If $$S = 1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}}......,$$ then

$${S_n} = \frac{{2n}}{{n + 1}}$$

$${S_n} = \frac{{2n}}{{n - 1}}$$

MCQ Questions for 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

0%
$-ln 3$
0%
$0$
0%
$-ln 2$
0%
$-ln 5$

solve that :-

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

0%
AP
0%
GP
0%
HP
0%
None of these

$[x]$ denotes the greates integer

$\le x$ , then

$\left[\dfrac{2}{3} \right] + \left[\dfrac{2}{3} + \dfrac{1}{99} \right] + \left[\dfrac{2}{3} + \dfrac{2}{99} \right] + .... + \left[\dfrac{2}{3} + \dfrac{98}{99} \right] =$

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

$=66$

$(1+ax)^{n }=1+8x+24x^{2}+...;$ then

$\cfrac {a-n}{a+n}$ is equal to

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

$(1+ax)^{ n }=1+8x+24x^{ 2 }+...$

$\Rightarrow an=8$

${ a }^{ 2 }=\cfrac { (n)(n-1) }{ 2 } =24$

$\Rightarrow { a }^{ 2 }\left( n \right) \left( n-1 \right) =48$

$\Rightarrow a\left( n-1 \right) =6$

$\Rightarrow \cfrac { n }{ n-1 } =\cfrac { 8 }{ 6 } =\cfrac { 4 }{ 3 }$

$\Rightarrow n=4,\Rightarrow a=2$

$\therefore \cfrac { a-n }{ a+n } =\cfrac { 2-4 }{ 2+4 } =\cfrac { -2 }{ 6 } =\cfrac { -1 }{ 3 }$

$S={1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}....$ upto

$n$ terms and

$n$ is even, then

$S$ equals _____

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

0%
$5$
0%
$10$
0%
$15$
0%
$20$

The series

${1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+.....+{99}^{2}-{100}^{2}=$ _____

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,

$S=(-1)^{100+1}\dfrac{100(100+1)}{2}$

$=(-1)\dfrac{10100}{2}$

$=-5050$

$x$ 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)

0%
$4$
0%
$2$
0%
$3$
0%
$5$

Explanation

Here, according to the question

$A=1$ , 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,

$x-y=5-3=2.$

$\begin{bmatrix} 12 & 47 & 21 \\ 10 & 52 & 4 \\ 64 & ? & 24 \end{bmatrix}$

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

$(B)$ .

1118273_1195981_ans_caaadc6d10234c1585ef265632453443.jpg

In a triangle

$ABC$ ,

$acos^2(\frac{C}{2})+c\;cos^2(\frac{A}{2})=\dfrac{3b}{2}$ , then the sides

$a,b,c$

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

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

0%
$11^{n}$
0%
$10.11^{n}$
0%
$11^{n+1}$
0%
$11^{n}-1$

Explanation

$\rightarrow 10(^{n}C_{0}+10.^{n}C_{1}+...+10^{n}$

$^{n}C_{n})$

$= 10\displaystyle(\sum_{r=1}^{n}$

$^{n}C_{r}(10)^{r}(1)^{n-r} )$

$= 10(1+10)^{n} = 10.11^{n} = (B)$

$1 + \left( {1 + a} \right)x + 1\left( {1 + a + {a^2}} \right){x^2} + \left( {1 + a + {a^2} + {a^3}} \right){x^3} + - - - - - \;{\text{where}}\;0 < a,x < 1,\;{\text{is}} -$

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

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 ________ .

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

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?

0%
$4^n-3^n+1$
0%
$4^n-3^n-1$
0%
$4^n-3^n+2$
0%
$4^n-3^n$

$| x| < 1$ , then the sum of series

$1+ 2x + 3x^2 + 4x^3 + .........\infty$ will be

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

$=$

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

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 :

0%
5 + 11 + 19 + 29 + 41 + .......
0%
3 + 7 + 14 + 24 + 37 +......
0%
6 + 9 + 16 + 27 + 42 + .....
0%
5 + 7 + 13 + 31 + 85 + ....

