**NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 9 Sequence and Series**

**NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 9 Sequence and Series**

NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 9 Sequence and Series is created here for the purpose of helping class 11 maths students in clearing their concept on the chapter 9 Sequence and Series. The NCERT Solutions of the Class 11 Maths Miscellaneous Exercise Chapter 9 Sequence and Series will clear all your doubts on the rest of exercises of the chapter 9. NCERT Solutions of maths are the best input study material for achieving excellent marks in the exam. Chapter 9 Sequence and Series contains 5 exercises, exercise 9.1, exercise 9.2, exercise 9.3, exercise 9.4, and Miscellaneous Exercise. The miscellaneous exercise of the chapter 9 Sequence and Series is the extract of all the exercises, therefore, every student is needed special attention to study this exercise by practicing it.

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**NCERT Solutions For Class 11 Maths Chapter 9 Sequence and Series**

**Exercise 9.1- Sequence and Series**

**Exercise 9.2 – Sequence and Series**

**Exercise 9.3 – Sequence and Series**

**Micellaneous Exercise- Sequence and Series**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 9 Sequence and Series**

**Q1. Show that the sum of (m + n) ^{th }and (m -n)^{th }terms of an AP is equal to twice the m ^{th }term.**

Solution. The p ^{th }term of an AP, a_{p }is given by

a_{p } = a + (p -1) d

Where a is the first term of an AP and d is the common difference

(m + n)^{th } term of the AP is

a_{m+n } = a + (m+n-1) d ….(i)

(m – n)^{th } term of the AP is

a_{m-n } = a + (m-n-1) d ….(ii)

The sum of (m + n)^{th }and (m -n)^{th }term is

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

a_{m+n }+ a_{m-n } = 2[ a + (m -1) d] …..(i)

Since m ^{th }term , a_{m} = a + (m-1) d

Substituring a_{m} = a + (m-1) d in equation (i)

**a _{m+n }+ a_{m-n } = 2 a_{m}**

Hence Proved

**Q2.If the sum of three numbers in AP is 24 and their product is 440, find the numbers.**

Solution.Let the three numbers in AP are (a -d), a and (a +d)

According to first condition

(a -d)+ a+(a +d) = 24

3a = 24 ⇒ a = 8

According to second condition

(a -d) a(a +d) = 24

Putting the value of a = 8

8( 8 – d)(8 + d) = 440

8 ² – d² = 55

d² = 64 -55 = 9

d = ± √9 = ±3

If d = 3, then numbers are (8 -3), 8,(8+ 3) = 5,8,11

If d = -3, then numbers are (8 +3,8,(8 -3)= 11,8,5

Hence,the required numbers of the AP are 5,8 and 11

**Q3.Let the sum of n,2n and 3n terms of an AP be S _{1 },S_{2 }and S_{3 }respectively ,show that S_{3 }=3(S_{2 }-S_{1})**

Solution. We are given sum of n ,2n and 3n terms of an AP be S_{1 },S_{2 }and S_{3 }

Applying the following formula for the sum of p terms of an AP

S_{p }= p/2[2a + (p-1)d]

S_{1 }= n/2[2a + (n-1)d]……(i)

S_{2 }= 2n/2[2a + (2n-1)d]

S_{2 }= n[2a + (2n-1)d]……(ii)

S_{3 }= 3n/2[2a + (3n-1)d]…(iii)

Subtracting the equation (i) from the equation (ii)

S_{2}-S_{1}= n[2a + (2n-1)d] – n/2[2a + (n-1)d]

= n/2[ 4a + 2d(2n -1) – (2a + (n -1)d]

= n/2[ 4a + 4dn -2d -2a – dn + d]

= n/2[2a + 3dn -d]

S_{2}-S_{1}= n/2[2a +(3n – 1)d]

Multiplying both sides by 3 and using equation 3

3(S_{2}-S_{1}) = 3n/2[2a +(3n – 1)d] = S_{3}

**3(S _{2}-S_{1}) = S_{3}**

**Hence Proved**

**Q4.Find the sum of all terms between 200 and 400 which are divisible by 7.**

Solution.

We can get least and highest number divisible by 7 between 200 and 400 by dividing 200 and 400 by 7.

