MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 5

The value of

$\displaystyle \sum_{n=1}^{10}\left (\sin \dfrac {2n\pi}{11}-\cos\dfrac {2n\pi}{11}\right )$ is equal to

0%
$2$
0%
$1$
0%
$0$
0%
$-1$

Explanation

Given series

$=\left (\sin \dfrac {2\pi}{11}-\cos\dfrac {2\pi}{11}\right )+\left (\sin \dfrac {4\pi}{11}-\cos\dfrac {4\pi}{11}\right )+\left (\sin \dfrac {6\pi}{11}-\cos\dfrac {6\pi}{11}\right )+....+\left (\sin \dfrac {20\pi}{11}-\cos\dfrac {20\pi}{11}\right )$

$=\left (\sin \dfrac {2\pi}{11}+\sin\dfrac {4\pi}{11}+....+\sin\dfrac {20\pi}{11}\right )$

$-\left (\cos \dfrac {2\pi}{11}+\cos\dfrac {4\pi}{11}+....+\cos\dfrac {20\pi}{11}\right )$

$=\dfrac {\sin \pi.\sin \dfrac {10.\pi}{11}}{\sin\dfrac {\pi}{11}}-\dfrac {\cos \pi.\sin \dfrac {10\pi}{11}}{\sin\dfrac {\pi}{11}}$

$=0+\dfrac {\sin\left (\pi-\dfrac {\pi}{11}\right )}{\sin \frac {\pi}{11}}=1$

For

$\frac {2^2+4^2+6^2+...+(2n)^2}{1^2+3^2+5^2+...+(2n-1)^2}$ to exceed 1.01, the maximum value of n is

0%
99
0%
100
0%
101
0%
150

Explanation

$\frac {2^2[1^2+2^2+...+n^2]}{[1^2+3^2+5^2+...+(2n-1)^2]}$

$1^2+2^2+3^2+....+(2n)^2=\frac {2n(2n+1)(4n+1)}{6}$

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

$=\frac {2n(2n+1)(4n+1)}{6}$

$S+4\frac {n(n+1)(2nn+1)}{6}=\frac {2n(2n+1)(4n+1)}{6}$

$S+\frac {2n(2n+1)(4n+1)}{6}-\frac {4n(n+1)(2n+1)}{6}$

$=2n\left (\frac {2n+1}{6}\right )[4n+1-2n-2]$

$=\frac {2n(2n+1)(2n-1)}{6}$

$Ratio =\frac {4n(n+1)(2n+1)}{6}\times \frac {6}{2n(2n+1)(2n-1)}=\frac {2n+2}{2n-1}$

$\frac {2n+2}{2n-1} > \frac {101}{100}$

$200n+200 > 202 n-101$

$2n < 301$

$n < \frac {301}{2}\Rightarrow$ maximum value

$=150$

The sum of the first n terms of the series

$\displaystyle 1^{2}+2.2^{2}+3^{3}+2.4^{2}+5^{2}+2.6^{2}+....$ is

$\displaystyle \frac{n\left ( n+1 \right )^{2}}{2}$ when n is even. When n is odd the sum is

0%
$\displaystyle \frac{n\left ( n+1 \right )^{2}}{4}$
0%
$\displaystyle \frac{n^{2}\left ( n+2 \right )}{4}$
0%
$\displaystyle \frac{n^{2}\left ( n+1 \right )}{4}$
0%
$\displaystyle \frac{n\left ( n+2 \right )^{2}}{4}$

Explanation

$n$ is even then sum

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

$n$ is odd, last term

$= n^2$

required sum

$= [1^2+ 2.2^2 + 3^2+ 2.4^2 +….. + 2(n-1)^2] + n^2$

$= \dfrac{(n-1)n^2}{2} +n^2$ [ Using sum for

$( n – 1 )$ terms to be even

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

required sum

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

The infinite sum

$1\, +\, \displaystyle {\frac{4}{7}\, +\, \frac{9}{7^2}\, +\, \frac{16}{7^3}\, +\, \frac{25}{7^4}\, +\, ......}$ equals

0%
$\displaystyle \frac{27}{14}$
0%
$\displaystyle \frac{21}{13}$
0%
$\displaystyle \frac{49}{27}$
0%
$\displaystyle \frac{256}{147}$

Explanation

Let

$x = 1+\displaystyle \frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+..$
Dividing both the sides by 7

$\displaystyle \frac{x}{7} = \frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^3}+\frac{16} {7^4}+...$
Subtracting iind from the 1st

