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NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

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NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

Edited By Vishal kumar | Updated on Nov 17, 2023 08:45 AM IST

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- Download Free PDF


NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- In this article, you will get NCERT solutions for Class 11 Maths Chapter 9 miscellaneous exercise. The miscellaneous exercise consists of mixed kinds of questions from all the topics covered in this chapter. You must have solved the previous exercise of this chapter. Now you can try to solve problems from the miscellaneous exercise chapter 9 Class 11.

As you have already familiar with the NCERT syllabus Class 11 Maths chapter 9 concepts, you can try to solve problems from miscellaneous exercise chapter 9 Class 11. These problems are a bit lengthy and complex as compared to the other exercise of this chapter but these problems will check your understanding of this chapter. You don't need to worry if you are not able to solve these problems by yourself at first. Class 11 Maths Chapter 9 miscellaneous exercise solutions are here to help you where you will get detailed miscellaneous solutions. The class 11 maths ch 9 miscellaneous exercise solutions Maths Miscellaneous Exercises are meticulously crafted by subject experts from Careers360. Each step is explained in detail, providing comprehensive and well-explained solutions. Furthermore, the availability of a PDF version of class 11 chapter 9 maths miscellaneous solutions allows students to access the solutions offline, providing flexibility and convenience to study according to their preferences. You can check NCERT Solutions link if you are looking for NCERT solutions for all the Classes at one place.

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** As per the CBSE Syllabus for the academic year 2023-24, it has been noted that miscellaneous exercise class 11 chapter 9 has been renumbered and is now recognized as Chapter 8.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise

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Access Sequences And Series Class 11 Chapter 9:Miscellaneous Exercise

Question:1 Show that the sum of ( m+n)^{th} and ( m-n)^{th} terms of an A.P. is equal to twice the m^{th}term.

Answer:

Let a be first term and d be common difference of AP.

Kth term of a AP is given by,

a_k=a+(k-1)d

\therefore a_{m+n}=a+(m+n-1)d

\therefore a_{m-n}=a+(m-n-1)d

a_m=a+(m-1)d

a_{m+n}+ a_{m-n}=a+(m+n-1)d+a+(m-n-1)d

=2a+(m+n-1+m-n-1)d

=2a+(2m-2)d

=2(a+(m-1)d)

=2.a_m

Hence, the sum of ( m+n)^{th} and ( m-n)^{th} terms of an A.P. is equal to twice the m^{th}term.

Question:2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Answer:

Let three numbers of AP are a-d, a, a+d.

According to given information ,

a-d+a+a+d=24

3a=24

\Rightarrow a=8

(a-d)a(a+d)=440

\Rightarrow (8-d)8(8+d)=440

\Rightarrow (8-d)(8+d)=55

\Rightarrow (8^2-d^2)=55

\Rightarrow (64-d^2)=55

\Rightarrow d^2=64-55=9

\Rightarrow d=\pm 3

When d=3, AP= 5,8,11 also if d=-3 ,AP =11,8,5.

Thus, three numbers are 5,8,11.

Question:3 Let the sum of n, 2n, 3n terms of an A.P. be S_1 , S_2 , S_3, respectively, show that S_3 = 3(S_2 - S_1)

Answer:

Let a be first term and d be common difference of AP.

S_n=\frac{n}{2}[2a+(n-1)d]=S_1..................................1

S_2n=\frac{2n}{2}[2a+(2n-1)d]=S_2..................................2

S_2_n=\frac{3n}{2}[2a+(3n-1)d]=S_3..................................3

Subtract equation 1 from 2,

S_2-S_1=\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d]

=\frac{n}{2}[4a+4nd-2d-2a-nd+d]

=\frac{n}{2}[2a+3nd-d]

=\frac{n}{2}[2a+(3n-1)d]

\therefore 3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1)d]=S_3

Hence, the result is proved.

Question:4 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Answer:

Numbers divisible by 7 from 200 to 400 are 203,210,.............399

This sequence is an A.P.

Here , first term =a =203

common difference = 7.

We know , a_n = a+(n-1)d

399 = 203+(n-1)7

\Rightarrow \, \, 196 = (n-1)7

\Rightarrow \, \, 28 = (n-1)

\Rightarrow \, \, n=28+1=29

S_n = \frac{n}{2}[2a+(n-1)d]

= \frac{29}{2}[2(203)+(29-1)7]

= \frac{29}{2}[2(203)+28(7)]

= 29\times 301

= 8729

The sum of numbers divisible by 7from 200 to 400 is 8729.

Question:5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer:

Numbers divisible by 2 from 1 to 100 are 2,4,6................100

This sequence is an A.P.

Here , first term =a =2

common difference = 2.