When we divide 200 by seven ,the remiender is 4, then first number divisible by 7 is 200+ (7-4)=203

When we divide 400 by seven ,the remiender is 1, then last number divisible by 7 is 400-1 =399

Therefore the AP is

203, 210,217…………399

Now, we have to determine the number of terms in AP from the following formula

a_{n } = a + (n -1) d, where a =203, d = 7, a_{n } =399

203 +(n -1)7 = 399

203 + 7n – 7 = 399

7n = 399 + 7 -203 = 406 – 203 = 203

n = 29

For calculating the sum of the AP,let’s apply the following formula

S_{n }= n/2[2a + (n-1)d]

= 29/2[2×203 + (29-1)7]

= 29/2[406 + 196]

= 29/2[602] = 29× 301 = 8729

S_{n }= 8729

Therefore required sum of the AP is 8729

**Q5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.**

Ans. We are given to find out the sum of integers from 1 to 100 that are divisible by 2 or 5

Sum of integers divisible by 2 or 5 = Sum of integers divisible by 2 + Sum of integers divisible by 5 – Sum of the integers divisible by both(i.e 10)…(i)

The number of terms which are divisible by 2 from 1 to 100 are following

2,4,6,8……100

Let’s find out the number of terms divisible by 2 by applying the following formula

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

Where a = 2, d=2,a_{n } = 100

100 = 2 + (n -1) 2

100 = 2 + 2n -2

n = 50

The sum of these 50 terms is determined by the following formula

S_{n }= n/2[2a + (n-1)d]

S_{50 }= 50/2[2×2 + (50-1)2]=2550

The number of terms which are divisible by 5 from 1 to 100 are following

5,10,15,20……100

Let’s find out the number of terms divisible by 2 by applying the following formula

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

Where a = 5, d=5,a_{n } = 100

100 = 5 + (n -1) 5

100 = 5 + 5n -5

n = 20

The sum of these 20 terms is determined by the following formula

S_{n }= 20/2[2×5 + (n-1)d]

S_{20 }= 20/2[2×5 + (20-1)5]=1050

The number of terms which are divisible by both from 1 to 100 are following

10,20,30,……100

Let’s find out the number of terms divisible by 10 by applying the following formula

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

Where a = 10, d=10,a_{n } = 100

100 = 10 + (n -1) 10

100 = 10 + 10n -10

n = 10

The sum of these 20 terms is determined by the following formula

S_{n }= n/2[2n+ (n-1)d]

S_{10 }= 10/2[2×10 + (10-1)10]=550

From (i), we have

Sum of integers divisible by 2 or 5 = 2550 + 1050 – 550 =3050

Therefore required sum is = 3050

**Q6. Find the sum of all two-digit numbers which when divided by 4 , yields as 1 remainder.**

Solution. The two digit numbers starts from 10 and finish at 99

Dividing 10 by 4 the remainder is 2, adding (2 +1=3) on 10 we can get the first such number as described in the question i.e 10 +3 = 13

Dividing 99 by 4, the remainder is 3, subtracting (4-3+1) from 99,we can get last two digit such number as described in the question i.e 99 -2 = 97

Therefore the AP is 13,17,21…….97

Let’s find out the number of terms of this AP by applying the following formula

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

Where a = 13, d=4,a_{n } = 97

97= 13 + (n -1) 4

97 = 13 + 4n -4

4n = 97 -9 = 88

n = 22

The sum of these 22 terms is determined by the following formula

S_{n }= n/2[2n+ (n-1)d]

S_{22 }= 22/2[2×13 + (22-1)4]=11(26 + 84) = 11 ×110 = 1210

Therefore the required sum of the AP is 1210

**Q7. If f is a function satisfying f(x + y) = f(x) f(y) for all x ,y ∈ N such that find the value of n**.

Solution. We are given that f(x + y) = f(x) f(y) for all x ,y ∈ N and f(1) =3

For x = 1, y = 1, f ( 1 +1) = f(2) = f(1) f(1) = 3×3 = 9

For x = 1, y =2, f(1 +2)= f(3) = f(1) f(2) = 3×9 = 27

For x = 1, y =3,f(1 + 3) = f(4) = f(1) f(3) = 3× 27 = 81

We are given

f(1) + f(2) + f(3) +…………f(n) = 120

3 + 9 + 81 + ….. = 120

The given series is GP, where a = 3, r = 9/3 = 3, a_{n}= 120

The sum of a GP is given by

3^{n}-1 = (120 ×2)/3 = 80

3^{n} = 81 =3^{4}

n = 4

Hence the value of n is 4

**Q8. The sum of some terms of GP is 315 whose first term and the common ratio are 5 and 2, respectively, find the last term and the number of terms.**

