MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 9 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 9
The value of $$\sum\limits_{n = 2}^\infty {\left( {1 - \frac{1}{{{n^2}}}} \right)} $$ equals
Report Question
0%
$$-ln 3$$
0%
$$0$$
0%
$$-ln 2$$
0%
$$-ln 5$$
solve that :-
Report Question
0%
$$14$$
0%
$$18$$
0%
$$11$$
0%
$$26$$
Explanation
Given,
Image $$A$$
$$4+2+5+3=14 \times 2=28$$
Image $$B$$
$$7+5+4+3=19 \times 2=38$$
$$\therefore$$ Image $$C$$
$$2+1+3+7=13 \times 2=26$$
The numbers $${\log _{180}}12,{\log _{2160}}12,{\log _{25920}}12$$ are in
Report Question
0%
AP
0%
GP
0%
HP
0%
None of these
If $$[x]$$ denotes the greates integer $$\le x$$, then $$\left[\dfrac{2}{3} \right] + \left[\dfrac{2}{3} + \dfrac{1}{99} \right] + \left[\dfrac{2}{3} + \dfrac{2}{99} \right] + .... + \left[\dfrac{2}{3} + \dfrac{98}{99} \right] =$$
Report Question
0%
$$99$$
0%
$$98$$
0%
$$66$$
0%
$$65$$
Explanation
$$\left [ \cfrac{2}{3} \right ]=0$$
$$\left [ \cfrac{2}{3}+\cfrac{1}{99} \right ]=0$$
$$\left [ \cfrac{2}{3}+\cfrac{2}{99} \right ]=0$$
$$...$$
$$...$$
$$\left [ \cfrac{2}{3}+\cfrac{33}{99} \right ]=\left [ \cfrac{2}{3}+\cfrac{1}{3} \right ]=\left [ 1 \right ]=1$$
$$\left [ \cfrac{2}{3}+\cfrac{34}{99} \right ]=\left [ 1+\cfrac{1}{99} \right ]=1$$
$$\left [ \cfrac{2}{3}+\cfrac{98}{99} \right ]=\left [ 1+\cfrac{65}{99} \right ]=1$$
So, the sum is
$$1+1+1+......... 66 $$ times
$$=66$$
If $$(1+ax)^{n }=1+8x+24x^{2}+...;$$ then $$\cfrac {a-n}{a+n}$$ is equal to
Report Question
0%
$$3$$
0%
$$\dfrac {-1}{3}$$
0%
$$-3$$
0%
$$\dfrac {1}{3}$$
Explanation
$$(1+ax)^{ n }=1+n\left( ax \right) +^{ n }{ C }_{ 2 }{ \left( ax \right) }^{ 2 }+^{ n }{ C }_{ 3 }{ \left( ax \right) }^{ 3 }....$$
According to question
$$(1+ax)^{ n }=1+8x+24x^{ 2 }+...$$
$$\Rightarrow an=8$$
$${ a }^{ 2 }=\cfrac { (n)(n-1) }{ 2 } =24$$
$$\Rightarrow { a }^{ 2 }\left( n \right) \left( n-1 \right) =48$$
$$\Rightarrow a\left( n-1 \right) =6$$
$$\Rightarrow \cfrac { n }{ n-1 } =\cfrac { 8 }{ 6 } =\cfrac { 4 }{ 3 } $$
$$\Rightarrow n=4,\Rightarrow a=2$$
$$\therefore \cfrac { a-n }{ a+n } =\cfrac { 2-4 }{ 2+4 } =\cfrac { -2 }{ 6 } =\cfrac { -1 }{ 3 } $$
If $$S={1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}....$$ upto $$n$$ terms and $$n$$ is even, then $$S$$ equals _____
Report Question
0%
$$\cfrac{n(n+1)}{2}$$
0%
$$\cfrac{n(n-1)}{2}$$
0%
$$\cfrac{-n(n+1)}{2}$$
0%
$$\cfrac{-n(n-1)}{2}$$
Explanation
$$S=1^{2}-2^{2}+3^{2}-4^{2}.........$$ upto $$n$$ term
$$=(1-2)(1+2)+(3-4)(3+4)+......+(n-1-n)(n-1+n)$$
$$=-(1+2+3+4+.....+n)$$
$$S=\dfrac{-n(n+1)}{2}$$
The sum $$\sum _{ i=0 }^{ m }{ \left( \begin{matrix} 10 \\ i \end{matrix} \right) \left( \begin{matrix} 20 \\ m-i \end{matrix} \right) }$$ (where $$\left( \begin{matrix} p \\ q \end{matrix} \right)=0$$ if $$p<q$$) is maximum where $$m$$ is
Report Question
0%
$$5$$
0%
$$10$$
0%
$$15$$
0%
$$20$$
The series $${1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+.....+{99}^{2}-{100}^{2}=$$ _____
Report Question
0%
$$-5050$$
0%
$$5050$$
0%
$$11000$$
0%
$$-11000$$
Explanation
The sum of nth term of the series is given as
$$S=(-1)^{n+1}\dfrac{n(n+1)}{2}$$
Here $$n=100$$
So,
$$S=(-1)^{100+1}\dfrac{100(100+1)}{2}$$
$$=(-1)\dfrac{10100}{2}$$
$$=-5050$$
If $$x$$ and $$y$$ are the number of possibilities that $$A$$ can assume such that the unit digit of A and $$A^3$$ are same and the unit digit of $$A^2$$ and $$A^3$$ are same respectively ,then the value of $$x-y$$ is (where $$A$$ is a single digit number)
Report Question
0%
$$4$$
0%
$$2$$
0%
$$3$$
0%
$$5$$
Explanation
Here, according to the question
if $$A=1$$, then $${ A }^{ 3 }={ 1 }$$,
We see that unit digit number of $$A$$ and $${ A }^{ 3 }$$ are same.
for, $$A=2\Rightarrow { A }^{ 3 }=8,$$, unit digits are not same
$$A=3\Rightarrow { A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 3 }=64,$$ unit digits are same
$$A=5\Rightarrow { A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 3 }=216,$$ unit digits are not same
$$A=7\Rightarrow { A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are same
So, there are five possible solutions where unit digits of $$A$$ and $${ A }^{ 3 }$$ are same.
