MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 3

Which of the following is

$GP?$

0%
$2, 4, 8, 16$
0%
$2, -2, 2, 3, 1$
0%
$0, 3, 6, 9, 12$
0%
$10, 20, 30, 40$

Explanation

The sequence

$2, 4, 8, 16$ is

$GP$

Because here common ration between the consecutive terms is same. That is illustrated below.

$\dfrac {4}{2}=2, \dfrac {8}{4}=2, \dfrac {16}{8}=2$

Here the common ratio is

$2$ .

Which of the following is a geometric sequence?

0%
{ $2, 4, 6, 8, 10$ }
0%
{ $-1, 2, 4, 8, -2$ }
0%
{ $2, -2, 2, -2, 2$ }
0%
{ $3, 13, 23, 33, 43$ }

Explanation

For a G.P., the ratio must be equal throughout

{

$2, -2, 2, -2, 2$ } is a geometric sequence because the common ratio is

$-1$ .

The difference of the squares of two consecutive even integers is divisible by which of the following integers?

0%
$3$
0%
$4$
0%
$6$
0%
$7$

Explanation

Let the two consecutive even integers be

$2n$ and

$(2n+2)$ . Then,

${(2n+2)}^{2}=(2n+2+2n)(2n+2-2n)$

$=2(4n+2)$

$=4(2n+1)$ , which is divisible by

$4$ .

The average IQ of

$4$ people is

$110$ . If three of these people each have an IQ of

$105$ , what is the IQ of the fourth person?

0%
110
0%
115
0%
120
0%
125

Explanation

Average IQ of

$4$ people

$=110$

$\therefore$ total IQ of

$4$ people

$=$

$110\times 4=440$

Average IQ of

$3$ people

$=105$

$\therefore$ total IQ of

$3$ people

$=$

$105\times 3=315$

Then IQ of forth person

$=$

$440-315=125$

$2, 3, 6, n, 42, . . .$
Find the value of

$n$ in the above series.

0%
$9$
0%
$12$
0%
$15$
0%
$18$
0%
$21$

Explanation

Given series is

$2,3,6,n,42.....$
we can decode the series, by multiplying the previous number by

$3$ and then subtracting

$3$ from it.
As we can see,

$2$ being the first number of series

$2\times 3=6$

$6-3=3$ (

$3$ is second number)

$3\times 3-3$

$=6$ (

$6$ is

$3^{rd}$ number)

$n$ will be

$6\times 3-3=18-3=15$

$n$ is

$15$

The 8th term of the sequence 1, 1, 2, 3, 5, 8 ............... is

0%
25
0%
24
0%
23
0%
21

Explanation

Actually it is Fibonacci sequence, in which next term is always sum of two previous terms.
As

$2=1+1$

$3=1+2$

$5=2+3$

$8=3+5$
Therefore, next (7th) term

$=5+8=13$
Hence, 7th term

$=8+13=21$
Option D is correct.

$\left( { 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+.......+{ 20 }^{ 2 } \right) =$ ?

0%
$770$
0%
$1155$
0%
$1540$
0%
$385\times 385$

Explanation

$\left( { 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+.......+{ 20 }^{ 2 } \right) ={ \left( 1\times 2 \right) }^{ 2 }+{ \left( 2\times 2 \right) }^{ 2 }+{ \left( 2\times 3 \right) }^{ 2 }+......+{ \left( 2\times 10 \right) }^{ 2 }\quad$

$=\left( { 2 }^{ 2 }\times { 1 }^{ 2 } \right) +\left( { 2 }^{ 2 }\times { 2 }^{ 2 } \right) +\left( { 2 }^{ 2 }\times { 3 }^{ 2 } \right) +....+\left( { 2 }^{ 2 }\times { 10 }^{ 2 } \right)$

$=(4\times 5\times 77)$

$=1540$

The common ratio is calculated in

0%
A.P.
0%
G.P.
0%
H.P.
0%
I.P.

Explanation

The common ratio is calculated in G.P.
For example:

$2,4,8,16,....$

Here the common ratio is

$2$ .

Complete the addition square by finding the missing numbers.
Find

$(A-B)-(C-D)$

$+$	$10,923$	$8,473$
$18,732$	$A$	$B$
$9,018$	$C$	$D$

0%
$0$
0%
$450$
0%
$1000$
0%
$300$

Explanation

From the given table, we have

$A=18,732+10,923=29,655$

$B=18,732+8,473=27,205$

$C=10,923+9,018=19,941$
and

$D=8473+9,018=17,491$
Now

$A-B=29,655-27,205=2,450$

$C-D=19,941-17,491=2,450$

$\therefore$

$(A-B)-(C-D)=2,450-2,450=0$

The models are shaded to show which of the following?