Solution. The sum of some terms of GP is 315, where the first term, a =5 and common ratio,r =2

The sum of a GP is given by

2^{n}-1 = 315/5 = 63

2^{n}=64 =2^{6}

n = 6

The term n^{th }of the GP is given by

a_{n}=ar^{n-1}= 5×2^{6-1 }=5×2^{5}= 5×32 =160

Therefore the last term of the GP is 160 and the number of terms in GP are 6

**Q9.The first term of a GP is 1. The sum of third term and fifth term is 90.Find the common ratio of the GP.**

Solutions. It is given to us that first term,a = 1

The term n^{th }of the GP is given by

a_{n}=ar^{n-1}

Third and fifth term of the GP are

a_{3}=1×r^{3-1 }=r^{2}

and a_{5}=1×r^{5-1 }=r^{4}

According to the question, the sum of the third and fifth term is 90

a_{3}+ a_{5}= 90

r^{2}+ r^{4}= 90

Arranging the equation as following

r^{4}+ r^{2}– 90 = 0

Let r^{2}= x

x² + x – 90 =0

Factorizing the LHS of the equation

x² + 10x -9x- 90 =0

x( x + 10) -9( x + 10) = 0

( x + 10) -9( x + 10) = 0

( x + 10) ( x -9) = 0

x = -10, x = 9

Since x = r^{2}= -10 is impossible,therefore neglecting the value x = -10

Taking x = 9 ,which gives

r^{2}= 9⇒ r = ± 3

Hence common ratio of the GP is ± 3

**Q10.The sum of three numbers in GP is 56. If we subtract 1,7,21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.**

Solution. We are given the sum of three numbers in GP is 56

Let the first term is a

Second term = ar

Third term = ar²

a + ar + ar² = 56

a( 1 + r + r²) = 56…..(i)

According to question

a-1, ar -7 and ar² -21 are in AP

Therefore

ar -7 -(a -1) = ar² -21 -( ar -7)

ar – 7 -a + 1 = ar² -21 -ar + 7

ar² -2ar + a-8 = 0…..(ii)

Putting the value of a = 56/( 1 + r + r²) from equation (i) in (ii) equation

56r²/( 1 + r + r²) – 2× 56r/( 1 + r + r²) + 56/( 1 + r + r²) -8 = 0

(56r² -112r +56 -8 -8r -8r²) = 0

48r² -120r + 48 = 0

6r² – 15r + 6 = 0

6r² – 12r -3r+ 6 = 0

6r (r -2) – 3( r -2) = 0

(r-2)(6r -3) = 0

r =2, r = 1/2

Putting the value of r = 2 in equation (i)

a( 1 + 2+ 2²) = 56⇒7a = 56 ⇒ a = 8

Putting r = 1/2 in equation (i)

a( 1 + 1/2+1/ 2²) = 56⇒7/4a = 56 ⇒ a = 32

When r = 2, the numbers are a=8, ar = 8×2 = 16 and ar²= 8×2² = 32

When r =1/2 ,the numbers are , a = 32, ar = 32× (1/2)= 16, ar² = 32 (1/2)² = 8

In both cases the numbers are same

Therefore the required numbers are 8,16 and 32

**Q11.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 it common ratio.**

Ans. Let the first term of the GP is a and its common ratio is r

Then the AP is

a, ar, ar²,ar³……… ar^{n-1}

The terms which are occupying odd places

a, ar²,ar^{4}……… ar^{n-2}

According to the question

a+ ar+ ar²+ar³……… ar^{n-1 }= 5(a+ ar²+ar^{4}……… +ar^{n-2})