$\sum _{ k=1 }^{ 2n+1 }{ (-1)^{k-1} }k^2=$

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

$= 1^2 + 2^2 + 3^2 + .... + (2n + 1) - 2(2^2) [1^2 + 2^2 + ..... + n^2]$

$= \dfrac{(2n + 1)(2n + 2)(4n + 3)}{6} - \dfrac{8 n (n + 1)(2n + 1)}{6}$

$= \dfrac{2(2n + 1)(n + 1)(4n + 3)}{6} - \dfrac{8 n(n + 1)(2n + 1)}{6}$

$= \dfrac{2}{6} (n + 1)(2n + 1) [4n + 3 - 4n]$

$= \dfrac{1}{3} (n + 1) (2n + 1) (3)$

$= (n + 1) (2n + 1)$

$S = 1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}}......,$ then

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

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

$= \dfrac {2n}{n + 1}$

$S_{\infty} = \displaystyle \lim_{n\rightarrow 0} \dfrac {2n}{n + 1} = \displaystyle \lim_{n\rightarrow \infty} = \dfrac {2}{1 + \dfrac {1}{n}}$

$S_{\infty} = 2$ .

$S = \tan ^ { - 1 } \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { n ^ { 2 } + 3 n + 3 } \right) + \ldots + \tan ^ { - 1 } \left( \frac { 1 } { 1 + ( n + 19 ) ( n + 20 ) } \right)$ then

$\tan S$ is equal to

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 ?

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

1203858_1249081_ans_10970632a8a14250a1f3fe91d24a6834.jpg

The value of the expression

$\sum _{ r=0 }^{ n }{ { (-1) }^{ r } } \left( \dfrac { ^nC_r }{^{r+3}C_r } \right)$ is

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?

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

$S = \sum_{r=0}^{19} tan^{-1} \left ( \frac{1}{1+(n+r)(n+r+1)} \right ) = tan^{-1}\left ( \frac{(n+r+1)-(n+r)}{1+(n+r)(n+r+1)} \right )$

$\Rightarrow S = \sum_{r=0}^{19} \left ( tan^{-1}(n+r+1)-tan^{-1}(n+r) \right )$

$=tan^{-1}(n+1) - tan^{-1}(n) = tan^{-1}(n+20)-tan^{-1}(n+19)$

$+ tan^{-1}(n+2) - tan^{-1}(n+1)$

$tan^{-1}(n+20) - tan^{-1}(n+19)$

$\Rightarrow tan \, S = \frac{(n+20)-(n)}{1+n(n+20)} = \frac{20}{n^{2}+20n+1}$

$\Rightarrow (C)$

1167324_1247594_ans_65897e2fb1b54e06a8ec2e626f4bcb16.jpg

13, 16, 22, 33, 51 ?

0%
89
0%
78
0%
102
0%
69
0%
none of these

Explanation

REF.Image

$13,16,22,33,51, ?$

1219100_1249292_ans_c5c3ca6fa710458ea60782db89b8013f.jpeg

Evaluate:-
If

$\sum\limits_{r - 0}^n {{{\left\{ {\frac{{^n{C_{r - 1}}}}{{^n{C_r}{ + ^n}{C_{r - 1}}}}} \right\}}^3} = \frac{{25}}{{24}}}$

0%
$3$
0%
$6$
0%
$4$
0%
$5$

655, 439, 314, 250, 223 ?

0%
215
0%
210
0%
195
0%
190
0%
none of these

Explanation

Solution

$\rightarrow$ Given terms are 655,439,314,250,223,?

$a_{1}=655$

$a_{2}= 439$

$a_{3}=314$

$a_{4}= 250$

$a_{5}= 223$

$a_{6}=?$

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

1207946_1249196_ans_d9030eb80a4342c1bc6b75d798b9631f.jpg

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

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

0%
$2007006$
0%
$1005004$
0%
$2000506$
0%
$None$

Find the next term

$210,209,205,196,180,?$

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

$20, 24, 33, 49, 74, 110, ?$

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

$1728, 2744, 4096, 5832, 8000, 10648, ?$

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

$4,6,12,30,90,315,?$

0%
$945$
0%
$1102$
0%
$1260$
0%
$1417.5$
0%
$None\ of\ these$

Explanation

$=315\times 4$

$=1260.$

Hence, the answer is

$1260.$

1380648_1263649_ans_9acbe6ce48324cbe93e13fd0aa5870ac.png

Find next term

$462,552,650,756,870,992,?$

0%
$1040$
0%
$1122$
0%
$1132$
0%
$105$

Find the missing number from the given alternatives.