0%
$\cfrac { 1 }{ 3 } =\cfrac { 2 }{ 4 }$
0%
$\cfrac { 1 }{ 4 } >\cfrac { 1 }{ 3 }$
0%
$\cfrac { 2 }{ 3 } <\cfrac { 2 }{ 4 }$
0%
$\cfrac { 2 }{ 4 } <\cfrac { 2 }{ 3 }$

Explanation

Fraction of shaded part in upper rectangle

$=2\times\dfrac{1}{3}=\dfrac{2}{3}$

Fraction of shaded part in lower rectangle

$=2\times\dfrac{1}{4}=\dfrac{2}{4}$

In upper rectangle shaded part is greater

So,

$\dfrac { 2 }{ 4 } <\dfrac { 2 }{ 3 }$

Option D is correct.

Fathom is a unit once used by sailors to measure the depth of water. If a sunken ship was located underwater at

$240$ feet, which expression would describe the location of the ship in fathoms?

${ 1\ fathom=6\ feet }$

0%
$240\times 6$
0%
$240\div 6$
0%
$240-6$
0%
$240+6$

Explanation

We have

$1fathom=6feet$

$1feet=\cfrac{1}{6}fathom$
Ship was located underwater at

$240$ feet

$\therefore$ location of ship underwater in fathoms

$=240\div6$

Consider the sequence of numbers

$2, 5, 5, 8, 8, 8, 11, 11, 11, 11, ....$ The

$150^{th}$ term of the sequence is

0%
$48$
0%
$50$
0%
$47$
0%
$53$

Explanation

$2$ is coming one time

$5$ is coming two times

$8$ is coming four times and so on.

We can see that

$2$ is

$1^{st}$ term,

$5$ is

$2^{nd}$ term,

$8$ is

$4^{th}$ term,

$11$ is

$7^{th}$ term and so on.

Hence,

$(3k+2)$ is the

$\left( \cfrac{k(k+1)}{2} +1 \right)^{th}$ term, where

$k\in I$

Take

$k=16 \Rightarrow \left( \cfrac{k(k+1)}{2} +1 \right)= 137 \\ k=17 \Rightarrow \left( \cfrac{k(k+1)}{2} +1 \right)=154$

Hence,

$150^{th}$ term will be for

$k=16$

The number will be

$3k+2=3\times 16+2=50$

Hence,

$150^{th}$ term is

$50$

$1 \, + \, 2 \, \cdot \, 2 \, + \, 3 \, \cdot \, 2^2 \, + \, 4 \, \cdot \, 2^3 \, + \, ..... \, + \, 100 \, \cdot \, 2^{99}$

0%
$S=50502^{100}$
0%
$S=5050.2^{99}$
0%
$S=5050.2^{101}$
0%
$S=5050.2^{98}$

Explanation

Consider the given series.

$1+2.2+3.2^2+4.2^3+5.2^4+.......+100.2^{99}$

The series mentioned above is a

$A.G.P$ series i.e arithmetic and geometric series.

The

$n^{th}$ term for this series will be

$=n.2^{n-1}$

Now,

$=\dfrac{n(n+1)}{2}.\dfrac{1.2^{n-1}}{2-1}$

Since,

$n=100$

Therefore,

$=\dfrac{100(100+1)}{2}.\dfrac{1.2^{100-1}}{2-1}$

$=\dfrac{100(101)}{2}.{1.2^{99}}$

$=50(101).2^{99}$

$=5050.2^{99}$

Hence, this is the answer.

A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in the two matrices given below. The columns and rows of Matrix I are numbered from

$0$ to

$4$ and that of Matrix

$I$ are numbered from

$5$ to

$9$ . A letter from these matrices can be represented first by its row and next by its column, e.g.

$'A'$ can be represented by

$00, 12, 23$ etc., and

$'P'$ can be represented by

$58, 69, 75$ etc. Similarly, you have to identify the set for the word 'POET'.

