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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 13 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 13
If $$(3x-1)^{}=a_{7}x^{7}+a_{6}x^{6}+....+a_{0}$$, then $$a_{7}+a_{6}+...+a_{0}$$ is
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$$128$$
0%
$$1$$
0%
$$64$$
0%
$$None\ of\ these$$
Explanation
$$\begin{array}{l} { \left( { 3x-1 } \right) ^{ 7 } }=\left[ { ^{ 7 }{ C_{ 0 } }.{ { \left( { 3x } \right) }^{ 7 } }{ -^{ 7 } }{ C_{ 1 } }{ { \left( { 3x } \right) }^{ 6 } }{ +^{ 7 } }{ C_{ 2 } }{ { \left( { 3x } \right) }^{ 5 } }{ +^{ 7 } }{ C_{ 3 } }{ { \left( { 3x } \right) }^{ 5 } }{ +^{ 7 } }{ C_{ 4 } }{ { \left( { 3x } \right) }^{ 4 } }{ -^{ 7 } }{ C_{ 5 } }{ { \left( { 3x } \right) }^{ 5 } }{ +^{ 7 } }{ C_{ 6 } }{ { \left( { 3x } \right) }^{ 6 } }{ -^{ 7 } }{ C_{ 7 } }{ { \left( { 3x } \right) }^{ 7 } } } \right] \\ \Rightarrow 2187{ x^{ 7 } }-5103{ x^{ 6 } }+5103{ x^{ 5 } }-2035{ x^{ 4 } }+945{ x^{ 3 } }-189{ x^{ 2 } }+21x-1\, \, \, \, \, \, \left[ \begin{array}{l} { ^{ n } }{ C_{ 1 } }=n \\ ^{ n }{ C_{ 0 } }{ =^{ n } }{ C_{ n } }=1 \end{array} \right] \\ \Rightarrow \therefore { a_{ 7 } }+{ a_{ 6 } }+{ a_{ 5 } }+{ a_{ 4 } }+............{ a_{ 0 } }=2187-5103+5103-2835+945-189+21-1=128\, \, \, \end{array}$$
If $$\left( 1+x+{ x }^{ 2 } \right) ={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+........{ a }_{ 2n }{ x }^{ 2n }$$, then the value of $${ a }_{ 0 }+{ a }_{ 3 }+{ a }_{ 6 }+.....$$ is
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$${ a }_{ 1 }+{ a }_{ 4 }+{ a }_{ 7 }+.....$$
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$${ a }_{ 2 }+{ a }_{ 5 }+{ a }_{ 8 }+.....$$
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$${ 3 }^{ n-1 }$$
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All(A,B,C)
If $$(1+x+x^{2})^{n}=a_{0}+a_{1}+a_{2}x^{2n}$$, then $$a_{0}+a_{2}+a_{4}+........+a_{2n}x^{2n},$$ is equal to :
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$$\dfrac{3^{n}+1}{2}$$
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$$\dfrac{3^{n}-1}{2}$$
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$$\dfrac{1-3^{n}}{2}$$
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$$3^{n}+\dfrac{1}{2}$$
Explanation
$$ { \left( { 1+x+{ x^{ 2 } } } \right) ^{ 4 } }={ a_{ 0 } }+{ a_{ 1 } }+{ a_{ 2 } }{ x^{ 2n } }+.........{ x^{ 4 } }+........{ a_{ 2x } }{ x^{ 2n } }\, \, \, \to \left( i \right) $$
$$\\ Put\, \, x=1\, \, in\, \, equation\, \left( i \right) \\ { 3^{ x } }={ a_{ 0 } }+{ a_{ 1 } }+{ a_{ 2 } }+....................\to \left( { ii } \right) $$
$$\\ Put\, \, x=-1\, \, \, in\, \, equation\, \, \left( i \right) \\ 1={ a_{ 0 } }-{ a_{ 1 } }+{ a_{ 2 } }-{ a_{ 3 } }+..............\to \left( { iii } \right) \\ $$
$$ On\, adding\, \, equation\, \, \left( { ii } \right) \, \, and\, \, \, \left( { iii } \right) \\ { 3^{ n } }\, +1=2\left( { { a_{ 0 } }+{ a_{ 2 } }+{ a_{ 4 } }+......... } \right) \\ \dfrac{ { 3^{ n } }+1}{2}={ a_{ 0 } }+{ a_{ 2 } }+{ a_{ 4 } }+............$$
Hence, this is the answer.