$\displaystyle \frac{6x}{7} = 1+\frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}...$
The RHS here is an arithmetic-Geometric series , Now divide the equation by 7 again

$\displaystyle \frac{6x}{49} = \frac{1}{7}+\frac{3}{7^2}+\frac{5}{7^3}+\frac{7}{7^4}$
Again subtracting if from the second one

$\Rightarrow \displaystyle \frac{36x}{49}-1 = \frac{\frac{2}{7}}{1-\frac{1}{7}}$

$\Rightarrow \displaystyle \frac{36x}{49}-1 = \frac{\frac{2}{7}}{\frac{6}{7}}$

$\Rightarrow \displaystyle \frac{36x}{49} = 1+\frac{1}{3} = \frac{4}{3}$

$\therefore x= \displaystyle \frac{49}{27}$

$\displaystyle x_{i}> 0,i=1,2,....,50\: \: and\: \: x_{1}+x_{2}+...+x_{50}=50$ then the minimum value of

$\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{50}}$ equals to

0%
$50$
0%
$\displaystyle \left ( 50 \right )^{2}$
0%
$\displaystyle \left ( 50 \right )^{3}$
0%
$\displaystyle \left ( 50 \right )^{4}$

Explanation

Using

$AM\ge HM$ , we get

$\dfrac{x_1+x_2+x_3+.......+x_{50}}{50}\ge \dfrac{50}{\dfrac{1}{x_1}+\dfrac{1}{x_2}+............+\dfrac{1}{x_{50}}}$

$\Rightarrow \dfrac{50}{50}\ge \dfrac{50}{\dfrac{1}{x_1}+\dfrac{1}{x_2}+............+\dfrac{1}{x_{50}}}$ , use the given values

$\therefore \dfrac{1}{x_1}+\dfrac{1}{x_2}+........+\dfrac{1}{x_{50}}\ge 50$

$\displaystyle \sum_{r=1}^{n}t_{r}=\frac{1}{12}n\left ( n+1 \right )\left ( n+2 \right )$ , then the value of

$\displaystyle \sum_{r=1}^{n}\frac{1}{t_{r}}$ is

0%
$\displaystyle \frac{2n}{n+1}$
0%
$\displaystyle \frac{n}{\left ( n+1 \right )}$
0%
$\displaystyle \frac{4n}{ n+1 }$
0%
$\displaystyle \frac{3n}{ n+1 }$

Explanation

If we take

${ t }_{ r }=\cfrac { 1 }{ 4 } n(n+1)$

${ t }_{ r }=\cfrac { 1 }{ 4\times 3 } n(n+1)[(n+2)-(n-1)]=\cfrac { 1 }{ 12 } [n(n+1)(n+2)-n(n+1)(n-1)]\\ { t }_{ 1 }=\cfrac { 1 }{ 12 } [1.2.3-0]\\ { t }_{ 2 }=\cfrac { 1 }{ 12 } [2.3.4-1.2.3]\\ { t }_{ 3 }=\cfrac { 1 }{ 12 } [3.4.5-2.3.4]\\ \underline { { t }_{ n }=\cfrac { 1 }{ 12 } [n(n+1)(n+2)-n(n+1)(n-1)] } \\ { \sum }_{ r=1 }^{ n }{ t }_{ r }=\cfrac { 1 }{ 12 } [n(n+1)(n+2)]\quad \quad Hence,\quad { t }_{ r }=\cfrac { n(n+1) }{ 4 } \\ \cfrac { 1 }{ { t }_{ r } } =\cfrac { 4 }{ n(n+1) } =4\left[ \cfrac { (n+1)-n }{ n(n+1) } \right] =4\left[ \cfrac { 1 }{ n } -\cfrac { 1 }{ (n+1) } \right] \\ \cfrac { 1 }{ { t }_{ 1 } } =4\left[ 1-\cfrac { 1 }{ 2 } \right] \\ \cfrac { 1 }{ { t }_{ 2 } } =4\left[ \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 3 } \right] \\ \underline { \cfrac { 1 }{ { t }_{ n } } =4\left[ \cfrac { 1 }{ n } -\cfrac { 1 }{ (n+1) } \right] \quad \quad \quad } \\ { \sum }_{ r=1 }^{ n }\cfrac { 1 }{ { t }_{ r } } =4\left[ 1-\cfrac { 1 }{ (n+1) } \right] =\cfrac { 4n }{ (n+1) }$