		Matrix I
	$0$	$1$	$2$	$3$	$4$
$0$	$A$	$R$	$S$	$N$	$C$
$1$	$N$	$C$	$A$	$R$	$S$
$2$	$S$	$N$	$C$	$A$	$R$
$3$	$R$	$S$	$N$	$C$	$A$
$4$	$C$	$A$	$R$	$S$	$N$

		Matrix II
	$5$	$6$	$7$	$8$	$9$
$5$	$O$	$E$	$L$	$P$	$T$
$6$	$T$	$O$	$E$	$L$	$P$
$7$	$P$	$T$	$O$	$E$	$L$
$8$	$L$	$P$	$T$	$O$	$E$
$9$	$E$	$L$	$P$	$T$	$O$

0%
$69, 88, 67, 65$
0%
$75, 56, 65, 67$
0%
$77, 88, 98, 78$
0%
$75, 66, 76, 78$

Explanation

From matrix,

$P\Rightarrow 58, 69, 75, 86, 97$

$O\Rightarrow 55, 66, 77, 88, 99$

$E\Rightarrow 56, 67, 78, 87, 98$

$T\Rightarrow 69, 88, 67, 65$

$\therefore POET\Rightarrow 69, 88, 67, 65$ .

Find the sum of

$1^3-2^3+3^3-4^3+....+9^3$

0%
$400$
0%
$425$
0%
$450$
0%
$475$

Explanation

$1^3-2^3+3^3-4^3+...+9^3$

$=1^3+2^3+3^3+...+9^3-2(2^3+4^3+6^3+8^3)$

$=[\dfrac{9 \times 10}{2}]^2-2.2^3[1^3+2^3+3^3+4^3]$

$=(45)^2-16[\dfrac{4 \times 5}{2}]^2$

$=2025-1600=425$

In this question, the sets of numbers given in the alternatives are represented by two classes of alphabets as in two matrices given below. The columns and rows of Matrix I are numbered from

$0$ to

$4$ and that of Matrix II are numbered from

$5$ to

$9$ . A letter from these matrices can be represented first by its row and next by its column, e.g.,

$'B'$ can be represented by

$00, 13,$ etc., and

$'A'$ can be represented by

$55, 69$ , etc. Similarly you have to identify the set for the word 'GIRL'.

		Matrix I
	$0$	$1$	$2$	$3$	$4$
$0$	$B$	$N$	$G$	$L$	$D$
$1$	$G$	$L$	$D$	$B$	$N$
$2$	$D$	$B$	$N$	$G$	$L$
$3$	$N$	$G$	$L$	$D$	$B$
$4$	$L$	$D$	$B$	$N$	$G$

		Matrix II
	$5$	$6$	$7$	$8$	$9$
$5$	$A$	$I$	$K$	$O$	$R$
$6$	$I$	$K$	$O$	$R$	$A$
$7$	$K$	$O$	$R$	$A$	$I$
$8$	$O$	$R$	$A$	$I$	$K$
$9$	$R$	$A$	$I$	$K$	$O$

0%
$02, 56, 97, 24$
0%
$31, 79, 68, 42$
0%
$23, 97, 77, 11$
0%
$11, 88, 95, 23$

Explanation

From matrix,

$G\Rightarrow 02, 10, 23, 31, 44$

$I\Rightarrow 56, 65, 79, 88, 97$

$R\Rightarrow 59, 68, 77, 86, 95$

$L\Rightarrow 03, 11, 24, 32, 40$

$\therefore GIRL\Rightarrow 23, 97, 77, 11$ .

The two sequence of number

${1,4,16,64,.......}$ and

${3,12,48,192,........}$ are mixed as follows:

${1,3,4,12,16,48,64,192,.......}$ . one of the numbers in the mixed in the mixed series is

$1048576$ , then number immediately preceding is-

0%
$3.4^9$
0%
$3.4^{10}$
0%
$4^9$
0%
$4^{10}$

Explanation

$1,4,16,64,....\equiv 1,4,4^2,4^3,..\equiv 4^{n-1}$

$3,12,48,192,... \equiv 3\times1,3\times4,3\times4^2,3\times4^3...\equiv 3\times4^{n-1}$

$1,3,4,12,16,48,64,192,.. \equiv 1,3\times1,4,3\times4,4^2,3\times4^2,4^3,3\times4^3,..$

$1048576=4^{10} is \space 4^{11-1}$

Number preceding

$3\times4^{10-1}=3.4^9$

Select a figure from the options which will replace the question mark to complete the given series.

