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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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 37

Result

Sequences and Series Test - 37

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Question 1

101010

999

888

none of these

Solution

Question 2

444

201620162016

24032^{{2^{4032}}}24032

222

Solution

Question 3

−1-1−1

000

−i-i−i

iii

Solution

Question 4

772,536\dfrac{7}{72},\dfrac{5}{36}727​,365​

1772,536\dfrac{17}{72},\dfrac{5}{36}7217​,365​

736,572\dfrac{7}{36},\dfrac{5}{72}367​,725​

572,1772\dfrac{5}{72},\dfrac{17}{72}725​,7217​

Solution

Question 5

(−1)rr!(n−r)!\dfrac{(-1)^r r !}{(n - r) !}(n−r)!(−1)rr!​

(−1)rr!(n−r)!\dfrac{(-1)^r}{r ! (n - r) !}r!(n−r)!(−1)r​

1r!(n−r)!\dfrac{1}{r ! (n - r) !}r!(n−r)!1​

r!(n−r)!\dfrac{r !}{(n - r) !}(n−r)!r!​

Question 6

1041\dfrac{{10}}{{41}}4110​

4041\dfrac{{40}}{{41}}4140​

2041\dfrac{{20}}{{41}}4120​

3041\dfrac{{30}}{{41}}4130​

Solution

Question 7

171717

333333

505050

147147147

Solution

Question 8

125125125

252525

625625625

156156156

Solution

Question 9

919\dfrac{9}{19}199​

1019\dfrac{10}{19}1910​

1021\dfrac{10}{21}2110​

1121\dfrac{11}{21}2111​

Solution

Question 10

116116116

129129129

108108108

111111111

Solution

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Question Analysis

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$10$

$9$

$8$

$4$

$2016$

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

$2$

$-1$

$0$

$-i$

$i$

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

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

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

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

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

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

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

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

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

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

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

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

$17$

$33$

$50$

$147$

$125$

$25$

$625$

$156$

$\dfrac{9}{19}$

$\dfrac{10}{19}$

$\dfrac{10}{21}$

$\dfrac{11}{21}$

$116$

$129$

$108$

$111$