$$\therefore x=5$$
Again, if we take $$A=1,{ A }^{ 2 }=1,{ A }^{ 3 }={ 1 }$$, unit digits are same
$$A=2\Rightarrow { A }^{ 2 }=4,{ A }^{ 3 }=8,$$ unit digits are not same
$$A=3\Rightarrow { A }^{ 2 }=9,{ A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 2 }=16,{ A }^{ 3 }=64,$$ unit digits are not same
$$A=5\Rightarrow { A }^{ 2 }=25,{ A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 2 }=36,{ A }^{ 3 }=216,$$ unit digits are same
$$A=7\Rightarrow { A }^{ 2 }=49,{ A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 2 }=64,{ A }^{ 3 }=512,$$ unit digits are not same
$$A=9\Rightarrow { A }^{ 2 }=81,{ A }^{ 3 }=729$$ unit digits are not same
Here, we can see that there are three possible solutions where unit digits of $${ A }^{ 2 },{ A }^{ 3 }$$ are same.
So, $$y=3$$
Hence, $$x-y=5-3=2.$$
$$\begin{bmatrix} 12 & 47 & 21 \\ 10 & 52 & 4 \\ 64 & ? & 24 \end{bmatrix}$$
Report Question
0%
$$40$$
0%
$$83$$
0%
$$62$$
0%
$$16$$
Explanation
$$\begin{bmatrix} 12 & 47 & 21\\ 10 & 52 & 4\\ 64 & ? & 24\end{bmatrix}$$
We can see that dividing third number in each row with second digit of middle number and multiplying with first digit gives first member.
Similarly, out of $$4$$ options, dividing $$24$$ by $$3$$ and multiplying by $$8$$ gives $$64$$.
$$\Rightarrow$$ Option $$(B)$$.
In a triangle $$ABC$$,
$$acos^2(\frac{C}{2})+c\;cos^2(\frac{A}{2})=\dfrac{3b}{2}$$, then the sides $$a,b,c$$
Report Question
0%
Satisfy $$a+b=c$$
0%
are in $$A.P.$$
0%
are in $$G.P.$$
0%
are in $$H.P.$$
Explanation
$$a\cos^{2}\left(\dfrac{C}{2}\right)+C\cos^{2}\left(\dfrac{A}{2}\right)=\dfrac{3b}{2}$$
$$a\left[\dfrac{1+\cos C}{2}\right]+C\left[\dfrac{1+\cos A}{2}\right]=\dfrac{3b}{2}$$
$$\dfrac{a}{2}+\dfrac{c}{2}+\dfrac{1}{2}\left[a\cos C+C \cos A\right]=\dfrac{3b}{2}$$
$$a+c+a\cos C+ c \cos A=3b$$
$$2R\sin A+2R\sin C+2R\sin A\cos C+2R\sin C\cos A =3(2R)(\sin B)$$
$$\sin A+\sin C+\sin (A+C)=3\sin (B)$$
$$A+B+C=\pi\Rightarrow \sin(A+C)=\sin B$$
$$\sin A+\sin C+\sin B=3\sin (B)$$
$$\sin A+\sin C=2\sin B$$
$$2R\sin A+2R\sin C=2R\sin B$$
$$a+c=2b$$
$$\therefore a,b,c$$ are an $$A.P$$
The value of $$\displaystyle \sum^{n-1}_{r=1}\sin^{2}\dfrac {r \pi}{n}$$ is equal to
Report Question
0%
$$n$$
0%
$$\dfrac {n}{2}$$
0%
$$n+1$$
0%
$$Zero$$
The sum of the series $$10.^{n}C_{0}+10^{2}.^{n}C_{1}+10^{3}.^{n}C_{2}+...10^{n+1}.^{n}C_{n}$$ is
Report Question
0%
$$11^{n}$$
0%
$$10.11^{n}$$
0%
$$11^{n+1}$$
0%
$$11^{n}-1$$
Explanation
$$ \rightarrow 10(^{n}C_{0}+10.^{n}C_{1}+...+10^{n}$$ $$^{n}C_{n})$$
$$ = 10\displaystyle(\sum_{r=1}^{n}$$ $$ ^{n}C_{r}(10)^{r}(1)^{n-r} )$$
$$ = 10(1+10)^{n} = 10.11^{n} = (B)$$
$$1 + \left( {1 + a} \right)x + 1\left( {1 + a + {a^2}} \right){x^2} + \left( {1 + a + {a^2} + {a^3}} \right){x^3} + - - - - - \;{\text{where}}\;0 < a,x < 1,\;{\text{is}} - $$
Report Question
0%
$$\frac{1}{{\left( {1 - x} \right)\left( {1 - a} \right)}}$$
0%
$$\frac{1}{{\left( {1 - x} \right)\left( {1 - ax} \right)}}$$
0%
$$\frac{1}{{\left( {1 - a} \right)\left( {1 - ax} \right)}}$$
0%
None of these
Explanation
We have,
$$\begin{matrix} 1+\left( { 1+a } \right) x+1\left( { 1+a+{ a^{ 2 } } } \right) { x^{ 2 } }+\left( { 1+a+{ a^{ 2 } }+{ a^{ 3 } } } \right) { x^{ 3 } }+.......... \\ \frac { 1 }{ { 1-a } } \left[ { \left( { 1+a } \right) +\left( { 1-{ a^{ 2 } } } \right) x+\left( { 1-{ a^{ 3 } } } \right) { x^{ 2 } }+\left( { 1-{ a^{ 4 } } } \right) { x^{ 3 } }.......... } \right] \\ \frac { 1 }{ { 1-a } } \left[ { 1+x+{ x^{ 2 } }+{ x^{ 3 } }+.........-\left( { a+{ a^{ 2 } }x+{ a^{ 3 } }{ x^{ 2 } }+{ a^{ 4 } }{ x^{ 3 } }+.... } \right) } \right] \\ \end{matrix}$$
$$\begin{matrix} \frac { 1 }{ { 1-a } } \left[ { \frac { 1 }{ { 1-x } } -\frac { a }{ { 1-ax } } } \right] \\ \frac { 1 }{ { 1-a } } \left[ { \frac { { 1-ax-a+ax } }{ { \left( { 1-x } \right) \left( { 1-ax } \right) } } } \right] \\ \frac { 1 }{ { \left( { 1-x } \right) \left( { 1-ax } \right) } } \\ \end{matrix}$$
Then,
Option $$A$$ is correct answer.