We know , a_n = a+(n-1)d

100= 2+(n-1)2

\Rightarrow \, \, 98 = (n-1)2

\Rightarrow \, \, 49 = (n-1)

\Rightarrow \, \, n=49+1=50

S_n = \frac{n}{2}[2a+(n-1)d]

= \frac{50}{2}[2(2)+(50-1)2]

= \frac{50}{2}[2(2)+49(2)]

= 25\times 102

= 2550

Numbers divisible by 5 from 1 to 100 are 5,10,15................100

This sequence is an A.P.

Here , first term =a =5

common difference = 5.

We know , a_n = a+(n-1)d

100= 5+(n-1)5

\Rightarrow \, \, 95 = (n-1)5

\Rightarrow \, \, 19 = (n-1)

\Rightarrow \, \, n=19+1=20

S_n = \frac{n}{2}[2a+(n-1)d]

= \frac{20}{2}[2(5)+(20-1)5]

= \frac{20}{2}[2(5)+19(5)]

= 10\times 105=1050

Numbers divisible by both 2 and 5 from 1 to 100 are 10,20,30................100

This sequence is an A.P.

Here , first term =a =10

common difference = 10

We know , a_n = a+(n-1)d

100= 10+(n-1)10

\Rightarrow \, \, 90 = (n-1)10

\Rightarrow \, \, 9 = (n-1)

\Rightarrow \, \, n=9+1=10

S_n = \frac{n}{2}[2a+(n-1)d]

= \frac{10}{2}[2(10)+(10-1)10]

= \frac{10}{2}[2(10)+9(10)]

= 5\times 110=550

\therefore Required \, \, sum=2550+1050-550=3050

Thus, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.

Question:6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer:

Numbers divisible by 4, yield remainder as 1 from 10 to 100 are 13,17,..................97

This sequence is an A.P.

Here , first term =a =13

common difference = 4.

We know , a_n = a+(n-1)d

97 = 13+(n-1)4

\Rightarrow \, \, 84 = (n-1)4

\Rightarrow \, \, 21 = (n-1)

\Rightarrow \, \, n=21+1=22

S_n = \frac{n}{2}[2a+(n-1)d]

= \frac{22}{2}[2(13)+(22-1)4]

= \frac{22}{2}[2(13)+21(4)]

= 11\times 110

=1210

The sum of numbers divisible by 4 yield 1 as remainder from 10 to 100 is 1210.

Question:7 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y \epsilon N such that f(1) = 3 and

\sum_{x=1}^{n} f(x) = 120 , find the value of n.

Answer:

Given : f (x +y) = f(x) f(y) for all x, y \epsilon N such that f(1) = 3

f(1) = 3

Taking x=y=1 , we have

f(1+1)=f(2)=f(1)*f(1)=3*3=9

f(1+1+1)=f(1+2)=f(1)*f(2)=3*9=27

f(1+1+1+1)=f(1+3)=f(1)*f(3)=3*27=81

f(1),f(2),f(3),f(4)..................... is 3,9,27,81,.............................. forms a GP with first term=3 and common ratio = 3.

\sum_{x=1}^{n} f(x) = 120=S_n

S_n=\frac{a(1-r^n)}{1-r}

120=\frac{3(1-3^n)}{1-3}

40=\frac{(1-3^n)}{-2}

-80=(1-3^n)

-80-1=(-3^n)

-81=(-3^n)

3^n=81

Therefore, n=4

Thus, value of n is 4.

Question:8 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Answer:

Let the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2

S_n=\frac{a(1-r^n)}{1-r}

315=\frac{5(1-2^n)}{1-2}

63=\frac{(1-2^n)}{-1}

-63=(1-2^n)

-63-1=(-2^n)

-64=(-2^n)

2^n=64

Therefore, n=6

Thus, the value of n is 6.

Last term of GP=6th term=a.r^{n-1}=5.2^5=5*32=160

The last term of GP =160

Question:9 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Answer:

Given: The first term of a G.P. is 1. The sum of the third term and fifth term is 90.

a=1

a_3=a.r^2=r^2 a_5=a.r^4=r^4

\therefore \, \, r^2+r^4=90

\therefore \, \, r^4+r^2-90=0

\Rightarrow \, \, r^2=\frac{-1\pm \sqrt{1+360}}{2}

r^2=\frac{-1\pm \sqrt{361}}{2}r^2=-10 \, or \, 9

r=\pm 3

Thus, the common ratio of GP is \pm 3.

Question:10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Answer:

Let three terms of GP be a,ar,ar^2.

Then, we have a+ar+ar^2=56

a(1+r+r^2)=56...............................................1

a-1,ar-7,ar^2-21 from an AP.

