Algebraic Expressions

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NCERT Solutions for Chapter 12 Algebraic Expressions Class 7 Maths

Book Solutions

1

Get the algebraic expressions in the following cases using variables, constants and arithmetic operations:
1. Subtraction of z from y.
2. One-half of the sum of numbers x and y.
3. The number z multiplied by itself.
4. One-fourth of the product of numbers p and q.
5. Numbers x and y both squared and added.
6. Number 5 added to three times the product of mm and n.
7. Product of numbers y and z subtracted from 10.
8. Sum of numbers a and b subtracted from their product.

Answer

(i) y−z
(ii) (x+y)/2
(iii) z2
(iv) pq/4
(v) x2+y2
(vi) 3mn+5
(vii) 10−yz
(viii) ab−(a+b)
Exercise 12.1 Page Number 234

2

(i) Identify the terms and their factors in the following expressions, show the terms and factors by tree diagram:
(a) x−3
(b) 1+x+x2
(c) y−y3
(d) 5xy2+7x2y
(e) −ab+2b2−3a2

(ii) Identify the terms and factors in the expressions given below:
(a) −4x+5
(b) −4x+5y
(c) 5y+3y2
(d) xy+2x2y2
(e) pq+q
(f) 1.2ab−2.4b+3.6a
(g) 3/4x+14
(h) 0.1p2+0.2q2

Answer

(i) (a) x−3


(b) 1+x+x2
Expression
 

(c) y−y3
Expression
 

(d) 5xy2+7x2y
Expression
 

(e) −ab+2b2−3a2
Expression
 

(ii) (a) −4x+5
Terms: −4x,5
Factors: −4,x ; 5

(b) −4x+5y
Terms: −4x,5y
Factors: −4,x ; 5,y

(c) 5y+3y2
Terms: 5y,3y2
Factors: 5,y ; 3,y,y

(d) xy+2x2y2
Terms: xy,2x2y2
Factors: x,y ; 2x,x,y,y

(e) pq+q
Terms: pq,q
Factors: p,q ; q

(f) 1.2ab−2.4b+3.6a
Terms: 1.2ab,−2.4b,3.6a
Factors: 1.2,a,b ; −2.4,b ; 3.6,a

(g) 3/4x+14
Terms: 3/4x,14
Factors: 3/4,x ; 14

(h) 0.1p2+0.2q2
Terms: 0.1p2,0.2q2
Factors: 0.1,p,p ; 0.2,q,q
Exercise 12.1 Page Number 234

3

Identify the numerical coefficients of terms (other than constants) in the following expressions:

(i)5−3t2

(ii)1+t+t2+t3

(iii)x+2xy+3y

(iv) 100m+1000n

(v)−p2q2+7pq

(vi) 1.2a+0.8b

(vii) 3.14r2

(viii) 2(l+b)

(ix)  0.1y+0.01y2

Answer

S.No.

Expression

Terms

Numerical Coefficient

(i)

5−3t2

−3t2

−3

(ii)

1+t+t2+t3

t

1

t2

1

t3

1

(iii)

x+2xy+3y

x

1

2xy

2

3y

3

(iv)

100m+1000n

100m

100

1000n

1000

(v)

−p2q2+7pq

−p2q2

−1

7pq

7

(vi)

1.2a+0.8b

1.2a

1.2

0.8b

0.8

(vii)

3.14r2

3.14r2

3.14

(viii)

2(l+b)=2l+2b

2l

2

2b

2

(ix)

0.1y+0.01y2

0.1y

0.1

0.01y2

0.01
Exercise 12.1 Page Number 235

4

(a) Identify terms which contain x and give the coefficient of x.

(i) y2x+y

(ii) 13y2−8yx

(iii) x+y+2

(iv) 5+z+zx

(v) 1+x+xy

(vi) 12xy2+25

(vii) 7x+xy2


(b) Identify terms which contain y2 and give the coefficient of y2.

(i) 8−xy2

(ii) 5y2+7x

(iii) 2x2y−15xy2+7y2

Answer

(a)

S.No.

Expression

Term with factor x

Coefficient of x

(i)

y2x+y

y2x

y2

(ii)

13y2−8yx

−8yx

−8y

(iii)

x+y+2

x

1

(iv)

5+z+zx

zx

z

(v)

1+x+xy

x

1

xy

y

(vi)

12xy2+25

12xy2

12y2

(vii)

7x+xy2

xy2

y2

7x

7

 

(b)

S.No.