The sum of the series $$\displaystyle \sum_{r = 1}^{n} (-1)^{r - 1}.^{n}C_{r} (a - r)$$ is
Report Question
0%
$$a$$
0%
$$0$$
0%
$$n.2^{n - 1} + a$$
0%
None of these
Explanation
$$\begin{array}{l} \sum _{ r=1 }^{ n }{ { { \left( { -1 } \right) }^{ r-1 } } }\, { \, ^{ n } }{ c_{ r } }\left( { a-r } \right) & \\ Put\, \, n=1 & \\ { \Rightarrow ^{ n } }{ c_{ 1 } }\left( { a-1 } \right) { -^{ n } }{ c_{ 2 } }\left( { a-2 } \right) { +^{ n } }{ c_{ 3 } }\left( { a-1 } \right) +.......{ \left( { -1 } \right) ^{ n-1 } }\, { \, ^{ n } }{ c_{ n } }\left( { a-n } \right) & \\ \Rightarrow a\left[ { ^{ n }{ c_{ 1 } }{ -^{ n } }{ c_{ 2 } }{ +^{ n } }{ c_{ 3 } }{ -^{ n } }{ c_{ 4 } }{ +^{ n } }{ c_{ 5 } }+.......{ -^{ n } }{ c_{ n } } } \right] { -^{ n } }{ c_{ 1 } }+{ 2^{ n } }{ c_{ 2 } }-{ 3^{ n } }{ c_{ 3 } }+{ 4^{ n } }{ c_{ 4 } }......... & \\ =a\left[ { ^{ n }{ c_{ 1 } }{ -^{ n } }{ c_{ 2 } }{ +^{ n } }{ c_{ 3 } }{ -^{ n } }{ c_{ 4 } }{ +^{ n } }{ c_{ 5 } }+...... } \right] -\left[ { ^{ n }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }+{ 4^{ n } }{ c_{ 4 } }+....... } \right] & \\ { \left( { 1-x } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ -^{ n } }{ c_{ 1 } }x{ +^{ n } }{ c_{ 2 } }{ x^{ 2 } }{ -^{ n } }{ c_{ 3 } }{ x^{ 3 } }....... & \left[ { According\, \, to\, \, this\, \, formula } \right] \\ \Rightarrow Add\, \, \& \, \, subtract\, { \, ^{ n } }{ c_{ 0 } }\, \, in\, a\, \, have\, \, equation\, \, \& \, \, taking\, \, out\, \, common\, \, \left( { -1 } \right) \, \, also. & \\ \Rightarrow a\left[ { -\left( { ^{ n }{ c_{ 0 } }{ -^{ n } }{ c_{ 1 } }{ +^{ n } }{ c_{ 2 } }{ -^{ n } }{ c_{ 3 } }+...... } \right) { +^{ n } }{ c_{ o } } } \right] -\left[ { ^{ n }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }-{ 4^{ n } }{ c_{ 4 } } } \right] & \\ Put\, \, x=1 & \\ { \left( { 1-1 } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ -^{ n } }{ c_{ 1 } }{ +^{ n } }{ c_{ 2 } }{ -^{ n } }{ c_{ 3 } }=0 & \\ So, & \\ a\left[ { -0{ +^{ n } }{ c_{ 0 } } } \right] -\left[ { ^{ n }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }-{ 4^{ n } }{ c_{ 4 } }+......... } \right] \to \left( i \right) & \\ { \left( { 1+n } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ +^{ n } }{ c_{ 1 } }x{ +^{ n } }{ c_{ 2 } }{ x^{ 2 } }{ +^{ n } }{ c_{ 3 } }{ x^{ 3 } }{ +^{ n } }{ c_{ 4 } }{ x^{ 4 } }+....... & \left[ { \leftarrow formula } \right] \\ Diff.\, \, w.r.t.\, \, n,\, we\, \, get & \\ n{ \left( { 1+x } \right) ^{ n-1 } }{ =^{ n } }{ c_{ 1 } }+2{ x^{ n } }{ c_{ 2 } }+3{ x^{ n } }{ c_{ 3 } }+4{ x^{ n } }{ c_{ 4 } }+.... & \\ Put\, \, x=-1\, \, So, & \\ x{ \left( { 1-x } \right) ^{ n-1 } }{ =^{ n } }{ c_{ 1 } }-2{ x^{ n } }{ c_{ 2 } }+3{ x^{ 2 } }^{ n }{ c_{ 3 } }+{ 4^{ n } }{ c_{ 4 } }+......... & \\ 0{ =^{ n } }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }-{ 4^{ n } }{ c_{ 4 } } & \end{array}$$
So, put the value in equation in (1)
$$\begin{array}{l} =a\left[ { ^{ n }{ c_{ 0 } } } \right] -\left[ 0 \right] \Rightarrow a\left( 1 \right) =a \\ \sum _{ r=1 }^{ n }{ { { \left( { -1 } \right) }^{ r-1 } }\, { \, ^{ n } }{ c_{ r } }\left( { a-r } \right) =a } \end{array}$$
$$1.3.4+2.5.8+3.6.9+$$ upto $$n$$ terms is equal to ________ .