\therefore ar-7-(a-1)=ar^2-21-(ar-7)

ar-7-a+1=ar^2-21-ar+7

ar-6-a=ar^2-14-ar

\Rightarrow ar^2-2ar+a=8

\Rightarrow ar^2-ar-ar+a=8
\Rightarrow a(r^2-2r+1)=8

\Rightarrow a(r^2-1)^2=8....................................................................2

From equation 1 and 2, we get

\Rightarrow 7(r^2-2r+1)=1+r+r^2

\Rightarrow 7r^2-14r+7-1-r-r^2=0

\Rightarrow 6r^2-15r+6=0

\Rightarrow 2r^2-5r+2=0

\Rightarrow 2r^2-4r-r+2=0

\Rightarrow 2r(r-2)-1(r-2)=0

\Rightarrow (r-2)(2r-1)=0

\Rightarrow r=2,r=\frac{1}{2}

If r=2, GP = 8,16,32

If r=0.2, GP= 32,16,8.

Thus, the numbers required are 8,16,32.

Question:11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Answer:

Let GP be A_1,A_2,A_3,...................A_2_n

Number of terms = 2n

According to the given condition,

(A_1,A_2,A_3,...................A_2_n)=5(A_1,A_3,...................A_{2_n-1})

\Rightarrow (A_1,A_2,A_3,...................A_2_n)-5(A_1,A_3,...................A_{2n-1})=0

\Rightarrow (A_2,A_4,A_6,...................A_2_n)=4(A_1,A_3,...................A_{2n-1})

Let the be GP as a,ar,ar^2,..................

\Rightarrow \frac{ar(r^n-1)}{r-1}=\frac{4.a(r^{n}-1)}{r-1}

\Rightarrow ar=4a

\Rightarrow r=4

Thus, the common ratio is 4.

Question:12 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Answer:

Given : first term =a=11

Let AP be 11,11+d,11+2d,11+3d,...........................................11+(n-1)d

Given: The sum of the first four terms of an A.P. is 56.

11+11+d+11+2d+11+3d=56

\Rightarrow 44+6d=56

\Rightarrow 6d=56-44=12

\Rightarrow 6d=12

\Rightarrow d=2

Also, The sum of the last four terms is 112.

11+(n-4)d+11+(n-3)d+11+(n-2)d+11+(n-1)d=112\Rightarrow 44+(n-4)2+(n-3)2+(n-2)2+(n-1)2=112

\Rightarrow 44+2n-8+2n-6+2n-4+2n-2=112

\Rightarrow 44+8n-20=112

\Rightarrow 24+8n=112

\Rightarrow 8n=112-24

\Rightarrow 8n=88

\Rightarrow n=11

Thus, the number of terms of AP is 11.

Question:13 If \frac{a+ bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} (x\neq 0 ) then show that a, b, c and d are in G.P.

Answer:

Given :

\frac{a+ bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} (x\neq 0 )

Taking ,

\frac{a+ bx}{a-bx} = \frac{b+cx}{b-cx}

\Rightarrow (a+ bx) (b-cx) = (b+cx) (a-bx)

\Rightarrow ab+ b^2x-bcx^2-acx) = ba-b^2x+acx-bcx^2

\Rightarrow 2 b^2x = 2acx

\Rightarrow b^2 = ac

\Rightarrow \frac{b}{a}=\frac{c}{b}..................1

Taking,

\frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}

\Rightarrow (b+cx)(c-dx)=(c+dx)(b-cx)

\Rightarrow bc-bdx+c^2x-cdx^2=bc-c^2x+bdx-cdx^2

\Rightarrow 2bdx=2c^2x

\Rightarrow bd=c^2

\Rightarrow \frac{d}{c}=\frac{c}{b}..............................2

From equation 1 and 2 , we have

\Rightarrow \frac{d}{c}=\frac{c}{b}=\frac{b}{a}

Thus, a,b,c,d are in GP.

Question:14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P^2 R^n = S ^n

Answer:

Ler there be a GP =a,ar,ar^2,ar^3,....................

According to given information,

S=\frac{a(r^n-1)}{r-1}

P=a^n \times r^{(1+2+...................n-1)}

P=a^n \times r^{\frac{n(n-1)}{2}}

R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+..................\frac{1}{ar^{n-1}}

R=\frac{r^{n-1}+r^{n-2}+r^{n-3}+..............r+1}{a.r^{n-1}}

R=\frac{1}{a.r^{n-1}}\times \frac{1(r^n-1)}{r-1}

R= \frac{(r^n-1)}{a.r^{n-1}.(r-1)}

To prove : P^2 R^n = S ^n

LHS : P^2 R^n

= a^{2n}.r^{n(n-1)}\frac{(r^n-1)^n}{a^n.r^{n(n-1)}.(r-1)^n}

= a^{n} \frac{(r^n-1)^n}{(r-1)^n}

= \left ( \frac{a(r^n-1)}{(r-1)} \right )^{n}

=S^n=RHS

Hence proved

Question:15 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that

(q - r )a + (r - p )b + (p - q )c = 0

Answer:

Given: The pth, qth and rth terms of an A.P. are a, b, c, respectively.