$$\dfrac { 7 } { 11 } : \dfrac { 336 } { 110 } : ? \quad : \quad \dfrac { 720 } { 272 }$$
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$$\dfrac { 9 } { 17 }$$
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$$\dfrac { 9 } { 13 }$$
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$$\dfrac { 11 } { 13 }$$
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$$\dfrac { 11 } { 17 }$$
$$ABCDEFGHIJKLMZYXWVUTSRQPON$$
From the above letter series find the letter which is at the $$6th$$ position to the right side of the letter which is at the centre position of the letters which are at the $$11th$$ place form the left and $$14th$$ place from the right.
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$$V$$
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$$F$$
0%
$$Y$$
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$$L$$
There is a specific relationship between the numbers that are given in the following figures. On the basis of the relationship choose the correct
alternative to replace the question mark.
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$$210$$
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$$266$$
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$$288$$
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$$318$$
If $${\left( {20} \right)^{19}} + 2\left( {21} \right){\left( {20} \right)^{18}} + 3{\left( {21} \right)^2}{\left( {20} \right)^{17}} + ... + 20{\left( {21} \right)^{19}} = k{\left( {20} \right)^{19}}$$
then $$k$$ is equal to
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0%
400
0%
100
0%
441
0%
420
I for $$n\in I, n> I0;1+(1+x)+(1+x)^{2}+....+(1+)^{n}=\displaystyle \sum_{k=0}^{n}a_{k}.x^{k},x\neq 0$$ then
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$$\displaystyle \sum_{k=0}^{n}a_{k}=2^{n+1}$$
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$$a_{n-2}=\dfrac {n(n+1)}{2}$$
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$$a_{p}> a_{p-1}$$ for $$p< \dfrac {n}{2}, p\in N$$
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$$(a_{9})^{2}-(a_{8})^{2}=^{n+2}C_{10}(^{n+1}C_{1}-^{n+1}C_{9})$$
If $${\left( {20} \right)^{19}} + 2\left( {21} \right){\left( {20} \right)^{18}} + 3{\left( {21} \right)^2}{\left( {20} \right)^{17}} + ... + 20{\left( {21} \right)^{19}} = k{\left( {20} \right)^{19}}$$ then k is equal to
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0%
400
0%
100
0%
441
0%
420
Let the $$n^{th}$$ terms of a series be given by
$$t_n$$ = $$\dfrac{n^2 -n -2}{n^2 + 3n}$$ , n 3 .