$n\in N>1$ , then the sum of real part of roots of

$z^n=(z+1)^n$ is equal to

0%
$\dfrac {n}{2}$
0%
$\dfrac {(n-1)}{2}$
0%
$-\dfrac {n}{2}$
0%
$\dfrac {(1-n)}{2}$

Explanation

${ z }^{ n }={ \left( z+1 \right) }^{ n }\\ { \left( \cfrac { z+1 }{ z } \right) }^{ n }=1\\ \therefore \cfrac { z+1 }{ z } ={ e }^{ i\cfrac { 2\pi m }{ n } },m=0,1,2,........,n-1\\ \cfrac { 1 }{ z } =-2\sin ^{ 2 }{ \left( \cfrac { \pi m }{ n } \right) } +i2\sin { \left( \cfrac { \pi m }{ n } \right) } \cos { \left( \cfrac { \pi m }{ n } \right) } =2i\sin { \left( \cfrac { \pi m }{ n } \right) } \left[ \cos { \left( \cfrac { \pi m }{ n } \right) } +i\sin { \left( \cfrac { \pi m }{ n } \right) } \right] \\ z=\cfrac { \cos { \left( \cfrac { \pi m }{ n } \right) } -i\sin { \left( \cfrac { \pi m }{ n } \right) } }{ 2i\sin { \left( \cfrac { \pi m }{ n } \right) } } =\cfrac { 1 }{ 2 } \left( -1-i\cot { \left( \cfrac { \pi m }{ n } \right) } \right)$

$\therefore$ Sum of real parts of roots =

$n\left( -\cfrac { 1 }{ 2 } \right) =-\cfrac { n }{ 2 }$ (

$\because$ n roots are there)

The value of

$\left(\displaystyle 1-\frac{1}{2}\right)\left(\displaystyle 1-\frac{1}{3}\right)\left(\displaystyle 1-\frac{1}{4}\right).........\left(\displaystyle 1-\frac{1}{10}\right)=$ ___________.

0%
$\displaystyle\frac{10}{11}$
0%
$\displaystyle\frac{1}{9}$
0%
$\displaystyle\frac{1}{10}$
0%
$\displaystyle\frac{1}{2}$

Explanation

$\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)..........\left(1-\dfrac{1}{10}\right)$

$\Rightarrow$

$\dfrac{1}{2}\times \dfrac{2}{3}\times \dfrac{3}{4}......\times \dfrac{8}{9}\times \dfrac{9}{10}$

$\Rightarrow$

$\dfrac{1}{10}$

Sum

$1+3x+5{ x }^{ 2 }+7{ x }^{ 3 }+9{ x }^{ 4 }+.....$ to infinity is

$\left( x<1 \right)$

0%
$\dfrac { 1+x }{ { \left( 1-x \right) }^{ 2 } }$
0%
$\dfrac { 1-x }{ { \left( 1+x \right) }^{ 2 } }$
0%
$\dfrac { 1-x }{ { \left( 1+x \right) } }$
0%
$\dfrac { 1+x }{ { \left( 1-x \right) } }$

Explanation

Let

$S=1+3x+5x^2+7x^3+9x^4+.......$

$xS = x + 3{x^2} + 5{x^3} + 7{x^4} + 9{x^5} + .......$

$S-xS = 1 + \left( {3x - x} \right) + \left( {5{x^2} - 3{x^2}} \right) + \left( {7{x^3} - 5{x^3}} \right) + \left( {9{x^4} - 7{x^4}} \right) + .......$

$S(1-x)=1+2x(1+x+x^2+x^3+....)$

$=1+2x \times \frac{1}{1-x}$

$=\dfrac{x+1}{1-x}$

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

Select the missing number from the given matrix:

5	2	4
4	4	7
2	5	3
18	30	?

0%
$43$
0%
$42$
0%
$33$
0%
$32$

Explanation

In first column $(5+4)\times 2=18$ In second column $(2+4)\times 5=30$

so in third column ans

$=(4+7)\times3$

$=11\times 3=33.$

What is the next term of the AP

$\sqrt{2}, \sqrt{8}, \sqrt{18},$ ____?

0%
$\sqrt{24}$
0%
$\sqrt{28}$
0%
$\sqrt{30}$
0%
$\sqrt{32}$

Explanation

We are given a series :

$\sqrt{2},\sqrt{8},\sqrt{18}$

$a = \sqrt{2}$

$a + d = \sqrt{8} = \sqrt{4.2} = 2\sqrt{2}$

so,

$d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$

so,

$4^{th}$ term will be

$a+3d = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2}$

$a+3d= \sqrt{4^{2}.2} = \sqrt{16.2} = \sqrt{32}$

so, the next term will be

$\sqrt{32}$

1180633_1068792_ans_445800f1816249c3aa6269f24faf99ed.jpg

$A$ is twice as fast as

$B$ is thrice as fast as

$C$ . The journey covered by

$C$ in

$42$ minutes, what will be covered by

$A$ is

0%
$21$ Min
0%
$64$ Min
0%
$17$ Min
0%
$40$ Min

Explanation

time taken ny C to complete the journey

$=$ 42 minutes

time taken by A to complete the journey

$=\dfrac{42}{2}$ minutes (

$\because$ A is twice as fast as C)

$=21$ minutes.