The value of

$\displaystyle\frac{1}{1\cdot2\cdot3}+\displaystyle\frac{1}{2\cdot3\cdot4}+\displaystyle\frac{1}{3\cdot4\cdot5}+\displaystyle\frac{1}{4\cdot5\cdot6}$ is equal to

0%
$\displaystyle\frac{7}{30}$
0%
$\displaystyle\frac{11}{30}$
0%
$\displaystyle\frac{13}{30}$
0%
$\displaystyle\frac{17}{30}$

Explanation

Given $\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}$

$\Rightarrow \dfrac{1}{1\times 2\times3 }+\dfrac{1}{2\times 3\times 4}+\dfrac{1}{3\times 4\times 5}+\dfrac{1}{4\times 5\times 6}$

$\Rightarrow \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+\dfrac{1}{120}$

$= \dfrac{20+5+2+1}{120}$

$= \dfrac{28}{120}$

$= \dfrac{7}{30}$

Value of

$9 + 99 + 999 + ....$ upto

$n$ terms is

0%
$\displaystyle \frac{10^{n}-9n-10}{9}$
0%
$\displaystyle \frac{10^{n}-9n-10}{81}$
0%
$\displaystyle \frac{10^{n+1}-9n-10}{81}$
0%
$\displaystyle \frac{10^{n+1}-9n-10}{9}$

Explanation

Given,

$9 + 99 + 999 + ....$ upto

$n$ terms
we can write this series in this form

$s=(10-1)+(100-1)+(1000-1).......$ up to

$n$ terms.

$= [(10+100+1000+......)+(-1-1-1-.......\text {up to n terms})]$

$= [(10+10^2+10^3+.....\text {up to n terms})-n]$ ........(i)
sum of n terms of GP.

$=\dfrac {a(r^n-1)}{r-1}$ , where

$a$ is first term and

$r$ is the common ratio of the GP.

$\text {We have the GP } 10+10^2+10^3+.... \text {up to n terms}$ , here first term

$a=10$ and common ratio

$r=10$
So,

$10+10^2+10^3+...... \text {n terms}= \dfrac {10(10^n-1)}{10-1} = \dfrac {10^{n+1}-10}{9}$ On Substituting

$s=10+10^2+10^3+...... \text {n terms}= \dfrac {10^{n+1}-10}{9}$ in equation (i). we get,

$=\dfrac {10^{n+1}-10}{9}-n = \dfrac {10^{n+1}-9n-10}{9}$

$\Rightarrow 9+99+999+.......\text {up to n terms} = \dfrac {10^{n+1}-9n-10}{9}$
Option D is correct.

In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)

$0,\,3,\,9,\,18,\,30,\,?$

0%
$48$
0%
$42$
0%
$36$
0%
$45$

Explanation

The difference between the consecutive numbers increases by

$3$ .

$3-0 = 3$

$9 - 3 = 6$

$18-9 = 9$

$30-18 = 12$ .

Pattern is adding multiples of 3

Next term is

$30+15=45$

Evaluate

$7 + 77 + 777 + .............$ upto

$n$ terms.

0%
$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n]$
0%
$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n\, -\, 10]$
0%
$\displaystyle \frac{7}{81}[10^n\, -\, 9n\, -\, 10]$
0%
$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, n\, -\, 10]$

Explanation

Let

$S=7+77+777+........+$ up to

$n$ terms

$\Rightarrow \dfrac{7}{9}\left [ 9+99+999+..... \right ]$

$\Rightarrow \dfrac{7}{9}\left [\left (10-1 \right )+\left (10^2-1 \right )+\left (10^3+1 \right )+........up\quad to\quad n\quad terms \right ]$

$\Rightarrow \dfrac{7}{9}\left [10+10^2+10^3+......+10^n-n \right ]$

$\Rightarrow \dfrac{7}{9}\left [ 10\frac{10^{n}-1}{9}-n \right ]$

$\Rightarrow \dfrac{7}{81}\left [ 10^{n+1}-9n-10 \right ]$

4, 9, 25, 49, ....

0%
64
0%
81
0%
121
0%
125

Find a wrong number in the series:
9, 19, 37, 75, 149, 297

0%
75
0%
37
0%
19
0%
149
0%
299

Explanation

Clearly the correct sequence is:
9, 2 x 9 + 1 = 19 ; 2 x 19 -1 = 37, 2 x 37 + 1 = 75 ; 2 x 75 - 1 = 149 ; 2 x 149 + 1 = 299 i.e. the logic is x 2 + 1.
Therefore

$299$ is the correct answer.