Comparing r + 1 = 5⇒ r = 4

**Hence the common ratio of the given GP is 4**

**Q12.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.**

Ans. Let the first term of the given AP is a and the common difference is d

Then the AP is a, a+d,a+2d,a + 3d,…….a +(n-1)d

According to question

a+ a+d+a+2d+a + 3d = 56

4a + 6d = 56

2a + 3d = 28

We are given a = 11

Them

22 + 3d = 28

d = 2

The sum of last 4 terms = 112

The last term is = a + (n -1) d

Therefore sum of last 4 terms

a + (n -4) d +a + (n -3) d +a + (n -2) d+a + (n -1) d = 112

4a + d(n-4 + n -3+ n -2 + n-1) = 112

4a + d(4n -10) = 112

4a + 2d(2n – 5) = 112

Putting a = 11 and d = 2

44 + 4(2n -5) = 112

8n -20 = 112 -44 = 68

n = 88/8 = 11

**Therefore the number of terms in the AP are 11**

**Then show that a,b,c and d are in GP.**

Ans. We are given that

Taking the following equation

(a + bx)( b – cx) = (b + cx)( a – bx)

ab – acx + b²x – bcx² = ab – b²x + acx – bcx²

2b²x = 2acx

b² = ac

b/c = a/b……..(i)

Taking the equation

(b + cx)( c – dx) = (c + dx)( b – cx)

bc – bdx + c²x – cdx² = bc – c²x + bdx – cdx²

2c²x = 2bdx

c² = bd

b/c = c/d…….(ii)

From equation (i) and equation (ii)

a/b = b/c = c/d

**Hence a,b and c are in GP.**

**Q14. 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}**

Ans. Let the first term of GP is a and common ratio is r

Then the GP is a, ar, ar², ar³……….ar^{n-1}

According to question S is the sum of the GP

According to question P is the product of the GP

P= a×ar×ar²× ar³……….ar^{n-1}

= **a ^{1+1+1…1 }**

**r**

^{1+2+3…..n-1}P= **a ^{n}**

**r**

^{n(n-1)/2}According to question R is the sum of reciprocals of *n* terms of the GP

R = 1/a + 1/ar + 1/ar² +……..1/ar^{n-1}

First term of the GP is 1/a and common ratio 1/r

Putting the value of P and R in the LHS of the equation **P ^{2}R^{n} = S^{n}**

= RHS Hence proved

**Q15. The p ^{th }**

**,q**

^{th }and r^{th }**terms of an AP are a,b,c respectively.**

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

Ans. We are given that p^{th },q^{th }and r^{th } terms of an AP are a,b,c respectively

Therefore determing the p^{th },q^{th }and r^{th } terms of an AP by applying the n^{th } term formula

Let the first term is A

A_{p} = A + (p-1)d

A +(p-1)d = a…….(i)

A_{q} = A + (q-1)d

A + (q-1)d = b…..(ii)

A_{r} = A + (r-1)d

A + (r-1)d = c…..(iii)

Subtracting equation (iii) from equation (ii) and then multiplying by a

[A + (q-1)d – A – (r-1)d ]a =ab -ac

d(q-1-r+1)a = ab-ac

d(q-r)a = ab -ac…..(iv)

Similarly Subtracting equation (i) from equation (iii) and then multiplying by b

We get

d(r-p)b = bc -ab…..(v)

Similarly Subtracting equation (ii) from equation (i) and then multiplying by c

d(p-q)c = ac-bc…..(vi)

Adding equation (iv),(v) and (vi)

d(q-r)a +d(r-p)b +d(p-q)c =ab -ac+ bc -ab + ac-bc

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

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

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

**are in AP ,prove that a,b and c are AP.**

Ans.We are given the terms in AP

Therefore

b²a +b²c -a²b -a²c = c²a + c²b – b²a -b²c

Arranging the terms

b²a -a²b + b²c -a²c = c²a – b²a +c²b-b²c

ab ( b -a) +c( b² -a²) =a ( c² – b²) +bc( c -b)

ab ( b -a) +c( b -a) (b+a)=a ( c – b)(c +b) +bc( c -b)

( b -a)(bc +ac+ab ) = ( c – b)(ac +ab + bc)

b -a = c-b

Therefore a,b and c will be an AP.