State whether the statement is True/False.

Two numbers between 3 and 81 such that the series

$3,G_{1},G_{2},81$ forms a GP are 9 and 27.

0%
True
0%
False

Explanation

Let

$G_{1}$ and

$G_{2}$ be two numbers between

$3$ and

$81$ such that the series

$3,G_{1},G_{2},81$ forms a GP.

Let

$3$ be the first term and

$r$ be the common ratio of the GP.

$a_4=3r^3$

$81=3r^3$

$r^3=27$

$r=3$

For

$r=3$ ,

$G_{1}=ar=3(3)=9$

$G_{2}=ar^2=3(3)^2=27$

Thus, the required two numbers are 9 and 27.

The first three terms of a sequence are 3,3,6 and each term after the second is the sum of two terms preceding it, the 8th term of the sequence is

0%
15
0%
24
0%
39
0%
63

The sum of the infinite series

$1+\left ( 1+\dfrac{1}{5} \right )\left ( \dfrac{1}{2} \right )+\left ( 1+\dfrac{1}{5}+\dfrac{1}{5^2} \right )\left ( \dfrac{1}{2^2} \right )+.......$

0%
$\dfrac{20}{9}$
0%
$\dfrac{10}{9}$
0%
$\dfrac{5}{9}$
0%
$\dfrac{5}{3}$

Explanation

Let

$\displaystyle S= 1+(1+\frac{1}{5})(\frac{1}{2})+(1+\frac{1}{5}+\frac{1}{5^{2}})(\frac{1}{2^{2}})+...$

$\displaystyle \Rightarrow \frac{S}{2}= \frac{1}{2}+(1+\frac{1}{5})(\frac{1}{2^{2}})+(1+\frac{1}{5}+\frac{1}{5^{2}})(\frac{1}{2^{3}})+...$

$\displaystyle S-\frac{S}{2}= 1+\frac{1}{5}(\frac{1}{2})+\frac{1}{5^{2}}(\frac{1}{2^{2}})+(\frac{1}{5^{3}})(\frac{1}{2^{3}})+...$

$\displaystyle\Rightarrow \frac{S}{2}= 1+\frac{1}{10}+\frac{1}{(10)^{2}}+\frac{1}{(10)^{3}}+...$

$\dfrac{S}{2}= \dfrac{1}{1-\dfrac{1}{10}}= \dfrac{10}{9}$ [infinite G.P sum

$S=\dfrac{a}{1-r}]$

$S= \dfrac{20}{9}$

Option A is correct

The sum of the series

$\sum_{r=0}6{88} \sec r^o\cdot \sec(r+1)^o$ is equal to

0%
$cosec\, 1^o\cdot \tan 1^o$
0%
$cosec\, 1^o\cdot \cot 1^o$
0%
$\sec\, 1^o\cdot \tan 1^o$
0%
$\sec\, 1^o\cdot \cot 1^o$

Let

$S_1=\displaystyle \sum_{K=1}^{n}K,S_2=\sum_{K=1}^{n}K^2$ and

$S_3=\displaystyle \sum_{K=1}^{n}K^3$ , then

$\dfrac{S_{1}^{4}S_{2}^{2}-S_{2}^{2}S_{3}^{2}}{S_{1}^{2}+S_{3}^{2}}$ is equal to

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

Explanation

$\displaystyle S_{1}= \sum k = \frac{n(n+1)}{2}, S_{2}=\sum k^{2}=\frac{n(n+1)(2n+1)}{6}, S_{3}=\sum k^{3} = \frac{n^{2}(n+1)^{2}}{4}$

$\displaystyle \frac{S_{1}^{4}S_{2}^{2}-S_{2}^{2}S_{3}^{2}}{S_{1}^{2}+S_{3}^{2}}$

$\displaystyle = \frac{ \dfrac{n^{4}(n+1)^{4}}{2^{4}}\,.\dfrac{n^{2}(n+1)^{2}(2n+1)^{2}}{36}-\dfrac{n^{2}(n+1)^{2}(2n+1)^{2}}{36}\,.\dfrac{n^{4}(n+1)^{4}}{4^{2}} }{S_{1}^{2}+S_{3}^{2}}$

$\displaystyle= 0$

$\therefore$ option D is correct

Choose the most appropriate option which follows the pattern

0%
0%
0%
0%
None of the above

Find the next term in the sequence.