The product $$t_3$$, $$t_4$$,.........$$t_{50}$$ equals-
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$$\dfrac{1}{5^2.7.13.53}$$
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$$\dfrac{1}{5.7^2.12.53}$$
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$$\dfrac{1}{5^2.7.12.51}$$
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$$\dfrac{1}{5.7^2.13.53}$$
If $$ (20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17} $$ $$ +\ldots+20(21)^{19}=k(20)^{19} $$ then $$ k $$ is equal to
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0%
$$400$$
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$$100$$
0%
$$441$$
0%
$$420$$
The sum of the series $$^{20}C_{0}-^{20}C_{1}+^{20}C_{2}-^{20}C_{3}+-..+^{20}C_{10}$$ is$$\dfrac{1}{2}^{20}C_{10}$$
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$$0$$
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$$-^{20}C_{10}$$
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$$^{20}C_{10}$$
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$$\dfrac{1}{2}^{20}C_{10}$$
Explanation
The sum of the series
$$\begin{array}{l} ^{ 20 }{ C_{ 0 } }{ -^{ 20 } }{ C_{ 1 } }{ +^{ 20 } }{ C_{ 2 } }{ -^{ 20 } }{ C_{ 3 } }+{ .....^{ 20 } }{ C_{ 10 } } \\ { \left( { 1+x } \right) ^{ 20 } }{ =^{ 20 } }{ C_{ 0 } }{ +^{ 20 } }{ C_{ 1 } }x{ +^{ 20 } }{ C_{ 2 } }{ x^{ 2 } }+{ ....^{ 20 } }{ C_{ 10 } }{ x^{ 10 } } \\ putx=-1 \\ { \left( { 1-1 } \right) ^{ 20 } }{ =^{ 20 } }{ C_{ 0 } }{ +^{ 20 } }{ C_{ 1 } }{ +^{ 20 } }{ C_{ 2 } }+......{ +^{ 20 } }{ C_{ 10 } } \\ 0{ =^{ 20 } }{ C_{ 0 } }{ -^{ 20 } }{ C_{ 1 } }{ +^{ 20 } }{ C_{ 2 } }+{ ........^{ 20 } }{ C_{ 10 } } \\ =2\left[ { ^{ 20 }{ C_{ 0 } }{ -^{ 20 } }{ C_{ 1 } }+{ { ..... }^{ 20 } }{ C_{ 9 } } } \right] { +^{ 20 } }{ C_{ 10 } } \\ ={ \frac { 1 }{ 2 } ^{ 20 } }{ C_{ 10 } } \end{array}$$
Hence, this is the answer.
$${ C }^{ 2n }{ C }_{ n }-{ { C }_{ 1 } }^{ 2n-2 }{ C }_{ n }+{ { C }_{ 2 } }^{ 2n-4 }{ C }_{ n }...$$ equals to
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$${ 2 }^{ n }$$
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$${ 2 }^{ n }(n+1)$$
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$${ 2 }^{ n-1 }$$
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$$\dfrac { (n+1){ 2 }^{ n-1 } }{ 2 } $$
If $${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ......... + {C_n}{x^n},n \in N$$. Then find the value of
$$\displaystyle C_0^2 + \frac{{C_1^2}}{2} + {2^3}\frac{{{C_2}}}{3} + {2^4}\frac{{{C_3}}}{4} + ...... + {2^{n }}\frac{{{C_n}}}{{n + 1}} $$
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0%
$$ \dfrac{{{3^{n - 1}} - 1}}{{n + 1}}$$
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$$ \dfrac{{{3^{n }} - 1}}{{n + 1}}$$
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$$ \dfrac{{{3^{n + 1}} - 1}}{{n + 1}}$$
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$$ \dfrac{{{3^{n + 1}} - 1}}{{n - 1}}$$
Let three matrices are $$A=\left[ \begin{matrix} 2 & 1 \\ 4 & 1 \end{matrix} \right] ;A=\left[ \begin{matrix} 2 & 4 \\ 2 & 3 \end{matrix} \right] $$ and $$C=\left[ \begin{matrix} 3 & -4 \\ -2 & 3 \end{matrix} \right] $$, then $${ t }_{ r }(A)+{ t }_{ r }\left( \frac { ABC }{ 2 } \right) +{ t }_{ r }\left( \frac { A{ (BC) }^{ 2 } }{ 4 } \right) +{ t }_{ r }\left( \frac { A{ (BC) }^{ 3 } }{ 8 } \right) +.......+\infty $$ is equal to -
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0%
6
0%
9
0%
12
0%
None
If $${ A }_{ 1 },{ A }_{ 2 }$$ are two A.M.S. between two numbers $$a$$ and $$b$$, then $$(2{ A }_{ 1 }-{ A }_{ 2 })(2{ A }_{ 2 }-{ A }_{ 1 })$$ is equal to
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$$a+b$$
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$$\frac { ab }{ a+b } $$
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$$ab$$
0%
none of these
The sum of the series $$\cfrac{9}{5^2.2.1}+\cfrac{13}{5^3.3.2}+\cfrac{17}{5^4.4.3}+....$$ to infinite terms, is
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$$\cfrac{2}{5}$$
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$$\cfrac{1}{5}$$