To prove : (q - r )a + (r - p )b + (p - q )c = 0

Let the first term of AP be 't' and common difference be d

a_p=t+(p-1)d=a...............................1

a_q=t+(q-1)d=b...............................2

a_r=t+(r-1)d=c...............................3

Subtracting equation 2 from 1, we get

(p-1-q+1)d=a-b

\Rightarrow (p-q)d=a-b

\Rightarrow d=\frac{a-b}{p-q}....................................4

Subtracting equation 3 from 2, we get

(q-1-r+1)d=b-c

\Rightarrow (q-r)d=b-c

\Rightarrow d=\frac{b-c}{q-r}....................................5

Equating values of d, from equation 4 and 5, we have

d=\frac{a-b}{p-q}=\frac{b-c}{q-r}.

\Rightarrow \frac{a-b}{p-q}=\frac{b-c}{q-r}.

\Rightarrow (a-b)(q-r)=(b-c)(p-q)

\Rightarrow aq-ar-bq+br=bp-bq-cp+cq

\Rightarrow aq-ar+br=bp-cp+cq

\Rightarrow aq-ar+br-bp+cp-cq=0

\Rightarrow a(q-r)+b(r-p)+c(p-q)=0

Hence proved.

Question:16 If a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b}) are in A.P., prove that a, b, c are in A.P.

Answer:

Given:a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b}) are in A.P.

\therefore \, \, \, b ( \frac{1}{c}+\frac{1}{a})-a (\frac{1}{b}+\frac{1}{c}) = c ( \frac{1}{a}+ \frac{1}{b})- b ( \frac{1}{c}+\frac{1}{a})

\therefore \, \, \, ( \frac{b(a+c)}{ac})- (\frac{a(c+b)}{bc}) = ( \frac{c(b+a)}{ab})- ( \frac{b(a+c)}{ac})

\therefore \, \, \, ( \frac{ab^2+b^2c-a^2c-a^2b}{abc}) = ( \frac{c^2b+c^2a-b^2a-b^2c}{abc})

\Rightarrow \, \, \, ab^2+b^2c-a^2c-a^2b= c^2b+c^2a-b^2a-b^2c

\Rightarrow \, \, \, ab(b-a)+c(b^2-a^2)= a(c^2-b^2)+bc(c-b)

\Rightarrow \, \, \, ab(b-a)+c(b-a)(b+a)= a(c-b)(c+b)+bc(c-b)

\Rightarrow \, \, \, (b-a)(ab+c(b+a))= (c-b)(a(c+b)+bc)

\Rightarrow \, \, \, (b-a)(ab+cb+ac)= (c-b)(ac+ab+bc)

\Rightarrow \, \, \, (b-a)= (c-b)

Thus, a,b,c are in AP.

Question:17 If a, b, c, d are in G.P, prove that (a^n + b^n), (b^n + c^n), (c^n + d^n) are in G.P.

Answer:

Given: a, b, c, d are in G.P.

To prove:(a^n + b^n), (b^n + c^n), (c^n + d^n) are in G.P.

Then we can write,

b^2=ac...............................1

c^2=bd...............................2

ad=bc...............................3

Let (a^n + b^n), (b^n + c^n), (c^n + d^n) be in GP

(b^n + c^n)^2 =(a^n + b^n) (c^n + d^n)

LHS: (b^n + c^n)^2

(b^n + c^n)^2 =b^{2n}+c^{2n}+2b^nc^n

(b^n + c^n)^2 =(b^2)^n+(c^2)^n+2b^nc^n

=(ac)^n+(bd)^n+2b^nc^n

=a^nc^n+b^nc^n+a^nd^n+b^nd^n

=c^n(a^n+b^n)+d^n(a^n+b^n)

=(a^n+b^n)(c^n+d^n)=RHS

Hence proved

Thus,(a^n + b^n), (b^n + c^n), (c^n + d^n) are in GP

Question:18 If a and b are the roots of x^2 -3 x + p = 0 and c, d are roots of x^2 -12 x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

Answer:

Given: a and b are the roots of x^2 -3 x + p = 0

Then, a+b=3\, \, \, \, and\, \, \, \, ab=p......................1

Also, c, d are roots of x^2 -12 x + q = 0

c+d=12\, \, \, \, and\, \, \, \, cd=q......................2

Given: a, b, c, d form a G.P

Let, a=x,b=xr,c=xr^2,d=xr^3

From 1 and 2, we get

x+xr=3 and xr^2+xr^3=12

\Rightarrow x(1+r)=3 xr^2(1+r)=12

On dividing them,

\frac{xr^2(1+r)}{x(1+r)}=\frac{12}{3}

\Rightarrow r^2=4

\Rightarrow r=\pm 2

When , r=2 ,

x=\frac{3}{1+2}=1

When , r=-2,

x=\frac{3}{1-2}=-3

CASE (1) when r=2 and x=1,

ab=x^2r=2\, \, \, and \, \, \, \, cd=x^2r^5=32

\therefore \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}

i.e. (q + p) : (q – p) = 17:15.