Find pair of consecutive odd natural numbers, both of which are larger than 13 such that their sum is less than 40

0%
(15, 17)
0%
(13, 17)
0%
(19, 21)
0%
(21, 17)

Explanation

Let the consecutive odd numbers be

$x, x + 2$

Given,

$x > 13$ -- (1)

Also,

$x + 2 > 13$

$x > 11$ -- (2)

And

$x + x + 2 < 40$

$2x + 2 < 40$

$2x < 38$

$x < 19$ -- (3)

From,

$1, 2, 3$

$13 < x < 19$
This means, the numbers are

$15, 17$

In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

0%
$18$
0%
$40$
0%
$21$
0%
$24$

The sum of 'n' terms of series

$1^2+(1^2+2^2)+(1^2+ 2^2+ 3^2)+(1^2+2^2+ 3^2+4^2)+ .......$ will be

0%
$\displaystyle \frac{1}{4} n (n + 1)^2 (n + 2)$
0%
$\displaystyle \frac{1}{3} n (n + 1)^2 (n + 2)$
0%
$\displaystyle \frac{1}{12} n (n + 1) (n + 2)^2$
0%
$\displaystyle \frac{1}{12} n (n + 1)^2 (n + 2)$

Explanation

The given series is

${ 1 }^{ 2 }+({ 1 }^{ 2 }+2^{ 2 })+({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 })+({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 }+4^{ 2 })+.....$

${ a }_{ n }=({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 }+.......+n^{ 2 })$

$=\dfrac { n(n+1)(2n+1) }{ 6 } \\ =\dfrac { n(2n^{ 2 }+3n+1) }{ 6 } \\ =\dfrac { 2n^{ 3 }+3n^{ 2 }+n }{ 6 } \\ =\dfrac { 1 }{ 3 } n^{ 3 }+\dfrac { 1 }{ 2 } n^{ 2 }+\dfrac { 1 }{ 6 } n$

Therefore, we have:

${ S }_{ n }=\sum _{ k=1 }^{ n }{ { a }_{ k } } \\ =\sum _{ k=1 }^{ n }{ \left( \dfrac { 1 }{ 3 } k^{ 3 }+\dfrac { 1 }{ 2 } k^{ 2 }+\dfrac { 1 }{ 6 } k \right) } \\ =\dfrac { 1 }{ 3 } \sum _{ k=1 }^{ n }{ k^{ 3 } } +\dfrac { 1 }{ 2 } \sum _{ k=1 }^{ n }{ k^{ 2 } } +\dfrac { 1 }{ 6 } \sum _{ k=1 }^{ n }{ k } \\ =\dfrac { 1 }{ 3 } \times \dfrac { n^{ 2 }+(n+1)^{ 2 } }{ 2^{ 2 } } +\dfrac { 1 }{ 2 } \times \dfrac { n(n+1)(2n+1) }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { n(n+1) }{ 2 }$

$=\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1) }{ 2 } +\dfrac { (2n+1) }{ 2 } +\dfrac { 1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n^{ 2 }+n+2n+1+1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1)+2(n+1) }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { (n+1)(n+2) }{ 2 } \right] \\ =\dfrac { n(n+1)^{ 2 }(n+2) }{ 12 }$

Hence, the sum is

$\dfrac { n(n+1)^{ 2 }(n+2) }{ 12 }$ .

$11, 26, 56, 101, 161, .....$

0%
$236$
0%
$226$
0%
$216$
0%
$206$

Explanation

Add

$15, 30, 45, 60, 75$ to each number Next number is

$161 + 75 = 236$

3, 15, 35, 63, 99, ...

0%
133
0%
143
0%
153
0%
165

Explanation

Product of two successive numbers

$1, 3, 5, 7, 9, 11, 13, 15,$ ... Next number is

$11 x 13 = 143$

10, 12, 16, 24, 40, ....

0%
60
0%
56
0%
70
0%
72

Find the Odd one among : 97, 77, 59, 43, 26, 17

0%
77
0%
59
0%
43
0%
26

Explanation

The rule is

$\displaystyle 10^{2}-3$ ,

$\displaystyle 9^{2}-4$ ,

$\displaystyle 8^{2}-5$ ,

$\displaystyle 7^{2}-6$ ,

$\displaystyle 6^{2}-7$ ,

$\displaystyle 5^{2}-8$ etc. 97,77, 59, 43, 29, 18 Thus 26 is the odd man out

- nmmn - mmnm - mnnm -

0%
nmmn
0%
mnnm
0%
nnmm
0%
nmnm

7th term of

$\displaystyle\frac{1}{2}$ ,

$\displaystyle\frac{1}{4}$ ,

$\displaystyle\frac{1}{6}$

$\dots\dots$ is

0%
$\displaystyle\frac{1}{10}$
0%
$\displaystyle\frac{1}{12}$
0%
$\displaystyle\frac{1}{14}$
0%
$\displaystyle\frac{1}{16}$