Report Question
0%
$$\dfrac {n(n+1)(n+2)}{6}$$
0%
$$\dfrac {n(n+1)(3n^{2}+23n+46)}{12}$$
0%
$$\dfrac {n(27n^{3}+90n^{2}+45n-5)}{4}$$
0%
$$\dfrac {n(n+1)(2n+1)}{6}$$
0%
$$None\ of\ these$$
The sum of the series
$$1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+...+1000.2^{999}$$ is
Report Question
0%
$$999.2^{999}-1$$
0%
$$999.2^{1000}-1$$
0%
$$999.2^{1000}+1$$
0%
$$999.2^{999}+1$$
Explanation
$$ S = 1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+..... +1000.2^{999}$$ __ (1)
Multiply with '2' on both sides
$$ 2S = 1.2+2.2^{2}+3.2^{2}+4.2^{4}+5.2^{5}+... +999.2^{999}+1000.2^{1000}$$ __ (2)
subtracting equation (1) & (2)
$$ (S-2S) = [ 1+(2.2-1.2)+(3.2^{2}-2.2^{2})+(4.2^{3}-3.2^{3})+(5.2^{4}-4.2^{4})+$$ ...
$$+(1000.2^{999}-999.2^{999})-1000.2^{1000}$$
$$(S-2S) = 1+1.2+1.2^{2}+1.2^{3}+1.2^{4}+.... +1.2^{999}-1000.2^{1000}$$
$$-S = 1+(2+2^{2}+2^{3}+2^{4}+...+2^{999})-1000.2^{1000}$$
$$ \left [ \because a+ar+ar^{2}+...+ar^{n-1} = \dfrac{a.(r^{n}-1)}{r-1} \right ]$$
$$ a = 2, r = 2, n = 999 $$
$$ -S= \dfrac{1+2.(2^{999}-1)}{2-1} - 1000.2^{1000}$$
$$ -S = 1+2^{1000}-2-1000.2^{1000}$$
$$ -S = -1+2^{1000}(1-1000)$$
$$ -S = -1-999.2^{1000}$$
$$ S = 999.2^{1000}+1$$
$$\displaystyle\sum^n_{r=1}\displaystyle\sum^{r-1}_{p=0}$$ $$^{n}C_r\cdot {^{r}C_p}\cdot 2^p$$ is equal to?
Report Question
0%
$$4^n-3^n+1$$
0%
$$4^n-3^n-1$$
0%
$$4^n-3^n+2$$
0%
$$4^n-3^n$$
If $$| x| < 1$$ , then the sum of series $$ 1+ 2x + 3x^2 + 4x^3 + .........\infty$$ will be
Report Question
0%
$$\dfrac{1}{1-x}$$
0%
$$\dfrac{1}{1+x}$$
0%
$$\dfrac{1}{(1+x)^2}$$
0%
$$\dfrac{1}{(1-x)^2}$$
Explanation
Let $$S=1+2x+3{x}^{2}+......\infty$$ $$\dots(1)$$
multiply by $$x$$ on both sides
$${x}{S}=x+2{x}^{2}+3{x}^{3}+.....\infty$$ $$\dots(2)$$
Subtract $$(1)$$ from $$(2)$$
$$S=1+2x+3{x}^{2}+......\infty$$
$${x}{S}=x+2{x}^{2}+3{x}^{3}+.....\infty$$
- - - -
--------------------------------------------------------
$$S(1-x)=1+x+{x}^{2}+{x}^{3}+.....\infty$$
$$S(1-x)=\cfrac{1}{1-x}$$ [sum of infinte G.P $$S_{\infty}=\dfrac{1}{1-r}]$$
$$S=\cfrac{1}{(1-x)^{2}}$$
$$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+..n$$ terms $$=$$
Report Question
0%
$$\dfrac{3n}{2(3n+2)}$$
0%
$$\dfrac{3n}{3n+2}$$
0%
$$\dfrac{n}{2(3n+2)}$$
0%
$$\dfrac{n}{3n+2}$$
Explanation
$$\cfrac { 1 }{ 2.5 } +\cfrac { 1 }{ 5.8 } +\cfrac { 1 }{ 8.11 } +.....+\cfrac { 1 }{ \left[ 2+3\left( n-1 \right) \right] .\left[ 5+3\left( n-1 \right) \right] } $$
$$=\sum _{ n=1 }^{ n }{ \cfrac { 1 }{ \left[ 2+3n-3 \right] .\left[ 5+3n-3 \right] } } $$
$$=\sum _{ n=1 }^{ n }{ \cfrac { 1 }{ \left( 3n-1 \right) .\left( 3n+2 \right) } } =\cfrac { 1 }{ 3 } \sum { \cfrac { \left[ 3n+2-\left( 3n-1 \right) \right] }{ \left( 3n-1 \right) \left( 3n+2 \right) } } $$
$$=\cfrac { 1 }{ 3 } \sum { \cfrac { 1 }{ 3n-1 } -\cfrac { 1 }{ 3n+2 } } $$
$$=\cfrac { 1 }{ 3 } \left[ \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 8 } -\cfrac { 1 }{ 11 } +...+\cfrac { 1 }{ 3n-4 } -\cfrac { 1 }{ 3n-1 } +\cfrac { 1 }{ 3n-1 } -\cfrac { 1 }{ 3n+2 } \right] $$
$$=\cfrac { 1 }{ 3 } \left[ \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 3n+2 } \right] =\cfrac { 3n }{ 3\left( 3n+2 \right) .2 } =\cfrac { n }{ 2\left( 3n+2 \right) } $$
The sum of the series $${5 \over {13}} + {{55} \over {{{13}^2}}} + {{555} \over {{{13}^3}}} + .........\infty $$ is
Report Question
0%
$${65\over {36}}$$
0%
$${65\over {32}}$$
0%
$${25\over {36}}$$
0%
none of these
Explanation
$$S=\dfrac{5}{13}+\dfrac{55}{13^2}+\dfrac{555}{13^3}+\dfrac{5555}{13^4}+.....$$ $$(1)$$
Now multiply by $$\dfrac{1}{13}$$, we get
$$\dfrac{1}{13}s=\dfrac{5}{13^2}+\dfrac{55}{13^3}+\dfrac{555}{13^4}+.....$$ $$(2)$$
Subtracting $$(2)$$ from $$(1)$$
$$s-\dfrac{1}{13}s\left[\dfrac{5}{13}+\dfrac{55}{13^2}+\dfrac{555}{13^3}+.....\right]-\left[\dfrac{5}{13^2}+\dfrac{55}{13^3}+..…\right]$$
$$\dfrac{12s}{13}=\dfrac{5}{13}+\left(\dfrac{55}{13^2}+\dfrac{5}{13^2}\right)+\left(\dfrac{555}{13^3}+\dfrac{55}{13^3}\right)+.....$$
$$\dfrac{12s}{13}=\dfrac{5}{13}+\dfrac{50}{13^2}+\dfrac{500}{13^3}+....$$
$$=\dfrac{\dfrac{5}{13}}{1-\dfrac{10}{13}}$$
$$\dfrac{12s}{13}=\dfrac{5}{3}$$
$$\Rightarrow s=\dfrac{65}{36}$$.