CASE (2) when r=-2 and x=-3,

ab=x^2r=-18\, \, \, and \, \, \, \, cd=x^2r^5=-288

\therefore \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-305}{-270}=\frac{17}{15}

i.e. (q + p) : (q – p) = 17:15.

Question:19 The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a: b = \left ( m + \sqrt {m^2 -n^2 }\right ) : \left ( m- \sqrt {m^2 - n^2} \right )

Answer:

Let two numbers be a and b.

AM=\frac{a+b}{2}\, \, \, and\, \, \, \, \, GM=\sqrt{ab}

According to the given condition,

\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}

\Rightarrow \frac{(a+b)^2}{4ab}=\frac{m^2}{n^2}

\Rightarrow (a+b)^2=\frac{4ab.m^2}{n^2}

\Rightarrow (a+b)=\frac{2\sqrt{ab}.m}{n}...................................................................1

(a-b)^2=(a+b)^2-4ab

We get,

(a-b)^2=\left ( \frac{4abm^2}{n^2} \right )-4ab

(a-b)^2=\left ( \frac{4abm^2-4abn^2}{n^2} \right )

\Rightarrow (a-b)^2=\left ( \frac{4ab(m^2-n^2)}{n^2} \right )

\Rightarrow (a-b)=\left ( \frac{2\sqrt{ab}\sqrt{(m^2-n^2)}}{n} \right ).....................................................2

From 1 and 2, we get

2a =\left ( \frac{2\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )

a =\left ( \frac{\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )

Putting the value of a in equation 1, we have

b=\left ( \frac{2.\sqrt{ab}}{n} \right ) m-\left ( \frac{\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )

b=\left ( \frac{\sqrt{ab}}{n} \right ) m-\left ( \frac{\sqrt{ab}}{n} \right )\left ( \sqrt{(m^2-n^2)} \right )

=\left ( \frac{\sqrt{ab}}{n} \right )\left (m- \sqrt{(m^2-n^2)} \right )

\therefore a:b=\frac{a}{b}=\frac{\left ( \frac{\sqrt{ab}}{n} \right )\left (m+ \sqrt{(m^2-n^2)} \right )}{\left ( \frac{\sqrt{ab}}{n} \right )\left (m- \sqrt{(m^2-n^2)} \right )}

=\frac{\left (m+ \sqrt{(m^2-n^2)} \right )}{\left (m- \sqrt{(m^2-n^2)} \right )}

a:b=\left (m+ \sqrt{(m^2-n^2)} \right ) : \left (m- \sqrt{(m^2-n^2)} \right )

Question:20 If a, b, c are in A.P.; b, c, d are in G.P. and 1/c , 1/d , 1/e are in A.P. prove that a, c, e are in G.P.

Answer:

Given: a, b, c are in A.P

b-a=c-b..............................1

Also, b, c, d are in G.P.

c^2=bd..............................2

Also, 1/c, 1/d, 1/e are in A.P

\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}...........................3

To prove: a, c, e are in G.P. i.e.c^2=ae

From 1, we get 2b=a+c

b=\frac{a+c}{2}

From 2, we get

d=\frac{c^2}{b}

Putting values of b and d, we get

\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}

\frac{2}{d}=\frac{1}{c}+\frac{1}{e}

\Rightarrow \frac{2b}{c^2}=\frac{1}{c}+\frac{1}{e}

\Rightarrow \frac{2(a+c)}{2c^2}=\frac{1}{c}+\frac{1}{e}

\Rightarrow \frac{(a+c)}{c^2}=\frac{e+c}{ce}

\Rightarrow \frac{(a+c)}{c}=\frac{e+c}{e}

\Rightarrow e(a+c)=c(e+c)

\Rightarrow ea+ec=ec+c^2

\Rightarrow ea=c^2

Thus, a, c, e are in G.P.

Question:21(i) Find the sum of the following series up to n terms: 5 + 55+ 555 + ....

Answer:

5 + 55+ 555 + .... is not a GP.

It can be changed in GP by writing terms as

S_n=5 + 55+ 555 + .... to n terms

S_n=\frac{5}{9}[9+99+999+9999+................]

S_n=\frac{5}{9}[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+................]

S_n=\frac{5}{9}[(10+10^2+10^3+........)-(1+1+1.....................)]

S_n=\frac{5}{9}[\frac{10(10^n-1)}{10-1}-(n)]

S_n=\frac{5}{9}[\frac{10(10^n-1)}{9}-(n)]

S_n=\frac{50}{81}[(10^n-1)]-\frac{5n}{9}

Thus, the sum is

S_n=\frac{50}{81}[(10^n-1)]-\frac{5n}{9}

Question:21(ii) Find the sum of the following series up to n terms: .6 +. 66 +. 666+…

Answer:

Sum of 0.6 +0. 66 + 0. 666+….................

It can be written as

S_n=0.6+0.66+0.666+.......................... to n terms

S_n=6[0.1+0.11+0.111+0.1111+................]