Explanation

We can see that the series has terms such that

$\dfrac {1}{2(1)}, \dfrac {1}{2(2)}, \dfrac {1}{2(3)}.....$

So, the

$7th$ term will be

$\dfrac {1}{2(7)} = \dfrac {1}{14}$

Find the missing letters
A, B, D, G, .....

0%
L
0%
M
0%
J
0%
K

Explanation

Pattern is

$\displaystyle ^{1}A,^{1}A+1,^{2}B+2,^{4}D+3,^{7}G+4,^{11}K$

$\displaystyle \therefore$ Missing letter = K

Find the Odd one among : 49, 121, 169, 225, 289, 361

0%
225
0%
121
0%
49
0%
361

In a series,

$T_n=2n+5$ , find

$S_4$ .

0%
40
0%
30
0%
20
0%
10

Explanation

$T_n=2n+5$

$T_1=2(1)+5=7$ ;
Similarly

$T_2=9$ ,

$T_3=11$ and

$T_4=13$

$\therefore S_4=T_1+T_2+T_3+T_4=40$ .

$\displaystyle \frac{1}{3^{2}-1}+\frac{1}{5^{2}-1}+\frac{1}{7^{2}-1}+......+\frac{1}{35^{2}-1}=$

0%
$\displaystyle \frac{4}{17}$
0%
$\displaystyle \frac{17}{72}$
0%
$\displaystyle \frac{19}{72}$
0%
$\displaystyle \frac{17}{36}$

Explanation

given series

$=\sum _{n=1} ^{17} \dfrac{1}{(2n+1)^2 -1}$

$=\sum \dfrac{1}{4n^4+4n} =\sum \dfrac{1}{4n} \times \dfrac{1}{n+1}$

solving this by integration, we will get

$=\dfrac{17}{72}$

There is no symmetry in symmetry in legs of Fig

The value of

$1^2+2^2+3^2+\cdots+ n^2$ is

0%
$\displaystyle\frac{n(n+1)}{2}$
0%
$\displaystyle\frac{n(n+1)(2n+1)}{6}$
0%
$\displaystyle\frac{n(n+1)(2n-1)}{6}$
0%
$\displaystyle\left[\frac{n(n+1)}{2}\right]^2$

Explanation

Let

${ S }_{ n }={ 1 }^{ 2 }+{ 2 }^{ 2 }+\dots +{ n }^{ 2 }$

Consider the identity

${ k }^{ 3 }-{ \left( k-1 \right) }^{ 3 }=3{ k }^{ 2 }-3k+1$

Putting

$k=1,2,....,n$ successively, we obtain

${ 1 }^{ 3 }-{ \left( 0 \right) }^{ 3 }=3{ \left( 1 \right) }^{ 2 }-3\left( 1 \right) +1$

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

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

$\vdots$

${ n }^{ 3 }-{ \left( n-1 \right) }^{ 3 }=3{ \left( n \right) }^{ 2 }-3\left( n \right) +1$

Adding both sides we get,

${ n }^{ 3 }-{ \left( 0 \right) }^{ 3 }=3{ \left( 1^{ 2 }+2^{ 2 }+\dots n^{ 2 } \right) }-3\left( 1+2+\dots +n \right) +n$

${ n }^{ 3 }=3{ \sum _{ k=1 }^{ n }{ k^{ 2 } } }-3\sum _{ k=1 }^{ n }{ k+n }$

Since

$\sum _{ k=1 }^{ n }{ k } =1+2+\dots +n=\dfrac { n\left( n+1 \right) }{ 2 }$

Therefore,

${ S }_{ n }={ \sum _{ k=1 }^{ n }{ k^{ 2 } } }=\dfrac { 1 }{ 3 } \left[ { n }^{ 3 }+\dfrac { 3n\left( n+1 \right) }{ 2 } -n \right]$

${ S }_{ n }={ \sum _{ k=1 }^{ n }{ k^{ 2 } } }=\dfrac { 1 }{ 6 } \left( { 2n }^{ 3 }+{ 3n }^{ 2 }+n \right)$

${ S }_{ n }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$

Hence option B is correct.

