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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

    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\;

    Option 4)

    K/4

    In the reaction,

    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 .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    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

    Bio Medical Engineer

    The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary. 

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    Data Administrator

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    Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

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    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. 

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    Cartographer

    How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

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    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.

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    Bank Probationary Officer (PO)

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    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.

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    Finance Executive

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    Treasurer

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    Product Manager

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    Also Read: Career as Nurse

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    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. 

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    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.

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    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.

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    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 cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

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    Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

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    Resource Links for Online MBA 

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    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. 

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    .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#. 

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    Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

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