S_n=\frac{6}{9}[0.9+0.99+0.999+0.9999+................]

S_n=\frac{6}{9}[(1-\frac{1}{10})+(1-\frac{1}{10^2})+(1-\frac{1}{10^3})+(1-\frac{1}{10^4})+................]

S_n=\frac{2}{3}[(1+1+1.....................n\, terms)-\frac{1}{10}(1+\frac{1}{10}+\frac{1}{10^2}+...................n \, terms)]

S_n=\frac{2}{3}[n-\frac{\frac{1}{10}(\frac{1}{10}^n-1)}{\frac{1}{10}-1}]

S_n=\frac{2n}{3}-\frac{2}{30}[\frac{10(1-10^-^n)}{9}]

S_n=\frac{2n}{3}-\frac{2}{27}(1-10^-^n)

Question:22 Find the 20th term of the series 2 \times 4+4\times 6+\times 6\times 8+....+n terms.

Answer:

the series = 2 \times 4+4\times 6+\times 6\times 8+....+n

\therefore n^{th}\, term=a_n=2n(2n+2)=4n^2+4n

\therefore a_2_0=2(20)[2(20)+2]

=40[40+2]

=40[42]

=1680

Thus, the 20th term of series is 1680

Question:23 Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…

Answer:

The series: 3+ 7 +13 +21 +31 +…..............

n th term = n^2+n+1=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k^{2}+k+1

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

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

=n\left ( \frac{(n+1)(2n+1)}{6}+\frac{n+1}{2}+1 \right )

=n\left ( \frac{(n+1)(2n+1)+3(n+1)+6}{6} \right )

=n\left ( \frac{2n^2+n+2n+1+3n+3+6}{6} \right )

=n\left ( \frac{2n^2+6n+10}{6} \right )

=n\left ( \frac{n^2+3n+5}{3} \right )

Question:24 If S_1 , S_2 , S_3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9 S ^2 _2 = S_3 ( 1+ 8 S_1)

Answer:

To prove : 9 S ^2 _2 = S_3 ( 1+ 8 S_1)

From given information,

S_1=\frac{n(n+1)}{2}

S_3=\frac{n^2(n+1)^2}{4}

Here ,RHS= S_3 ( 1+ 8 S_1)

\Rightarrow S_3 ( 1+ 8 S_1)=\frac{n^2(n+1)^2}{4}\left ( 1+8\frac{n(n+1)}{2} \right )

=\frac{n^2(n+1)^2}{4}\left ( 1+4.n(n+1) \right )

=\frac{n^2(n+1)^2}{4}\left ( 1+4n^2+4n \right )

=\frac{n^2(n+1)^2}{4}\left ( 2n+1 \right )^2

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

Also, RHS=9S_2^2

\Rightarrow 9S_2^2=\frac{9\left [ n(n+2)(2n+1) \right ]^2}{6^2}

=\frac{9\left [ n(n+2)(2n+1) \right ]^2}{36}

=\frac{\left [ n(n+2)(2n+1) \right ]^2}{4}..........................................2

From equation 1 and 2 , we have

9S_2^2=S_3(1+8S_1)=\frac{\left [ n(n+2)(2n+1) \right ]^2}{4}

Hence proved .

Question:25 Find the sum of the following series up to n terms: \frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ...

Answer:

n term of series :

\frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ........=\frac{1^3+2^3+3^3+..........n^3}{1+3+5+...........(2n-1)}

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

Here, 1,3,5............(2n-1) are in AP with first term =a=1 , last term = 2n-1, number of terms =n


1+3+5............(2n-1)=\frac{n}{2}\left [ 2(1)+(n-1)2 \right ]

=\frac{n}{2}\left [ 2+2n-2 \right ]=n^2

a_n=\frac{n^2(n+1)^2}{4n^2}

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

=\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4}

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{(k+1)^2}{4}

=\frac{1}{4}\sum _{k=1}^{n} k^2+\frac{1}{2}\sum _{k=1}^{n} k+\sum _{k=1}^{n}\frac{1}{4}

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

=n\left ( \frac{(n+1)(2n+1)}{24}+\frac{n+1}{4}+\frac{1}{4} \right )

=n\left ( \frac{(n+1)(2n+1)+6(n+1)+6}{24} \right )

=n\left ( \frac{2n^2+n+2n+1+6n+6+6}{24} \right )

=n\left ( \frac{2n^2+9n+13}{24} \right )

Question:26 Show that \frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}

Answer:

To prove :

\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}

the nth term of numerator =n(n+1)^2=n^3+2n^2+n

nth term of the denominator =n^2(n+1)=n^3+n^2

RHS:\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)}..........................1

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

Numerator :

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

=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{2.n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}

=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}

=\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+1)

=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+8n+4+6}{6} \right )

=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+11n+10}{6} \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+11n+10 \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+6n+5n+10 \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n(n+2)+5(n+2) \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ((n+2)(3n+5) \right )

=\frac{n(n+1)(n+2)(3n+5)}{12}........................................................2

Denominator :

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

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

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

=\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{2n+1}{3})

=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+4n+2}{6} \right )

=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+7n+2}{6} \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+7n+2 \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+6n+n+2 \right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n(n+2)+1(n+2)\right )

=\left [ \frac{n(n+1)}{12} \right ] \left ( (n+2)(3n+1)\right )

=\left [ \frac{n(n+1)(n+2)(3n+1)}{12} \right ].....................................3

From equation 1,2,3,we have

\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} =\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}

=\frac{3n+5}{3n+1}

Hence, the above expression is proved.