Sum to n terms the following series :
Report Question
0%
5 + 11 + 19 + 29 + 41 + .......
0%
3 + 7 + 14 + 24 + 37 +......
0%
6 + 9 + 16 + 27 + 42 + .....
0%
5 + 7 + 13 + 31 + 85 + ....
$$\sum _{ k=1 }^{ 2n+1 }{ (-1)^{k-1} }k^2=$$
Report Question
0%
$$(n+1)(2n+1)$$
0%
$$(n+1)(2n-1)$$
0%
$$(n-1)(2n-1)$$
0%
$$(n-1)(2n+1)$$
Explanation
$$\displaystyle \sum_{k = 1}^{2n + 1} (-1)^{k -1} \, k^2$$
$$= 1^2 - 2^2 + 3^2 - 4^2 + 5^2 ...... - (2n)^2 + (2n + 1)^2$$
$$= 1^2 + 3^2 + 5^2 + ..... + (2n + 1)^2 - [2^2 + 4^2 + ... + (2n)^2]$$
$$= 1^2 + 2^2 + 3^2 + ..... + (2n + 1)^2 - 2 [2^2 + 4^2 + .... (2n)^2]$$
Adding and subtract even terms to complete the series.
Using identify $$1^2 + 2^2 + .... + n^2 = \dfrac{n (n + 1) (2n + 1)}{6}$$
The addition will become
$$= 1^2 + 2^2 + 3^2 + .... + (2n + 1) - 2(2^2) [1^2 + 2^2 + ..... + n^2]$$
$$= \dfrac{(2n + 1)(2n + 2)(4n + 3)}{6} - \dfrac{8 n (n + 1)(2n + 1)}{6}$$
$$= \dfrac{2(2n + 1)(n + 1)(4n + 3)}{6} - \dfrac{8 n(n + 1)(2n + 1)}{6}$$
$$= \dfrac{2}{6} (n + 1)(2n + 1) [4n + 3 - 4n]$$
$$= \dfrac{1}{3} (n + 1) (2n + 1) (3)$$
$$= (n + 1) (2n + 1)$$
If $$S = 1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}}......,$$ then
Report Question
0%
$${S_n} = \frac{{2n}}{{n + 1}}$$
0%
$${S_n} = \frac{{2n}}{{n - 1}}$$
0%
$${S_\infty } = 2$$
0%
$${S_\infty } = 1$$
Explanation
$$S_{n} = \dfrac {1}{1} + \dfrac {1}{1 + 2} + \dfrac {1}{1 + 2 + 3} + \dfrac {1}{1 + 2 + 3 + 4} + ......$$
$$T_{n} = \dfrac {1}{1 + 2 + 3 + ...n} = \dfrac {1}{n\dfrac {(n + 1)}{2}}$$
$$T_{n} =\dfrac {2}{n(n + 1)} = 2\left [\dfrac {(n + 1) - n}{n(n + 1)}\right ]$$
$$= 2\left [\dfrac {1}{n} - \dfrac {1}{n + 1}\right ]$$
$$T_{1} = 2\left [\dfrac {1}{1} - \dfrac {1}{2}\right ]$$
$$T_{2} = 2\left [\dfrac {1}{2} - \dfrac {1}{3}\right ]$$
$$.$$
$$.$$
$$.$$
$$.$$
$$T_{n} = 2\left [\dfrac {1}{n} - \dfrac {1}{n + 1}\right ]$$
Add the above terms
$$S_{n} = 2\left [1 - \dfrac {1}{n + 1}\right ] = 2\left [\dfrac {n + 1 - 1}{n + 1}\right ]$$
$$= \dfrac {2n}{n + 1}$$
$$S_{\infty} = \displaystyle \lim_{n\rightarrow 0} \dfrac {2n}{n + 1} = \displaystyle \lim_{n\rightarrow \infty} = \dfrac {2}{1 + \dfrac {1}{n}}$$
$$S_{\infty} = 2$$.
If $$S = \tan ^ { - 1 } \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { n ^ { 2 } + 3 n + 3 } \right) + \ldots + \tan ^ { - 1 } \left( \frac { 1 } { 1 + ( n + 19 ) ( n + 20 ) } \right)$$ then $$\tan S$$ is equal to
Report Question
0%
$$\frac { 20 } { 401 + 20 n }$$
0%
$$\frac { n } { n ^ { 2 } + 20 n + 1 }$$
0%
$$\frac { 20 } { n ^ { 2 } + 20 n + 1 }$$
0%
$$\frac { n } { 401 + 20 n }$$
Explanation
$$\begin{array}{l} S={ \tan ^{ -1 } }\left( { \frac { 1 }{ { { n^{ 2 } }+n+1 } } } \right) +{ \tan ^{ -1 } }\left( { \frac { 1 }{ { { n^{ 2 } }+3n+3 } } } \right) +....+{ \tan ^{ -1 } }\left( { \frac { 1 }{ { 1+\left( { n+19 } \right) \left( { n+20 } \right) } } } \right) \\ S={ \tan ^{ -1 } }\left( { \frac { { n+1-n } }{ { 1+n\left( { n+1 } \right) } } } \right) +{ \tan ^{ -1 } }\left[ { \frac { { \left( { n+2 } \right) -\left( { n-1 } \right) } }{ { 1+\left( { n+1 } \right) \left( { n+2 } \right) } } } \right] +.....+{ \tan ^{ -1 } }\left[ { \frac { { \left( { n+20 } \right) -\left( { n+19 } \right) } }{ { 1+\left( { n+19 } \right) \left( { n+20 } \right) } } } \right] \\ =\left[ { { { \tan }^{ -1 } }\left( { n+1 } \right) -{ { \tan }^{ -1 } }\left( n \right) } \right] +\left[ { { { \tan }^{ -1 } }\left( { n+2 } \right) -{ { \tan }^{ -1 } }\left( { n+1 } \right) } \right] +......+\left[ { { { \tan }^{ -1 } }\left( { n+20 } \right) -{ { \tan }^{ -1 } }\left( { n+19 } \right) } \right] \\ ={ \tan ^{ -1 } }\left( { n+20 } \right) -ta{ n^{ -1 } }\left( n \right) \\ ={ \tan ^{ -1 } }\left( { \frac { { 20 } }{ { 1+{ n^{ 2 } }+20n } } } \right) \\ { { tans } }=\frac { { 20 } }{ { { n^{ 2 } }+20n+1 } } \\ Hence,\, the\, option\, C\, is\, correct\, answer. \end{array}$$
7, 11, 23, 51, 103 ?
