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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, ais 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 ,Sand Srespectively ,show that S3 =3(S2 -S1)

Solution. We are given sum of n ,2n and 3n terms of an AP be S1 ,Sand S

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

S= 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                          $\fn_cm f\left ( 1 \right )=3\: and\: \sum_{x=1}^{n}f\left ( x \right )=120$                                                                                                            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

$\fn_cm \sum_{x=1}^{n}f\left ( x \right )=f\left ( 1 \right )+f\left ( 2 \right )+f\left ( 3 \right )+....f\left ( n \right )$

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

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

$\fn_cm \frac{3\left ( 3^{n}-1 \right )}{3-1}=120$

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

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

$\fn_cm \frac{5\left ( 2^{n} -1\right )}{2-1}=315$

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)

$\fn_cm \frac{a\left ( r^{n} -1\right )}{r-1}$    $\fn_cm \: =\frac{5a\left ( r^{2n} -1\right )}{r^{2}-1}$

$\fn_cm \frac{a\left ( r^{n} -1\right )}{r-1}=\frac{5a\left [ \left ( r^{n} \right )^{2}-1 \right ]}{\left ( r+1 \right )\left ( r-1 \right )}$

$\fn_cm \frac{a\left ( r^{n} -1\right )}{r-1}=\frac{5a\left ( r^{n}+1 \right )\left ( r^{n}-1 \right )}{\left ( r+1 \right )\left ( r-1 \right )}$

$\fn_cm \frac{r^{n}+1}{r+1}=\frac{1}{5}$

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

$\fn_cm Q13.If\: \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}\left ( x\neq 0 \right )$

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

Ans. We are given that

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

Taking the following equation

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

(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

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

(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

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

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

$\fn_cm \therefore R=\frac{\left [ \left ( 1/a \right )^{n}-1 \right ]}{a\left [ \left ( 1/r \right )-1 \right ]}$

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

$\fn_cm =\frac{\left ( 1-r^{n} \right )}{\left ( 1-r \right )ar^{n-1}}$

Putting the value of P and R in the LHS of the equation P2Rn = Sn

$\fn_cm =\left ( a^{n} r^{n\left ( n-1 \right )/2}\right )^{2}\frac{\left ( 1-r^{n} \right )^{n}}{\left ( 1-r \right )^{n}a^{n}r^{n\left ( n-1 \right )}}$

$\fn_cm =\frac{a^{2n}r^{n\left ( n-1 \right )}}{a^{n}r^{n\left ( n-1 \right )}}.\left (\frac{r^{n}-1}{r-1} \right )^{n}$

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

= 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)

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

$\fn_cm Q16.If\: a\left ( \frac{1}{b} +\frac{1}{c}\right ),b\left ( \frac{1}{c} +\frac{1}{a}\right ),c\left ( \frac{1}{a} +\frac{1}{b}\right )$

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

Ans.We are given the terms in AP

$\fn_cm a\left ( \frac{1}{b} +\frac{1}{c}\right ),b\left ( \frac{1}{c} +\frac{1}{a}\right ),c\left ( \frac{1}{a} +\frac{1}{b}\right )$

Therefore

$\fn_cm b\left ( \frac{1}{c} +\frac{1}{a}\right ) -a\left ( \frac{1}{b} +\frac{1}{c}\right )=c\left ( \frac{1}{a} +\frac{1}{b}\right )-b\left ( \frac{1}{c} +\frac{1}{a}\right )$

$\fn_cm b\left ( \frac{a+c}{ac} \right )-a\left ( \frac{b+c}{bc} \right )=c\left ( \frac{a+b}{ab} \right )-b\left ( \frac{a+c}{ac} \right )$

$\fn_cm \frac{b^{2}a+b^{2}c-a^{2}b-a^{2}c}{abc}=\frac{c^{2}a+c^{2}b-b^{2}a-b^{2}c}{abc}$

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

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 c+bnd+bncn+bncn

From equation (iii)

= an c+bnd+bncn+andn

Arranging the terms

=an c+bnc+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 are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, cd, 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

### 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