$117, 389, 525, 593, 627 ?$

0%
644
0%
640
0%
634
0%
630
0%
none of these

958, 833, 733, 658, 608 ?

0%
577
0%
583
0%
567
0%
573
0%
none of these

Explanation

$958, 833, 733, 658, 608, -$

$a_{0}= 958$

$a_{1}= 833 = a_{0}- 125$

$a_{2}= 733=a_{1}-100$

$a_{3}= 658=a_{2}-75$

$a_{4}= 608 = a_{3}-50$

These numbers are in A.P.

Where common difference = +25

Answer =

$a_{4}-25= 608-25=583$

1202751_1249241_ans_10e943ed69c24454a9f9d3688cd034fb.jpg

The value of the expression

$\left(1 + \dfrac{1}{\omega}\right) \left(1 + \dfrac{1}{\omega^2}\right) + \left(2 + \dfrac{1}{\omega}\right) \left(2 + \dfrac{1}{\omega^2}\right) + \left(3 + \dfrac{1}{\omega^2}\right) \left(3 + \dfrac{1}{\omega^2}\right) +.............+ \left(n + \dfrac{1}{\omega}\right) \left(n + \dfrac{1}{\omega^2}\right)$

(

$\omega$ is the root of unity) is

0%
$\dfrac{n(n^2 + 2)}{3}$
0%
$\dfrac{n(n^2 - 2)}{3}$
0%
$\dfrac{n(n^2 + 1)}{3}$
0%
None of these

Explanation

$a_n = \left ( n +\dfrac {1}{\omega } \right )\left ( n +\dfrac {1}{\omega^2} \right )$

$\sum a_n= \sum \left ( n+\dfrac {1}{\omega } \right )\left ( n+\dfrac {1}{\omega^2 } \right )$

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

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

$= \sum n^2 + (\omega+ \omega^2)n+1 \,\,\,\, .............. [\omega^3=1]$

$= \sum n^2 +(-1)n+1 \,\,\,\, ................ [1+\omega+\omega^2=0]$

$= \sum n^2 - n +1$

$= \sum n^2 - \sum n + \sum 1$

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

$= n \left [ \dfrac {2n^2+2n+n+1-3n -3+6}{6} \right ]$

$= n \left [ \dfrac {2n^2+4}{6} \right ]$

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

Find the number suitable for blank in given series.

1,2,3,5,8,13,21,___,55

0%
35
0%
33
0%
36
0%
34

33, 43, 65, 99, 145 ?

0%
201
0%
203
0%
205
0%
211
0%
none of these

Explanation

REF.Image

So Correct Answer is

$145+58= 203$

1207902_1249185_ans_741a3c9b532f484cac2df2ff80365a9a.jpg

Find the missing number

0%
74
0%
62
0%
91
0%
97

Explanation

$\textbf{Step 1: Observing the difference between the adjacent numbers}$

$\text{If we notice the adjacent numbers, we see } 4\times 2 -1 =7$

$\text{Similarly, } 7 \times 2-1=13$

$\text{Similarly, } 13\times2 -1=25$

$\text{Similarly, } 25\times2-1=49$

$\textbf{Step 2: Calculating the missing number}$

$\text{Thus, looking at the above observations, we can conclude the missing number will be:}$

$\Rightarrow 49\times 2-1=97$

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

If D=4 and COVER = 63, then BASIC is equal to

0%
49
0%
50
0%
34
0%
55

$5\times 9=144; 7\times 8=151:4\times 6=102,$ then

$2\times 5=?$

0%
$73$
0%
$77$
0%
$37$
0%
$97$

Explanation

$5\times 9=\underset{(5+9)}{14}\underset{(9-5)}{4}$ ;

$7\times 8=\underset{(7+8)}{15}\underset{(8-7)}{1}$ So,

$2\times 5=\underset{(2+5)}{7}\underset{(5-3)}{3}$

Last digit in

$2^{2^n} +1, n \displaystyle \in N, n \neq 1$ is

0%
$7$
0%
$3$
0%
$5$
0%
$1$

Explanation

$2^{2^n}+1$

$=(4)^n+1$

The last digit of

$4^n$ can be either

$4$ or

$6$ .