Report Question
0%
186
0%
188
0%
185
0%
187
0%
none of these
Explanation
$$7,11,23,51,103,?$$
$$7+4\times 1= 11 $$
$$11+4\times 3=23$$ $$\leftarrow $$ difference of 2
$$23+4\times 7=51$$ $$\leftarrow $$ difference of 4
$$51+4\times 13= 103$$ $$\leftarrow $$ difference of 6
$$103+4\times 21= 187$$ $$\leftarrow $$ difference will be of 8
The value of the expression $$\sum _{ r=0 }^{ n }{ { (-1) }^{ r } } \left( \dfrac { ^nC_r }{^{r+3}C_r } \right) $$ is
Report Question
0%
$$\dfrac{n(n+1)}{2}$$
0%
$$\dfrac{n+3}{3}$$
0%
$$\dfrac{3}{n+3}$$
0%
$$\dfrac{n+2}{2}$$
$$S=\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right)+\tan^{-1}\left(\dfrac{1}{n^2+3n+3}\right)+.....+\tan^{-1}\left(\dfrac{1}{1+(n+19)(n+20)}\right)$$, then $$\tan S$$ is equal to?
Report Question
0%
$$\dfrac{20}{401+20n}$$
0%
$$\dfrac{n}{n^2+20n+1}$$
0%
$$\dfrac{20}{n^2+20n+1}$$
0%
$$\dfrac{n}{401+20n}$$
Explanation
$$ S = \sum_{r=0}^{19} tan^{-1} \left ( \frac{1}{1+(n+r)(n+r+1)} \right ) = tan^{-1}\left ( \frac{(n+r+1)-(n+r)}{1+(n+r)(n+r+1)} \right )$$
$$ \Rightarrow S = \sum_{r=0}^{19} \left ( tan^{-1}(n+r+1)-tan^{-1}(n+r) \right )$$
$$ =tan^{-1}(n+1) - tan^{-1}(n) = tan^{-1}(n+20)-tan^{-1}(n+19) $$
$$ + tan^{-1}(n+2) - tan^{-1}(n+1)$$
$$ tan^{-1}(n+20) - tan^{-1}(n+19)$$
$$ \Rightarrow tan \, S = \frac{(n+20)-(n)}{1+n(n+20)} = \frac{20}{n^{2}+20n+1}$$
$$\Rightarrow (C)$$
13, 16, 22, 33, 51 ?
Report Question
0%
89
0%
78
0%
102
0%
69
0%
none of these
Explanation
REF.Image
$$ 13,16,22,33,51, ?$$
Evaluate:-
If $$\sum\limits_{r - 0}^n {{{\left\{ {\frac{{^n{C_{r - 1}}}}{{^n{C_r}{ + ^n}{C_{r - 1}}}}} \right\}}^3} = \frac{{25}}{{24}}} $$
Report Question
0%
$$3$$
0%
$$6$$
0%
$$4$$
0%
$$5$$
655, 439, 314, 250, 223 ?
Report Question
0%
215
0%
210
0%
195
0%
190
0%
none of these
Explanation
Solution $$\rightarrow $$ Given terms are 655,439,314,250,223,?
As $$a_{1}=655$$
$$a_{2}= 439$$
$$a_{3}=314$$
$$a_{4}= 250$$
$$a_{5}= 223$$
$$a_{6}=?$$
So, $$a_{1}-a_{2}=655- 439= 216=6^{3}$$
$$a_{3}-a_{2}= 439- 314= 125=5^{3}$$
$$a_{3}-a_{4}= -250+314= 64=4^{3}$$
$$a_{4}-a_{5}=-223+250= 27=3^{3}$$
So, we can observe difference b/w terms are cubes
of 6,5,4,3 so, is similar way
$$a_{5}-a_{6}=2^{3}$$
$$223-a_{6}= 8$$
$$a_{6} = 215$$
So, we find next term is 215
If the expansion of $$\left( x+a \right) ^{ n }$$ if the sum of odd terms be P & sum of even terms be Q, prove that
Report Question
0%
$${ P }^{ 2 }-{ Q }^{ 2 }=({ x }^{ 2 }-{ a }^{ 2 })^{ n }$$
0%
$$4PQ=(x+a)^{ 2n }-(x-a)^{ 2n }$$
0%
$$P^2-Q^2=(x^2+a^2)^{n}$$
0%
$$None\ of\ these$$
Sum of the series
$$S=1^{2}-2^{2}+3^{2}-4^{2}+..... -2000^{2}+2003^{2}$$ is
Report Question
0%
$$2007006$$
0%
$$1005004$$
0%
$$2000506$$
0%
$$None$$
Find the next term
$$210,209,205,196,180,?$$
Report Question
0%
$$138$$
0%
$$77$$
0%
$$155$$
0%
$$327$$
Explanation
$$T_1=210$$
$$,T_2=209,$$
$$T_3=205,$$
$$T_4=196,$$
$$T_5=180,$$
$$T_6=?$$
$$T_1-T_2=1^2\\$$
$$T_2-T_3=2^2\\$$
$$T_3-T_4=3^2\\$$
$$T_4-T_5=4^2\\$$
$$T_5-T_6=5^2\\$$
$$\implies T_6=T_5 - 5^2=180-25=155$$
Determine the next term $$20, 24, 33, 49, 74, 110, ?$$
Report Question
0%
$$133$$
0%
$$147$$
0%
$$159$$
0%
$$163$$
0%
$$171$$
Explanation
$$20,24,33,49,74,110,...$$
Let the number be $$x$$
difference $$24-20=4,33-24=9,49-33=16,74-49=25,110-74=36,x-110=49$$
$$\implies24-20=2^2, 33-24= 3^2, 49-33= 4^2, \,\, 74-49= 5^2\,\, 110-74=6^2,\,\,\, x-110= 7^2$$
So, $$110+49=159$$
ANSWER IS C
Find the next term of the series $$1728, 2744, 4096, 5832, 8000, 10648, ?$$
Report Question
0%
$$2167$$
0%
$$13824$$
0%
$$15625$$
0%
$$9261$$
0%
$$17576$$
Explanation
$$1728,2744,4096,5832,8000,10648.....$$
$$12^3\rightarrow 1728$$
$$14^3\rightarrow 2744$$
$$16^3\rightarrow 4096$$
$$18^3\rightarrow 5832$$
$$20^3\rightarrow 8000$$
$$22^3\rightarrow 10648$$
$$24^3\rightarrow 13824$$
ANSWER IS B
$$4,6,12,30,90,315,?$$
Report Question
0%
$$945$$
0%
$$1102$$
0%
$$1260$$
0%
$$1417.5$$
0%
$$None\ of\ these$$
Explanation
$$=315\times 4$$
$$=1260.$$
Hence, the answer is $$1260.$$
Find next term
$$462,552,650,756,870,992,?$$
Report Question
0%
$$1040$$
0%
$$1122$$
0%
$$1132$$
0%
$$105$$
Find the missing number from the given alternatives.