The product

$\displaystyle \left ( 1-\frac{1}{n} \right )\left ( 1-\frac{1}{n+1} \right )\left ( 1-\frac{1}{n+2} \right )..\left ( 1-\frac{1}{2n} \right )$ Equals

0%
$\displaystyle \frac{n-1}{2n}$
0%
$\displaystyle \frac{1}{2n}$
0%
$\displaystyle \frac{2n}{n-1}$
0%
$\displaystyle \frac{1}{n}$

Explanation

Given product
$\displaystyle =\left ( 1-\dfrac{1}{n} \right )\left ( 1-\dfrac{1}{n+1} \right )\left ( 1-\dfrac{1}{n+2} \right )\cdot \cdot \cdot \left ( 1-\dfrac{1}{2n} \right )$
$\displaystyle =\dfrac{\left ( n-1 \right )}{n}\times \dfrac{n}{\left ( n+1 \right )}\times \dfrac{n+1}{\left ( n+2 \right )}\cdot \cdot \cdot \left ( \dfrac{2n-1}{2n} \right )=\dfrac{\left ( n-1 \right )}{2n}$

Find the sum of first three terms of the following sequence whose

$n^{th}$ term is

$5n^2-2$ .

0%
$68$
0%
$86$
0%
$69$
0%
$82$

Explanation

Given,

$T_n = 5n^2 - 2$
So,

$T_1= 5-2 = 3$

$T_2 = 5(2^2) -2 = 18$

$T_3 = 5(3^2) -2 = 47$

So, Sum of first three terms

$= 3+18+47 = 68$

In a certain language '

$\displaystyle +\div ?$ ' means 'where are you' , '

$\displaystyle @-\div$ ' means ' we are here' , and '

$\displaystyle +@\times$ ' means ' you come here' What is the code for ' Where' ?

0%
+
0%
$\displaystyle \div$
0%
?
0%
@

Explanation

s.no code sentence
1.

$\displaystyle '+\div ?'$ Where are you
2.

$\displaystyle '@-\div '$ We are here
3.

$\displaystyle '+@\times$ you come here

As we can see that where is only in sentence

$1$ [Where is the word for which we have to find the code ] Therefore we need to gather the codes for

$are$ &

$you$ to find out the code for where sentence 1 & 2 have the word

$are$ in common and the symbol

$\displaystyle \div$ in common Therefore

$\displaystyle \div$ is the symbol for

$are$
Sentence 1 & 3 have the word you in common and the symbol + in common therefore + stands for

$you$

thus, leaving only ?, which stands for

$where$

How is 'red written in that code ?

0%
hee
0%
silk
0%
be
0%
cannot be determined

Explanation

Type Direct coding (Jumbled fashion)
S.no. code sentence
1. il be pee roses are blue
2. silk hee red flowers
3. pee mit hee Flowers are Vegetables
common word in sentences 1 & 3

$\displaystyle \rightarrow$ 'are' and code

$\displaystyle \rightarrow$ 'pee'
common word in sentences 2 & 3

$\displaystyle \rightarrow$ flowers and code

$\displaystyle \rightarrow$ 'hee'
Therefore
Codes words
pee are
hee flowers
silk red
niit vegetables
il roses/ blue
be blue./ rose

Image

0%
70
0%
75
0%
65
0%
80

Explanation

Suppose, the apple is

$x$ and the pear is

$y$

Given in the question,

$2x+y=230$ eq—1

And,

$y-x=5$ eq—2

So, from —2

$y= 5+x$ eq—3

Putting eq—3 in eq—1

$2x+5+x=230$

$3x+5=230$

$3x=225$

$x=\dfrac{225}{3}$

$x=75$

Now from eq—3

$y=5+75$

$y=80$

So the value of a pear is

$80$

Hence Option D is the correct answer.

How is 'vegetables are red flowers' written in this code ?

0%
Pec silk mit hee
0%
silk peehee be
0%
il silk mit hee
0%
none

Explanation

Type Direct coding (Jumbled fashion)
S.no. code sentence
1. il be pee roses are blue
2. silk hee red flowers
3. pee mit hee Flowers are Vegetables
common word in sentences 1 & 3

$\displaystyle \rightarrow$ 'are' and code

$\displaystyle \rightarrow$ 'pee'
common word in sentences 2 & 3

$\displaystyle \rightarrow$ flowers and code

$\displaystyle \rightarrow$ 'hee'
Therefore
Codes words
pee are
hee flowers
silk red
niit vegetables
il roses/ blue
be blue./ rose
nit silk hee pee

Find the sum of odd numbers between

$2$ and

$76$ .

