfrac{1}{3} \right )+\dfrac{1}{2}\left ( \dfrac{1}{3} -\dfrac{1}{5}\right )+\dfrac{1}{2}\left ( \dfrac{1}{5} -\dfrac{1}{7}\right )+..............\dfrac{1}{2}\left ( \dfrac{1}{19} -\dfrac{1}{21}\right )$$

$$=\dfrac{1}{2}\left ( 1-\dfrac{1}{21} \right )$$

$$=\dfrac{10}{21}$$
Question 10
1 / -0

$$67,84,95,.,133,158$$

A
$$116$$

B
$$129$$

C
$$108$$

D
$$111$$

Solution

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Sequences and Series Test - 37

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Sequences and Series Test - 37

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SHARING IS CARING

Question 1

$$10$$

$$9$$

$$8$$

none of these

Solution

Question 2

$$4$$

$$2016$$

$$^{{2^{4032}}}$$

$$2$$

Solution

Question 3

$$-1$$

$$0$$

$$-i$$

$$i$$

Solution

Question 4

$$\dfrac{7}{72},\dfrac{5}{36}$$

$$\dfrac{17}{72},\dfrac{5}{36}$$

$$\dfrac{7}{36},\dfrac{5}{72}$$

$$\dfrac{5}{72},\dfrac{17}{72}$$

Solution

Question 5

$$\dfrac{(-1)^r r !}{(n - r) !}$$

$$\dfrac{(-1)^r}{r ! (n - r) !}$$

$$\dfrac{1}{r ! (n - r) !}$$

$$\dfrac{r !}{(n - r) !}$$

Question 6

$$\dfrac{{10}}{{41}}$$

$$\dfrac{{40}}{{41}}$$

$$\dfrac{{20}}{{41}}$$

$$\dfrac{{30}}{{41}}$$

Solution

Question 7

$$17$$

$$33$$

$$50$$

$$147$$

Solution

Question 8

$$125$$

$$25$$

$$625$$

$$156$$

Solution

Question 9

$$\dfrac{9}{19}$$

$$\dfrac{10}{19}$$

$$\dfrac{10}{21}$$

$$\dfrac{11}{21}$$

Solution

Question 10

$$116$$

$$129$$

$$108$$

$$111$$

Solution

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