Report Question
0%
$$39,116$$
0%
$$52,156$$
0%
$$30,117$$
0%
$$31,116$$
Explanation
$$ 13\quad\quad 39\\26\quad\quad 78\\?\quad\quad ?$$
The pattern satisfies
$$x\quad\quad 3x$$
Therefore correct option is
$$52\quad\quad 156$$
$$8, 31, 122, 485, 1936, 7739, ?$$
Report Question
0%
$$30950$$
0%
$$46430$$
0%
$$34650$$
0%
$$42850$$
0%
$$38540$$
The sum of infinite series $$\begin{vmatrix} 1 & 2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 2 } & 2 \\ 2 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 4 } & 2 \\ \frac { 2 }{ 3 } & 4 \end{vmatrix}+.........$$
Report Question
0%
$$-10$$
0%
$$0$$
0%
$$10$$
0%
$$\infty$$
If $$\dfrac{\pi}{4}-1+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{13}+=0$$ then value of $$\dfrac{1}{1\times3}+\dfrac{1}{5\times7}+\dfrac{1}{9\times11}+\dfrac{1}{13\times15}+..$$ is
Report Question
0%
$$\dfrac{\pi}{8}$$
0%
$$\dfrac{\pi}{6}$$
0%
$$\dfrac{\pi}{4}$$
0%
$$\dfrac{\pi}{34}$$
Let $$t_{r}=\frac{r}{1+r^{2}+r^{4}}$$ then, $$\lim_{n\rightarrow \infty }\sum_{r=1}^{n}t_{r}$$ equals
Report Question
0%
$$\frac{1}{4}$$
0%
1
0%
$$\frac{1}{2}$$
0%
None of these
Explanation
$$t_r=\dfrac{r}{1+r^2+r^4}$$
$$=\dfrac{r}{r^4+2r^2+1-r^2}$$
$$=\dfrac{r}{(r^2+1)^2-r^2}=\dfrac{r}{(r^2+1-r)(r^2+1+r)}$$
$$=\dfrac{1}{2}\dfrac{2r}{(r^2+1-r)(r^2+r+1)}$$
$$=\dfrac{1}{2}\dfrac{(r^2+r+1)-(r^2-r+1)}{(r^2-r+1)(r^2+r+1)}$$
$$=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}\right]$$
$$=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{(r+1)^2-(r+1)+1}\right]$$
$$\displaystyle\sum^n_{r=1}t_r=\dfrac{1}{2}\left[\dfrac{1}{1-1+1}-\dfrac{1}{(n+1)^2-(n+1)+1}\right]$$
$$=\dfrac{1}{2}\left[1-\dfrac{1}{n^2+n+1}\right]=\dfrac{1}{2}\left[\dfrac{n^2+n+1-1}{n^2+n+1}\right]$$
$$\displaystyle\sum t_n=\dfrac{1}{2}\left[\dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\right]$$
$$\underset{n\rightarrow \infty}{lt}\displaystyle\sum t_r=\dfrac{1}{2}\left[\dfrac{1+0}{1+0+0}\right]$$
$$=\dfrac{1}{2}$$.
The sum to $$n$$ terms of the series
$$\dfrac {3}{1^{2}}+\dfrac {5}{1^{2}+2^{2}}+\dfrac {7}{1^{2}+2^{2}+3^{2}}+.........$$ is
Report Question
0%
$$\dfrac {6n}{n+1}$$
0%
$$\dfrac {9n}{n+1}$$
0%
$$\dfrac {12n}{n+1}$$
0%
$$\dfrac {3n}{n+1}$$
Explanation
$$\begin{array}{l} Let \\ { T_{ r } }=\dfrac { { 2r+1 } }{ { { 1^{ 2 } }+{ 2^{ 2 } }+{ r^{ 2 } } } } \\ =\dfrac { { \left( { 2r+1 } \right) } }{ { r\left( { r+1 } \right) \left( { 2r+1 } \right) } } \\ =\dfrac { 6 }{ { r\left( { 2+1 } \right) } } \\ \sum { { T_{ r } } } =\sum { \dfrac { 6 }{ { r\left( { r+1 } \right) } } } \\ =6\sum _{ r=1 }^{ n }{ \dfrac { { \left( { r+1 } \right) -r } }{ { r\left( { r+1 } \right) } } } \\ =6\sum _{ r=1 }^{ n }{ \left[ { \dfrac { 1 }{ r } -\dfrac { 1 }{ { r+1 } } } \right] } \\ =6\left[ { \left( { 1-\dfrac { 1 }{ 2 } } \right) +\left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) +.......+\left( { \dfrac { 1 }{ n } -\dfrac { 1 }{ { n+1 } } } \right) } \right] \\ =6\left[ { 1-\dfrac { 1 }{ { n+1 } } } \right] \\ =\dfrac { { 6n } }{ { n+1 } } \\ Hence,\, option\, A\, is\, \, the\, correct\, answer. \end{array}$$
$$499, 622, 868, 1237, 1729, 2344, ?$$
Report Question
0%
$$3205$$
0%
$$3082$$
0%
$$2959$$
0%
$$3462$$
0%
$$2876$$
Explanation
$$499, 622, 868, 1237, 1729, 2344,$$ ______
$$499+1\times 123=499+123=622$$
$$622+2\times 123=622+146=868$$
$$868+3\times 123=868+369=1237$$
$$1237+4\times 123=1237+492=1729$$
$$1729+5\times 123=1729+615=2344$$
$$2344+6\times 123=2344+738=3082$$.