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

$52\quad\quad 156$

$8, 31, 122, 485, 1936, 7739, ?$

0%
$30950$
0%
$46430$
0%
$34650$
0%
$42850$
0%
$38540$

The sum of infinite series

$\begin{vmatrix} 1 & 2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 2 } & 2 \\ 2 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 4 } & 2 \\ \frac { 2 }{ 3 } & 4 \end{vmatrix}+.........$

0%
$-10$
0%
$0$
0%
$10$
0%
$\infty$

$\dfrac{\pi}{4}-1+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{13}+=0$ then value of

$\dfrac{1}{1\times3}+\dfrac{1}{5\times7}+\dfrac{1}{9\times11}+\dfrac{1}{13\times15}+..$ is

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

0%
$\frac{1}{4}$
0%
1
0%
$\frac{1}{2}$
0%
None of these

Explanation

$t_r=\dfrac{r}{1+r^2+r^4}$

$=\dfrac{r}{r^4+2r^2+1-r^2}$

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

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

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

$=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}\right]$

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

$\displaystyle\sum^n_{r=1}t_r=\dfrac{1}{2}\left[\dfrac{1}{1-1+1}-\dfrac{1}{(n+1)^2-(n+1)+1}\right]$

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

$\displaystyle\sum t_n=\dfrac{1}{2}\left[\dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\right]$

$\underset{n\rightarrow \infty}{lt}\displaystyle\sum t_r=\dfrac{1}{2}\left[\dfrac{1+0}{1+0+0}\right]$

$=\dfrac{1}{2}$ .

1192651_1281350_ans_4dcd4516020c47938f902c73af559a29.jpg

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

0%
$\dfrac {6n}{n+1}$
0%
$\dfrac {9n}{n+1}$
0%
$\dfrac {12n}{n+1}$
0%
$\dfrac {3n}{n+1}$

Explanation

$\begin{array}{l} Let \\ { T_{ r } }=\dfrac { { 2r+1 } }{ { { 1^{ 2 } }+{ 2^{ 2 } }+{ r^{ 2 } } } } \\ =\dfrac { { \left( { 2r+1 } \right) } }{ { r\left( { r+1 } \right) \left( { 2r+1 } \right) } } \\ =\dfrac { 6 }{ { r\left( { 2+1 } \right) } } \\ \sum { { T_{ r } } } =\sum { \dfrac { 6 }{ { r\left( { r+1 } \right) } } } \\ =6\sum _{ r=1 }^{ n }{ \dfrac { { \left( { r+1 } \right) -r } }{ { r\left( { r+1 } \right) } } } \\ =6\sum _{ r=1 }^{ n }{ \left[ { \dfrac { 1 }{ r } -\dfrac { 1 }{ { r+1 } } } \right] } \\ =6\left[ { \left( { 1-\dfrac { 1 }{ 2 } } \right) +\left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) +.......+\left( { \dfrac { 1 }{ n } -\dfrac { 1 }{ { n+1 } } } \right) } \right] \\ =6\left[ { 1-\dfrac { 1 }{ { n+1 } } } \right] \\ =\dfrac { { 6n } }{ { n+1 } } \\ Hence,\, option\, A\, is\, \, the\, correct\, answer. \end{array}$

$499, 622, 868, 1237, 1729, 2344, ?$

0%
$3205$
0%
$3082$
0%
$2959$
0%
$3462$
0%
$2876$

Explanation

$499, 622, 868, 1237, 1729, 2344,$ ______

$499+1\times 123=499+123=622$

$622+2\times 123=622+146=868$

$868+3\times 123=868+369=1237$

$1237+4\times 123=1237+492=1729$

$1729+5\times 123=1729+615=2344$

$2344+6\times 123=2344+738=3082$ .

1192684_1267327_ans_6dba0cee72914a23955da7b762580a24.jpg

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

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 :

$12,15,21,24,30,33 , ? , ?$

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 :

0%
$125$
0%
$90$
0%
$105$
0%
$225$

Explanation

$\textbf{Step 1: Observe the numbers opposite to each other}$

$\text{We can see that the number opposite to 315 is 21. Also, } \dfrac{315}{21}=15$

$\text{Hence, we can say the missing number has at least one factor of 15 }$

$\text{We see the missing number is opposite to 15}$

$\text{Also, } 15\times15=225$

$\textbf{Step 2: Verification}$

$\text{Let the missing number be x.}$

$\text{ATQ, }$

$\Rightarrow 21+15+x=261$

$\Rightarrow x=261-21-15$

$x=225$

$\text{Hence, it is verified that x=225}$

$\textbf{Thus, the missing number is D 225.}$

$23,29,47,75 , ?$

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

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

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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