0%
$1,350$
0%
$1,443$
0%
$1,342$
0%
$1,360$

Explanation

The sum of first n terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ .............. (1)
The list of odd numbers between

$2$ and

$76$ is

$3, 5, 7, 9, ........... 75$
Where

$n =$ number of terms

$= ?$

$a =$ first odd number

$\Rightarrow$

$3$

$d =$ common difference of A.P.

$\Rightarrow$

$2$

$T_n =$ Last term

$\Rightarrow$

$75$

$T_n = a+ (n - 1) d$

$75 = 3 + (n - 1)2$

$75 = 3 + 2n - 2$

$75 - 3 + 2 = 2n$

$74 = 2n$

$n = 37$
Apply the given data in the formula (1),

$S_{37} = \dfrac{37}{2} [2 \times 3 + (37 - 1)2]$

$= 18.5 [6 + (36)2]$

$= 18.5 [6 + 72]$

$= 18.5 [78]$

$= 1,443$
So, the sum of odd numbers between

$2$ and

$76$ is

$1,443$ .

Find the next number of the series.

$7, 10, 8, 11, 9, 12,....$

0%
9
0%
10
0%
8
0%
7

Explanation

The next number is

$10$ .
The first

$2$ numbers difference is

$3$ .
The second

$2$ numbers difference is

$2$ .
So, this is simple alternating addition of

$3$ and subtraction of

$2$ .

In a certain sequence of 8 numbers, each number after the first is 1 more than the previous number .If the first number is -5, how many of the numbers in the sequence are positive?

0%
None
0%
One
0%
Two
0%
Three
0%
Four

Find the value of

$'?'$ in the series:

$12, 17, 15, ?, 18, 23, 21,....$

0%
20
0%
21
0%
22
0%
24

Explanation

The series alternates the addition of

$5$ with the subtraction of

$2$ .
So, the number is

$20$ .

Find the value of

$'x'$ in the series:

$144, 169, x, 225, 256, 289$ .

0%
190
0%
192
0%
196
0%
195

Explanation

The series is a perfect square in increasing order of

$12, 13, 14, 15...$
So, the value of

$x$ is

$196$ .

Find the missing number in the pattern:

$2, 5, 8, 11,$ __

$, 17, 20, 23, 26$

0%
11
0%
12
0%
13
0%
14

Explanation

The series adds

$3$ to each number to get the next number.
So, the next number is

$11 + 3 = 14$

Complete the pattern:

$3, 5, 8, 13,$ ___

0%
21
0%
22
0%
23
0%
24

Explanation

The next number is

$21$ .
We get the next number by adding the previous two numbers.

Which pair of numbers comes next?

$25, 23, 5, 20, 16, 5,....$

0%
$11, 5$
0%
$5, 6$
0%
$10, 5$
0%
$9, 5$

Explanation

This is an alternating subtraction series with the interpolation of a random number,

$5$ , as every third number.
In the subtraction series,

$2$ is subtracted, then

$3$ then

$4$ , and so on.
The next pair of numbers are

$11$ and

$5$ .

Which number comes next?

$5, 9, 13, 17, 21, 25, 29,...$

0%
$31$
0%
$32$
0%
$30$
0%
$33$

Explanation

The series adds

$4$ to each number to get the next number.
So, the next number is

$29 + 4 = 33$

$2, 3,$ __

$, 4, 4, 5, 6, 6, 6, 7, 7, 7...$ What number should fill the blank?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Which pair of numbers comes next?

$10, 18, 12, 21, 15 ,25, ,....$

0%
$19, 30$
0%
$19, 27$
0%
$30, 19$
0%
$27, 19$

Explanation

This is an alternating addition series

First series =

$10, 10+2=12, 12+3=15$ .

$\therefore$ next number=

$15+4=19$
Second series =

$18, 18+3=21, 21+4=25$

$\therefore$ next number =

$25+5=30$
The next pair of numbers are

$19, 30$ .

Fill in the blank:

$62, 66, 63, 66, 64,$ __

$, 65,....$

0%
60
0%
61
0%
62
0%
63

Explanation

The series alternates the addition of

$4$ with the subtraction of

$3$ .
So, the missing number is

$61$ .

Look at this series:

$32, 31, 33, 32, 34, 33,...$ . What is the next number?

0%
34
0%
32
0%
35
0%
36

Explanation

The next number is

$35$ .
Here first

$1$ is subtracted and then

$2$ is added, and so on.