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$$1$$
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None of these
The average expenditure of Sharma for the January to June is Rs. 4200 and he spent Rs. 1200 in January and Rs.1500 in July. The average expenditure for the months of February to July is:
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4250
0%
6520
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2320
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9999
Suppose n be integer than 1, let $$a_n=\frac 1{\log _n2002}$$. Suppose $$b=a_2+a_3+a_4+a_5$$ and $$c=a_{10}+a_{11}+a_{13}+a_{14}$$, Then (b-c) equals
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$$\frac 1{1001}$$
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$$\frac 1{1002}$$
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$$-1$$
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$$-2$$
If $$\frac { 48 }{ (2)(3) } +\frac { 47 }{ (3)(4) } +\frac { 46 }{ (4)(5) } +........+\frac { 2 }{ (48)(49) } +\frac { 1 }{ (49)(50) } =\frac { 51 }{ 2 } +k(1+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +.....+\frac { 1 }{ 50 } )$$, then K equals
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$$-1$$
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$$-\frac { 1 }{ 2 } $$
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$$1$$
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$$2$$
There are
n
A.M.'s between 3 and 29 such that $$6^{th} mean:(n-1)^{th} mean = 3:5$$, then the value of
n
, is
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0%
10
0%
11
0%
12
0%
none of these
$$\text{Find the value of } \log \sin 1^{\circ} . \log \sin 2^{\circ} \ldots \ldots \log \sin 179^{\circ} $$
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1
0%
0
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-1
0%
2
Explanation
$$\textbf{Step-1: Apply standard angle of trigonometry function to get the required unknown.}$$
$$\text{We have, }$$
$$\log \sin 1^{\circ} . \log \sin 2^{\circ} \ldots \ldots \log \sin 179^{\circ} $$
$$\text{Above expression can be rewritten as}$$
$$\log \sin 1^{\circ} . \log \sin 2^{\circ} \ldots \log\sin 90^{\circ}\ \ldots \log \sin 179^{\circ} $$
$$\Rightarrow$$ $$0$$ $$\textbf{[As log 1 = 0 and sin}$$ $$\boldsymbol{90^{\circ} = 0]}$$
$$\textbf{Hence, option- B is correct answer}$$
The $$20^{th}$$ term of the series $$2\frac{1}{2}+1\frac{7}{13}+1\frac{1}{9}+\frac{20}{23}+......$$ is
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20/103
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20/97
0%
10/113
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none of these
21,25,52,68,193,?
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0%
229
0%
242
0%
257
0%
409
49,64,?,100,121
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0%
74
0%
80
0%
75
0%
81
1001,1004,1012,1027,?
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0%
1051
0%
1050
0%
259
0%
269
For $$x\in R$$, let $$[x]$$ denote the greatest integer $$\leq x$$, then the sum of the series $$\left[-\dfrac{1}{3}\right]+\left[-\dfrac{1}{3}-\dfrac{1}{100}\right]+\left[-\dfrac{1}{3}-\dfrac{2}{100}\right]+...+\left[-\dfrac{1}{3}-\dfrac{99}{100}\right]$$ is?
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$$-153$$
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$$-133$$
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$$-131$$
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$$-135$$
In the following question, select the number (s) from the given options for completing the given series.
1,1,2,8,64?, 65536
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1024
0%
2556
0%
4096
0%
1088
0.15,0.3,?,1.2,2.4
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0%
0.6
0%
0.9
0%
0.06
0%
4.8
The sum of the series $$i+i^{2}+i^{3}+.....$$ upto 1000 terms is _____________
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0%
i
0%
-i
0%
0
0%
1
Set of even numbers form a progression.
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0%
True
0%
False
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