Question:27 A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

Answer:

Given : Farmer pays Rs 6000 cash.

Therefore , unpaid amount = 12000-6000=Rs. 6000

According to given condition, interest paid annually is

12% of 6000,12% of 5500,12% of 5000,......................12% of 500.

Thus, total interest to be paid

=12\%of\, 6000+12\%of\, 5500+.............12\%of\, 500

=12\%of\, (6000+ 5500+.............+ 500)

=12\%of\, (500+ 1000+.............+ 6000)

Here, 500, 1000,.............5500,6000 is a AP with first term =a=500 and common difference =d = 500

We know that a_n=a+(n-1)d

\Rightarrow 6000=500+(n-1)500

\Rightarrow 5500=(n-1)500

\Rightarrow 11=(n-1)

\Rightarrow n=11+1=12

Sum of AP:

S_1_2=\frac{12}{2}\left [ 2(500)+(12-1)500 \right ]

S_1_2=6\left [ 1000+5500 \right ]

=6\left [ 6500 \right ]

=39000

Thus, interest to be paid :

=12\%of\, (500+ 1000+.............+ 6000)

=12\%of\, ( 39000)

=Rs. 4680

Thus, cost of tractor = Rs. 12000+ Rs. 4680 = Rs. 16680

Question:28 Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Answer:

Given: Shamshad Ali buys a scooter for Rs 22000.

Therefore , unpaid amount = 22000-4000=Rs. 18000

According to the given condition, interest paid annually is

10% of 18000,10% of 17000,10% of 16000,......................10% of 1000.

Thus, total interest to be paid

=10\%of\, 18000+10\%of\, 17000+.............10\%of\, 1000

=10\%of\, (18000+ 17000+.............+ 1000)

=10\%of\, (1000+ 2000+.............+ 18000)

Here, 1000, 2000,.............17000,18000 is a AP with first term =a=1000 and common difference =d = 1000

We know that a_n=a+(n-1)d

\Rightarrow 18000=1000+(n-1)1000

\Rightarrow 17000=(n-1)1000

\Rightarrow 17=(n-1)

\Rightarrow n=17+1=18

Sum of AP:

S_1_8=\frac{18}{2}\left [ 2(1000)+(18-1)1000 \right ]

=9\left [ 2000+17000 \right ]

=9\left [ 19000 \right ]

=171000

Thus, interest to be paid :

=10\%of\, (1000+ 2000+.............+ 18000)

=10\%of\, ( 171000)

=Rs. 17100

Thus, cost of tractor = Rs. 22000+ Rs. 17100 = Rs. 39100

Question:29 A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

Answer:

The numbers of letters mailed forms a GP : 4,4^2,4^3,.............4^8

first term = a=4

common ratio=r=4

number of terms = 8

We know that the sum of GP is

S_n=\frac{a(r^n-1)}{r-1}

=\frac{4(4^8-1)}{4-1}

=\frac{4(65536-1)}{3}

=\frac{4(65535)}{3}

=87380

costs to mail one letter are 50 paise.

Cost of mailing 87380 letters

=Rs. \, 87380\times \frac{50}{100}

=Rs. \,43690

Thus, the amount spent when the 8th set of the letter is mailed is Rs. 43690.

Question:30 A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

Answer:

Given : A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

=\frac{5}{100}\times 10000=Rs.500

\therefore Interest in fifteen year 10000+ 14 times Rs. 500

\therefore Amount in 15 th year =Rs. 10000+14\times 500

=Rs. 10000+7000

=Rs. 17000

\therefore Amount in 20 th year =Rs. 10000+20\times 500

=Rs. 10000+10000

=Rs. 20000

Question:31 A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Answer:

Cost of machine = Rs. 15625

Machine depreciate each year by 20%.

Therefore, its value every year is 80% of the original cost i.e. \frac{4}{5} of the original cost.

\therefore Value at the end of 5 years

=15625\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}

=5120

Thus, the value of the machine at the end of 5 years is Rs. 5120

Question:32 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on.

Answer:

Let x be the number of days in which 150 workers finish the work.