Since,

$1$ is added to the number, the last digit of the number is also increased by

$1$ .

Hence, the last digit of

$4^n+1$ can be either

$5$ or

$7$

Last digit of

$2^{2^n}+1, n\in N, n\neq 1$ is

$5$ or

$7$ .

1188600_1332681_ans_b65ce0785647419c80a4c23950168691.jpg

Find the sum of the following geometric series:

$1,-a,a^2,-a^3,...$ to n terms

$\left ( a\neq 1 \right )$

0%
$\dfrac{1+\left ( -a \right )^n}{1+a}$
0%
$\dfrac{1-\left ( a \right )^n}{1+a}$
0%
$\dfrac{1-\left ( -a \right )^n}{1+a}$
0%
$\dfrac{1+\left ( a \right )^n}{1+a}$

Explanation

Let

$S_n$ denote the sum of n terms of the G.P.

$1-a, a^2, -a^3,...$

Cleraly, the given series is a

$G.P.$ with first term

$=a=1$ and common ratio

$=r=-a$

Then,

$S_n = 1 \left \{ \dfrac{\left ( -a \right )^n-1}{-a-1} \right \} = \dfrac{1-\left ( -1 \right )^n a^n}{1+a}$

In a sequence,

$a_{n}=n^{2}-1$ then

$a_{n+1}$ is equal to

0%
$a^{2}-5n$
0%
$n^{2}-2n$
0%
$a^{2}+10n$
0%
$n^{2}+2n$

Explanation

Given,

$a_n=n^2-1$

$a_{n+1}=(n+1)^2-1$

$=n^2+2n+1-1$

$=n^2+2n$

$1.4+2.5+...+\mathrm{n}(\mathrm{n}+3)=$

0%
$\displaystyle \frac{n(n+3)(n+5)}{9}$
0%
$\displaystyle \frac{n(n+1)(n+5)}{3}$
0%
$\displaystyle \frac{n(n+5)(n+7)}{6}$
0%
$\displaystyle \frac{n(n+3)(n+9)}{12}$

Explanation

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

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

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

$=\dfrac{n(n+1)(n+5)}{3}$

$1+ 3 + 6 + 10 + ...+\displaystyle \frac{(n-1)n}{2}+\frac{n(n+1)}{2}=$

0%
$\displaystyle \frac{n(n+1)(n+2)}{3}$
0%
$\displaystyle \frac{(n+1)(n+2)}{6}$
0%
$\displaystyle \frac{n(n+1)(n+2)}{6}$
0%
$\displaystyle \frac{(n+2)(n+1)^{2}}{3}$

Explanation

We can see that

${ T }_{ n }=\dfrac { n\left( n+1 \right) }{ 2 }$
hence

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

Sum of the series

$S=1+\dfrac{1}{2} \left ( 1+2 \right )+\dfrac{1}{3}\left ( 1+2+3 \right )+\dfrac{1}{4}\left ( 1+2+3+4 \right )+...$ upto

$20$ terms is

0%
$110.5$
0%
$111$
0%
$115$
0%
$116.5$

Explanation

The general term for the above series is

$T_{n}=\dfrac{\sum n}{n}$

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

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

Hence,

$S_{20}= \displaystyle \sum_{n=1} ^{n=20} T_{n}$

$\displaystyle \sum_{n=1} ^{n=20} \left (\dfrac{n+1}{2}\right)$

$\Rightarrow \left [1+\dfrac{3}{2}+\dfrac{4}{2}+.....+\dfrac{21}{2} \right ]$

$\Rightarrow \left [\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+.....+\dfrac{21}{2} \right ]$

It is an A.P. with

$a=\dfrac{2}{2}, d=\dfrac{1}{2}, \ and \ n=20$

We know that the sum of

$n$ terms of A.P. is

$S_n=\dfrac{n}{2}\left [2a+(n-1)d \right]$

$\therefore \ S_{20}=\dfrac{20}{2}\left [2\times \dfrac{2}{2}+(20-1)\dfrac{1}{2} \right]$

$\Rightarrow S_{20}=10\left[2+\dfrac{19}{2}\right]=\dfrac{230}{2}=115$

Hence, the answer is

$115$

$S_1,S_2$ and

$S_3$ . are the sums of first n natural numbers, their squares and their cubes respectively, then