**Q17. If a,b,c and d are in GP,prove that (a ^{n }+b^{n} ),(b^{n }+c^{n} ),(c^{n }+d^{n} ) are in GP.**

Ans. We are given that a,b,c and d are in GP

Therefore

b² = ac …..(i)

c² = bd …..(ii)

a/b = c/d

ad = bc…..(iii)

We have to prove that (a^{n }+b^{n} ),(b^{n }+c^{n} ),(c^{n }+d^{n} ) are in GP. for this we are required to prove

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

Taking LHS

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

From equation (i),(ii)

= (ac)^{n }+(bd)^{n} +2b^{n }c^{n}

= a^{n }c^{n }+b^{n}d^{n }+b^{n}c^{n}+b^{n}c^{n}

From equation (iii)

= a^{n }c^{n }+b^{n}d^{n }+b^{n}c^{n}+a^{n}d^{n}

Arranging the terms

=a^{n }c^{n }+b^{n}c^{n }+a^{n}d^{n}+b^{n}d^{n}

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

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

Hence

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

**Therefore (a ^{n }+b^{n} ),(b^{n }+c^{n} ),(c^{n }+d^{n} ) are in GP**

**Q18.If ****a**** and ****b ****are the roots of ****x**^{2}** – 3x + p = 0**** and c, d are roots of ****x**^{2}** – 12x + q = 0****, where ****a****,**** b****,**** c****, ****d****, form a G.P. Prove that (****q**** + ****p****): (****q**** – ****p****) = 17:15.**

Ans. We are given a,b roots of ** ***x*^{2}* – 3x + p = 0 and c,d are roots of x ^{2} – 12x + q = 0,*

Therefore a +b = -(-3)/1 = 3 and ab = p/1 = p, c+d = -(-12)/1 = 12, cd = q/1 = q

a,b,c and d are given in GP

b = ar, c= ar², d = ar³ (where r is common ratio)

a + ar = 3⇒ a(1 +r) = 3 ….(i) and aar = p⇒ a²r = p…..(ii)

ar² + ar³ = 12⇒ ar²(1 + r) = 12….(iii) and ar².ar³ =q⇒ a²r^{5} = q…..(iv)

From equation (i) and (iii)

ar²(1 + r) /a(1 +r)= 12/3

r² = 4 ⇒ ±2

∴ (q +p)/(q-p) = (a²r^{5}+a²r )/(a²r^{5}– a²r)

=a²r(r^{4}+1)/a²r(r^{4} -1) =(r^{4}+1)/(r^{4}-1) = [(±2)^{4}+1]/[(±2)^{4}-1] =17/15

(q +p): (q-p) = 17 : 15. Hence proved

**Q19.The ratio of the A.M and G.M. of two positive numbers ****a**** and ****b****, is ****m****:**** n****. Show that a : b =m + √(m² -n²) : m – √(m² -n²).**

Ans. The given positive numbers are a and b

The AM of the given positive numbers is = (a+b)/2 and GM = √(ab)

The ratio of AM to GM given to us = m:n

(a +b)/[2 √(ab)] = m/n

Squaring both sides

(a +b)²/(4ab) = m²/n²

(a + b)² = 4ab m²/n²

a + b = 2√(ab)m/n…….(i)

Since (a -b)² = ( a +b)² – 4ab

(a -b)² = 4ab m²/n² – 4ab = (4abm² – 4abn²)/n² =4ab(m²- n²)/n²

a -b = 2√(ab)√(m²- n²)/n …(ii)

Adding both equation (i) and equation (ii)

2a = 2√(ab)m/n + 2√(ab)√(m²- n²)/n

2a= 2√(ab/n[ m + √(m²- n²)]

a = √(ab/n[ m + √(m²- n²)]

Putting a = √(ab/n[ m + √(m²- n²)] in equation (i),we get

b=√(ab/n[ m -√(m²- n²)]

∴a : b = √(ab/n[ m + √(m²- n²)] : √(ab/n[ m -√(m²- n²)]

a : b = [ m + √(m²- n²)] : [ m -√(m²- n²)], Hence proved

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