According to the given information, we have

150x=150+146+142+............(x+8)terms

Series 150x=150+146+142+............(x+8)terms is a AP

first term=a=150

common difference= -4

number of terms = x+8

\Rightarrow 150x=\frac{x+8}{2}\left [ 2(150)+(x+8-1)(-4) \right ]

\Rightarrow 300x=x+8\left [ 300-4x-28 \right ]

\Rightarrow 300x= 300x-4x^2-28x+2400-32x+224

\Rightarrow x^2+15x-544=0

\Rightarrow x^2+32x-17x-544=0

\Rightarrow x(x+32)-17(x+32)=0

\Rightarrow (x+32)(x-17)=0

\Rightarrow x=-32,17

Since x cannot be negative so x=17.

Thus, in 17 days 150 workers finish the work.

Thus, the required number of days = 17+8=25 days.

More About NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-

Class 11 Maths chapter 9 miscellaneous solutions consist of questions related to finding the nth term of the arithmetic progression, the sum of the terms of arithmetic progression, arithmetic mean, nth term of the geometric progression, the sum of the terms of geometric progression, geometric mean, etc. There are few solved examples given before the miscellaneous exercise chapter 9 Class 11 that you can try to solve.

Topic Covered in Class 11 Chapter 9 Maths Miscellaneous Solutions

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 9 - Sequences and Series covers the following topics:

  1. Introduction
  2. Sequences
  3. Series
  4. Arithmetic Progression (A.P.)
  5. Geometric Progression (G.P.)
  6. Relationship Between A.M. and G.M.
  7. Sum to n terms of Special Series
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By engaging with the concepts presented in NCERT Solutions for Class 11 Maths, students can address any doubts related to these topics, establishing a strong foundation for their understanding of Class 12 Maths.

Also Read| Sequences And Series Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-

  • There are 32 questions in the miscellaneous exercise chapter 9 Class 11 which you can try to solve by yourself.
  • You can use Class 11 Maths chapter 9 miscellaneous exercise solutions when you are finding it difficult to solve miscellaneous problems.
  • Class 11 Maths chapter 9 miscellaneous solutions are designed by subject matter experts in a detailed manner that could be understood by an average student also.

Key Features of Class 11 Maths ch 9 Miscellaneous Exercise Solutions

  • Comprehensive Approach: The miscellaneous exercise class 11 chapter 9 offers a thorough explanation of topics, ensuring a well-rounded understanding of Sequences and Series.
  • Stepwise Clarity: Each problem in Class 11 maths miscellaneous exercise chapter 9 is solved with step-by-step solutions, facilitating a logical and clear progression of concepts for Class 11 students.
  • Accessible Language: The class 11 chapter 9 miscellaneous exercise solutions are presented in clear and concise language, making intricate mathematical concepts more accessible for students in Chapter 9 of Class 11 Maths.
  • Variety of Problems: The Miscellaneous Exercise in Class 11 Maths Chapter 9 provides a diverse range of problems, catering to different difficulty levels and scenarios, allowing for a comprehensive understanding.
  • Free PDF Access: Solutions for Chapter 9 of Class 11 Maths Miscellaneous Exercise are accessible in a free PDF format, providing students with easy access to study materials according to their convenience.

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. What is the meaning of sequence ?

Sequence means an arrangement of numbers in a definite order according to some rule.

2. What is the meaning of finite sequence ?

The finite sequence is s sequence containing a finite number of terms.

3. What is the meaning of infinite sequence ?

The infinite sequence is s sequence containing an infinite number of terms.

4. What is the meaning of series ?

If the terms of a sequence are expressed as the sum of terms then it is called a series.

5. What is the sum of A.P. if the first term is 'a' and the last term of the A.P. is 'l' and the number of terms are 'n' ?

Sum of A.P. = n(a+l)/2

6. What is the arithmetic mean of two numbers 5, 9 ?

Arithmetic mean (A.M.) = (5+9)/2 = 7

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

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2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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3 Jobs Available
Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction. 

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions. 

3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.  

3 Jobs Available
Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Credit Manager

Credit Management refers to the process of granting credit, setting the terms it’s granted on, recovering the credit when it’s due, and confirming compliance with the organization's credit policy, among other credit-related operations. Individuals who opt for a career as Credit Manager should have hands-on experience with accounting software, a solid understanding of lending procedures, excellent analytical skills with the ability to create and process financial spreadsheets, negotiation skills, and a bachelor’s or master’s degree in a field relevant to finance or accounting. Ultimately, Credit Management job is to help organizations minimize bad debts and increase revenues from the loan.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Healthcare Social Worker

Healthcare social workers help patients to access services and information about health-related issues. He or she assists people with everything from locating medical treatment to assisting with the cost of care to recover from an illness or injury. A career as Healthcare Social Worker requires working with groups of people, individuals, and families in various healthcare settings such as hospitals, mental health clinics, child welfare, schools, human service agencies, nursing homes, private practices, and other healthcare settings.  

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Fashion Blogger

Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns. 

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Advertising Manager

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
Merchandiser
2 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
ITSM Manager
3 Jobs Available
.NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

2 Jobs Available
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