Choose the missing number in the series:

$2, 2, 3, 4, 4,$ __

$, 6, 6, 7, 8, 8, 9....$

0%
5
0%
6
0%
7
0%
8

Explanation

This is a continuous series of adding

$1$ to the previous number, in which every third number is not repeated.
So, the missing number is

$5$ .

The value of

$\displaystyle \tan { \alpha } +2\tan { \left( 2\alpha \right) } +4\tan { \left( 4\alpha \right) } +...+{ 2 }^{ n-1 }\tan { \left( { 2 }^{ n-1 }\alpha \right) } +{ 2 }^{ n }\cot { \left( { 2 }^{ n }\alpha \right) }$ is

0%
$\displaystyle \cot { \left( { 2 }^{ n }\alpha \right) }$
0%
$\displaystyle { 2 }^{ n }\tan { \left( { 2 }^{ n }\alpha \right) }$
0%
$\displaystyle 0$
0%
$\displaystyle \cot { \alpha }$

Explanation

Now,

$\displaystyle { 2 }^{ n }\tan { \left( { 2 }^{ n }a \right) } +{ 2 }^{ n }\cot { \left( { 2 }^{ n }a \right) }$

$\displaystyle ={ 2 }^{ n-1 }\left[ \frac { \sin { { 2 }^{ n-1 }a } }{ \cos { { 2 }^{ n-1 }a } } +2\frac { \cos { { 2 }^{ n }a } }{ \sin { { 2 }^{ n }a } } \right]$

$\displaystyle ={ 2 }^{ n-1 }\left[ \frac { \cos { { 2 }^{ n }a } \cos { { 2 }^{ n-1 }a } +\sin { { 2 }^{ n }a\sin { { 2 }^{ n-1 }a+\cos { { 2 }^{ n }a } \cos { { 2 }^{ n-1 }a } } } }{ \sin { { 2 }^{ n }a\cos { { 2 }^{ n-1 }a } } } \right]$

$\displaystyle ={ 2 }^{ n-1 }\left[ \frac { \cos { { 2 }^{ n-1 }a } \left( 1+\cos { { 2 }^{ n }a } \right) }{ \sin { { 2 }^{ n }a } \cos { { 2 }^{ n-1 }a } } \right]$

$\displaystyle ={ 2 }^{ n-1 }\cot { { 2 }^{ n-1 }a }$
Proceeding in similar way in last, we get

$\displaystyle \tan { a } +2\cot { 2a }$

$\displaystyle =\frac { \sin { a } }{ \cos { a } } +2\frac { \cos { 2a } }{ \sin { 2a } }$

$\displaystyle =\frac { \cos { 2a\cos { a } +\sin { 2a } \sin { a } +\cos { 2a\cos { a } } } }{ \sin { 2a\cos { a } } }$

$\displaystyle =\frac { \cos { a } \left( 1+\cos { 2a } \right) }{ 2\sin { a } { cos }^{ 2 }a } =\frac { 2{ cos }^{ 2 }a }{ 2\sin { a } }$

$\displaystyle =\frac { \cos { a } }{ \sin { a } } =\cot { a }$

What is the missing number in the sequence

$1, 5, 10, 16, 23, 31,$ ____?

0%
$37$
0%
$38$
0%
$39$
0%
$40$
0%
$41$

Explanation

We can see that the difference between the consecutive numbers of the series is increasing by

$1$

$5 - 1 = 4$

$10 - 5 = 5$

$16 - 10 = 6$

$23 - 16 = 7$

$31 - 23 = 8$

Hence, Next term

$- 31 = 9$
=> Next term

$= 31 + 9 = 40$

The sum of

$24$ terms of the following series

$2+4+6.....$

0%
$100$
0%
$200$
0%
$600$
0%
$250$

Explanation

$\overline { 2 } +\overline { 8 } +\overline { 18 } +\overline { 32 } ......24$ terms

$=\sqrt { 2 } +\sqrt { 8 } +\sqrt { 18 } +\sqrt { 32 } .....24$ terms

$\\ =\sqrt { 2 } (1+\sqrt { 4 } +\sqrt { 9 } +\sqrt { 16 } +........24$ terms)

$=\sqrt { 2 } (1+2+3+4+.....24)$

Sum of natural number

$=\cfrac { n(n+1) }{ 2 }$

$=\sqrt { 2 } \cfrac { (24)(25) }{ 2 } \\ =300\sqrt { 2 } =300\overline { 2 }$

Answer

$(C)$

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.