The sum $$\dfrac{1}{1+1^{2}+1^{4}}+\dfrac{2}{1+2^{2}+2^{4}}+\dfrac{3}{1+3^{2}+3^{4}}+...+\dfrac{99}{1+99^{2}+99^{4}}$$ lies between
Report Question
0%
$$0.46$$ and $$0.47$$
0%
$$0.52$$ and $$1.0$$
0%
$$0.48$$ and $$0.49$$
0%
$$0.49$$ and $$0.50$$
Explanation
We have,
$$\dfrac { 1 }{ { 1+{ 1^{ 2 } }+{ 1^{ 4 } } } } +\dfrac { 2 }{ { 1+{ 2^{ 2 } }+{ 2^{ 4 } } } } +\dfrac { 3 }{ { { 1^{ 3 } }+{ 2^{ 2 } }+{ 3^{ 4 } } } } +.........+\dfrac { { 99 } }{ { 1+{ { 99 }^{ 2 } }+{ { 99 }^{ 4 } } } } \\ =\sum _{ n=1 }^{ 99 }{ \dfrac { x }{ { 1+{ x^{ 2 } }+{ x^{ 4 } } } } } \\ =\sum _{ n=1 }^{ 99 }{ \dfrac { x }{ { { { \left( { { x^{ 2 } }+1 } \right) }^{ 2 } }-{ x^{ 2 } } } } } \\ =\sum _{ n=1 }^{ 99 }{ \dfrac { { \left( { { x^{ 2 } }+x+1 } \right) -\left( { { x^{ 2 } }-x+1 } \right) } }{ { 2\left( { { x^{ 2 } }+x+1 } \right) \left( { { x^{ 2 } }-x+1 } \right) } } } \\ =\dfrac { 1 }{ 2 } \sum _{ n=1 }^{ 99 }{ \left( { \dfrac { 1 }{ { { x^{ 2 } }-x+1 } } -\dfrac { 1 }{ { { x^{ 2 } }+x+1 } } } \right) } \\ =\dfrac { 1 }{ 2 } \left( { \dfrac { 1 }{ 1 } -\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } -\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 7 } +.........+\dfrac { 1 }{ { { { 99 }^{ 2 } }+99+1 } } } \right) \\ =\dfrac { 1 }{ 2 } \left( { 1-\dfrac { 1 }{ { 9901 } } } \right) \\ =\dfrac { { 4950 } }{ { 9901 } } \\ \approx \dfrac { 1 }{ 2 } $$
$$ \\ Hence,\, option\, D\, is\, correct\, answer.$$
Find : $$12,15,21,24,30,33 , ? , ?$$
Report Question
0%
$$39,42$$
0%
$$37,42$$
0%
$$38,47$$
0%
$$39,51$$
Explanation
$$12,15,21,24,30,33,?,?$$
$$12$$
$$12 + 3 = 15$$
$$15+3\times 2=21$$
$$21+3=24$$
$$24+3\times 2=30$$
$$30+3=33$$
$$33+3\times 2=39$$
$$39+3=42$$
Find the missing number :
Report Question
0%
$$125$$
0%
$$90$$
0%
$$105$$
0%
$$225$$
Explanation
$$\textbf{Step 1: Observe the numbers opposite to each other}$$
$$ \text{We can see that the number opposite to 315 is 21. Also, } \dfrac{315}{21}=15$$
$$\text{Hence, we can say the missing number has at least one factor of 15 }$$
$$\text{We see the missing number is opposite to 15}$$
$$\text{Also, } 15\times15=225$$
$$\textbf{Step 2: Verification}$$
$$\text{Let the missing number be x.}$$
$$\text{ATQ, }$$
$$\Rightarrow 21+15+x=261$$
$$\Rightarrow x=261-21-15$$
$$x=225$$
$$\text{Hence, it is verified that x=225}$$
$$\textbf{Thus, the missing number is D 225.}$$
$$23,29,47,75 , ?$$
Report Question
0%
$$87$$
0%
$$93$$
0%
$$110$$
0%
$$117$$
Explanation
$$2\underbrace { 3\quad ,\quad 2 } \underbrace { 9\quad ,\quad 4 } \underbrace { 7\quad ,\quad 7 } \underbrace { 5\quad ,\quad ? } \quad =75+35$$
$$6$$ $$18$$ $$28$$ $$35$$
$$\left( 2\times 3 \right) \uparrow $$ $$\left( 2\times 9 \right) \uparrow $$ $$\left( 4\times 7 \right) \uparrow $$ $$\left( 7\times 5 \right) \uparrow $$
$$=110$$
The value of $$\dfrac{1}{6.10}+\dfrac{1}{10.14}+\dfrac{1}{14.18}+....\infty$$ equals to
Report Question
0%
$$\dfrac{1}{(24)^2}$$
0%
$$\dfrac{1}{6}$$
0%
$$\dfrac{1}{24}$$
0%
$$\dfrac{1}{(24)^3}$$
Explanation
$$S=\frac{1}{6 \cdot 10}, \frac{1}{10 \cdot 14}+\frac{1}{1419}+.....\infty$$
$$S=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{10}\right)+\frac{1}{4}\left(\frac{1}{10}-\frac{1}{14}\right)+\frac{1}{4}\left(\frac{1}{14}-\frac{1}{18}\right)+.....\infty$$
Now take $$\frac{1}{4}$$ common
$$S=\frac{1}{4}\left[\frac{1}{6}-\frac{1}{16}+\frac{1}{10}-\frac{1}{1 / 4}+\frac{1}{14}-\frac{1}{18}+1,-\infty\right]$$
$$S=\frac{1}{4} \times \frac{1}{6}=\frac{1}{24}$$
option $$c$$ is correct
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Engineering Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
