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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, ap is given by
ap = 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
am+n = a + (m+n-1) d ….(i)
(m – n)th term of the AP is
am-n = a + (m-n-1) d ….(ii)
The sum of (m + n)th and (m -n)th term is
am+n + am-n
= a + (m+n-1) d + a + (m-n-1) d
= 2a + (m +n -1 + m-n -1) d
= 2a + (2m -2) d
am+n + am-n = 2[ a + (m -1) d] …..(i)
Since m th term , am = a + (m-1) d
Substituring am = a + (m-1) d in equation (i)
am+n + am-n = 2 am
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 S1 ,S2 and S3 respectively ,show that S3 =3(S2 -S1)
Solution. We are given sum of n ,2n and 3n terms of an AP be S1 ,S2 and S3
Applying the following formula for the sum of p terms of an AP
Sp = p/2[2a + (p-1)d]
S1 = n/2[2a + (n-1)d]……(i)
S2 = 2n/2[2a + (2n-1)d]
S2 = n[2a + (2n-1)d]……(ii)
S3 = 3n/2[2a + (3n-1)d]…(iii)
Subtracting the equation (i) from the equation (ii)
S2-S1= 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]
S2-S1= n/2[2a +(3n – 1)d]
Multiplying both sides by 3 and using equation 3
3(S2-S1) = 3n/2[2a +(3n – 1)d] = S3
3(S2-S1) = S3
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
an = a + (n -1) d, where a =203, d = 7, an =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
Sn = n/2[2a + (n-1)d]
= 29/2[2×203 + (29-1)7]
= 29/2[406 + 196]
= 29/2[602] = 29× 301 = 8729
Sn = 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
an = a + (n -1) d
Where a = 2, d=2,an = 100
100 = 2 + (n -1) 2
100 = 2 + 2n -2
n = 50
The sum of these 50 terms is determined by the following formula
Sn = n/2[2a + (n-1)d]
S50 = 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
an = a + (n -1) d
Where a = 5, d=5,an = 100
100 = 5 + (n -1) 5
100 = 5 + 5n -5
n = 20
The sum of these 20 terms is determined by the following formula
Sn = 20/2[2×5 + (n-1)d]
S20 = 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
an = a + (n -1) d
Where a = 10, d=10,an = 100
100 = 10 + (n -1) 10
100 = 10 + 10n -10
n = 10
The sum of these 20 terms is determined by the following formula
Sn = n/2[2n+ (n-1)d]
S10 = 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
an = a + (n -1) d
Where a = 13, d=4,an = 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
Sn = n/2[2n+ (n-1)d]
S22 = 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, an= 120
The sum of a GP is given by
3n-1 = (120 ×2)/3 = 80
3n = 81 =34
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
2n-1 = 315/5 = 63
2n=64 =26
n = 6
The term nth of the GP is given by
an=arn-1= 5×26-1 =5×25= 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 nth of the GP is given by
an=arn-1
Third and fifth term of the GP are
a3=1×r3-1 =r2
and a5=1×r5-1 =r4
According to the question, the sum of the third and fifth term is 90
a3+ a5= 90
r2+ r4= 90
Arranging the equation as following
r4+ r2– 90 = 0
Let r2= 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 = r2= -10 is impossible,therefore neglecting the value x = -10
Taking x = 9 ,which gives
r2= 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……… arn-1
The terms which are occupying odd places
a, ar²,ar4……… arn-2
According to the question
a+ ar+ ar²+ar³……… arn-1 = 5(a+ ar²+ar4……… +arn-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 P2Rn = Sn
Ans. Let the first term of GP is a and common ratio is r
Then the GP is a, ar, ar², ar³……….arn-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……….arn-1
= a1+1+1…1 r1+2+3…..n-1
P= anrn(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/arn-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 P2Rn = Sn
= RHS Hence proved
Q15. The pth ,qth and rth 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 pth ,qth and rth terms of an AP are a,b,c respectively
Therefore determing the pth ,qth and rth terms of an AP by applying the nth term formula
Let the first term is A
Ap = A + (p-1)d
A +(p-1)d = a…….(i)
Aq = A + (q-1)d
A + (q-1)d = b…..(ii)
Ar = 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 (an +bn ),(bn +cn ),(cn +dn ) 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 (an +bn ),(bn +cn ),(cn +dn ) are in GP. for this we are required to prove
(bn +cn )² = (an +bn )(cn +dn )
Taking LHS
(bn +cn )² = b2n +c2n +2bn cn
From equation (i),(ii)
= (ac)n +(bd)n +2bn cn
= an cn +bndn +bncn+bncn
From equation (iii)
= an cn +bndn +bncn+andn
Arranging the terms
=an cn +bncn +andn+bndn
=cn(an +bn ) +dn(an +bn )
=(an +bn )(cn +dn )
Hence
(bn +cn )² =(an +bn )(cn +dn )
Therefore (an +bn ),(bn +cn ),(cn +dn ) are in GP
Q18.If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 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 x2 – 3x + p = 0 and c,d are roots of x2 – 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⇒ a²r5 = q…..(iv)
From equation (i) and (iii)
ar²(1 + r) /a(1 +r)= 12/3
r² = 4 ⇒ ±2
∴ (q +p)/(q-p) = (a²r5+a²r )/(a²r5– a²r)
=a²r(r4+1)/a²r(r4 -1) =(r4+1)/(r4-1) = [(±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
CERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions of class 9 maths
| Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |
| Chapter 2-Polynomial | Chapter 10-Circles |
| Chapter 3- Coordinate Geometry | Chapter 11-Construction |
| Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |
| Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |
| Chapter 6-Lines and Angles | Chapter 14-Statistics |
| Chapter 7-Triangles | Chapter 15-Probability |
| Chapter 8- Quadrilateral |
NCERT Solutions of class 9 science
CBSE Class 9-Question paper of science 2020 with solutions
CBSE Class 9-Sample paper of science
CBSE Class 9-Unsolved question paper of science 2019
NCERT Solutions of class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT solutions of class 10 science
Solutions of class 10 last years Science question papers
CBSE Class 10 – Question paper of science 2020 with solutions
CBSE class 10 -Latest sample paper of science
NCERT solutions of class 12 maths
| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
| Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
| Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
| Chapter 7- Integrals | |
| Chapter 8-Application of Integrals |