$S_3\left (1+8S_1 \right )=$

0%
$S_{2}^{2}$
0%
$9S_2$
0%
$9S_{2}^{2}$
0%
None

Explanation

$S_1\; =\; \dfrac{n(n+1)}{2}$

$1+ 8S_1 = 4n(n+1) +1 = (2n+1)^2$

$S_2 = \dfrac{n(n+1)(2n+1)}{6}$

$\displaystyle S_3 = \left(\frac{n(n+1)}{2} \right)^2$

Now,

$S_3 \times (1+ 8 S_1)$

$= \dfrac{(2n+1)^2(n^2)(n+1)^2}{4}$

$= 9 \times \left (\dfrac{n(n+1)(2n+1)}{6} \right)^2$

$= 9 (S_2)^2$
Hence, option C is correct.

$3.6+6.9+9.12+...+3\mathrm{n}(3\mathrm{n}+3)=$

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

Explanation

$\sum _{ k=1 }^{ n }{3k(3k+3)}=9\sum _{ k=1 }^{ n }{k^2} +9\sum _{ k=1 }^{ n }{k}$

$=9\times \dfrac{n(n+1)(2n+1)}{6}+9\times \dfrac{n(n+1)}{2}$

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

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

$2\cdot1^{2}+3\cdot2^{2}+4\cdot3^{2}+\dots$ up to

$n$ terms

$=$

0%
$\displaystyle \frac{n(n+1)(n+2)(3n+1)}{12}$
0%
$\displaystyle \frac{n(n+2)(n+3)(n+11)}{12}$
0%
$\displaystyle \frac{n(n+1)(n+2)(3n-2)}{6}$
0%
$\displaystyle \frac{n(n+2)(n+5)(n+8)}{6}$

Explanation

Let

$S$ be the given sequence.
General term,

$t_{r}=(r+1)\cdot r^2=r^3+r^2$
Thus, the sum would be,

$S=\sum r^3 +\sum r^2$

$\therefore S=\dfrac{n^2(n+1)^2}{4} + \dfrac{n(n+1)(2n+1)}{6}$
Hence,

$S=\dfrac{n(n+1)(3n^2+7n+2)}{12}$
or

$S=\dfrac{n(n+1)(n+2)(3n+1)}{12}$

The sum to infinity of

$\displaystyle\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+...$ is

0%
$\dfrac{1}{5}$
0%
$\dfrac{7}{24}$
0%
$\dfrac{5}{48}$
0%
$\dfrac{3}{16}$

Explanation

$\dfrac{1}{7}+\dfrac{2}{7^{2}}+\dfrac{1}{7^{3}}+\dfrac{2}{7^{4}}...\infty$

$=(\dfrac{1}{7}+\dfrac{1}{7^{3}}+\dfrac{1}{7^{5}}+...\infty)+(\dfrac{2}{7^{2}}+\dfrac{2}{7^{4}}+\dfrac{2}{7^{6}}+...\infty)$

$=\dfrac{\dfrac{1}{7}}{1-\dfrac{1}{7^{2}}}+2\left[\dfrac{\dfrac{1}{7^{2}}}{1-\dfrac{1}{7^{2}}}\right]$

$[$ Sum of infinite GP

$=\cfrac { a }{ 1-r }]$

$=\dfrac{7}{48}+\dfrac{2}{48}$

$=\dfrac{9}{48}$

$=\dfrac{3}{16}$

$1.3+3.5+5.7+...+(2\mathrm{n}-1)(2\mathrm{n}+1)=$

0%
$\displaystyle \frac{n(4n^{2}+6n-1)}{3}$
0%
$\displaystyle \frac{n(3n^{2}+5n+1)}{3}$
0%
$\displaystyle \frac{n(5n^{2}+7n-1)}{3}$
0%
$\displaystyle \frac{n(7n^{2}-5n+1)}{3}$

Explanation

General term :

$(2k-1)(2k+1)$

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

$=\sum_{k=1}^{n}(4k^2-1)$

$=\sum_{k=1}^{n}(4k^2)-\sum_{k=1}^{n}(1)$

By using the formula,

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

On simplifying, we get the desired result !

$1.4+2.7+3.10+...+\mathrm{n}(3\mathrm{n}+1)=$

0%
$n(2n+1)^{2}$
0%
$2\mathrm{n}(\mathrm{2n}+1)^{2}$
0%
$\mathrm{n}(\mathrm{n}+1)^{2}$
0%
$(\mathrm{n}+2)(\mathrm{n}+3)$

Explanation

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

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

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

$=n(n+1)^2$

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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