Expression

Term contains y2

Coefficient of y2

(i)

8−xy2

−xy2

−x

(ii)

5y2+7x

5y2

5

(iii)

2x2y−15xy2+7y2

−15xy2

−15x

7y2

7
Exercise 12.1 Page Number 235

5

Classify into monomials, binomials and trinomials:

(i) 4y−7x

(ii) y2

(iii) x+y−xy

(iv) 100

(v) ab−a−b

(vi) 5−3t

(vii) 4p2q−4pq2

(viii) 7mn

(ix) z2−3z+8

(x) a2+b2

(xi) z2+z

(xii) 1+x+x2

Answer

S.No.

Expression

Type of Polynomial

(i)

4y−7z

Binomial

(ii)

y2

Monomial

(iii)

x+y−xy

Trinomial

(iv)

100

Monomial

(v)

ab−a−b

Trinomial

(vi)

5−3t

Binomial

(vii)

4p2q−4pq2

Binomial

(viii)

7mn

Monomial

(ix)

z2−3z+8

Trinomial

(x)

a2+b2

Binomial

(xi)

z2+z

Binomial

(xii)

1+x+x2

Trinomial
Exercise 12.1 Page Number 235

6

State whether a given pair of terms is of like or unlike terms:

(i) 1, 100
(ii) −7x,5/2x
(iii) −29x,−29y
(iv)  14xy,42yx
(v)  4m2p,4mp2
(vi) 12xz,12x2z2

Answer

S.No.

Pair of terms

Like / Unlike terms

(i)

1, 100

Like terms

(ii)

−7x,5/2x

Like terms

(iii)

−29x,−29y

Unlike terms

(iv)

14xy,42yx

Like terms

(v)

4m2p,4mp2

Unlike terms

(vi)

12xz,12x2z2

Unlike terms
Exercise 12.1 Page Number 235

7

Identify like terms in the following:
(a) -xy2, -4yx2, 8x2, 2xy2, 7y, -11x2 -100x, -11yx, 20x2y, - 6x2, y, 2xy, 3x
(b) 10pq, 7p, 8q, -p2q2, -7qp, -100q, -23, 12q2p2, -5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2

Answer

(a) Like terms are:
(i) −xy2,2xy2
(ii) −4yx2,20x2y
(iii) 8x2,−11x2,−6x2
(iv) 7y,y
(v) −100x,3x
(vi) −11yx,2xy

(b) Like terms are:
(i) 10pq,−7pq,78pq
(ii) 7p,2405p
(iii) 8q,−100q
(iv) −p2q2,12p2q2
(v) −12,41
(vi) −5p2,701p2
(vii) 13p2q,qp2

Exercise 12.1 Page Number 235

1

Simplify combining like terms:
1. 21b−32+7b−20b
2. −z2+13z2−5z+7z3−15z
3. p−(p−q)−q−(q−p)
4. 3a−2b−ab−(a−b+ab)+3ab+b−a
5. 5x2y−5x2+3yx2−3y2+x2−y2+8xy2−3y2
6. (3y2+5y−4)−(8y−y2−4)

Answer

(i) 21b−32+7b−20b=21b+7b−20b−32
= 28b−20b−32 = 8b−32

(ii) −z2+13z2−5z+7z3−15z
=7z3+(−z2+13z2)−(5z+15z)
= 7z3+12z2−20z

(iii) p−(p−q)−q−(q−p)=p−p+q−q−q+p
= p−p+p+q−q−q = p−q

(iv) 3a−2b−ab−(a−b+ab)+3ab+b−a
=3a−2b−ab−a+b−ab+3ab+b−a
= 3a−a−a−2b+b+b−ab−ab+3ab
= (3a−a−a)−(2b−b−b)−(ab+ab−3ab)
= a−0−(−ab)
= a+ab

(v) 5x2y−5x2+3yx2−3y2+x2−y2+8xy2−3y2
=5x2y+3yx2+8xy2−5x2+x2−3y2−y2−3y2
= (5x2y+3x2y)+8xy2−(5x2−x2)−(3y2+y2+3y2)
= 8x2y+8xy2−4x2−7y2

(vi) (3y2+5y−4)−(8y−y2−4)=3y2+5y−4−8y+y2+4
= (3y2+y2)+(5y−8y)−(4−4)
= 4y2−3y−0 = 4y2−3y
Exercise 12.2 Page Number 239

2

Add:
1. 3mn,−5mn,8mn−4mn
2. t−8tz,3tz−z,z−t
3. −7mn+5,12mn+2,9mn−8,−2mn−3
4. a+b−3,b−a+3,a−b+3
5. 14x+10y−12xy−13,18−7x−10y+8xy,4xy
6. 5m−7n,3n−4m+2,2m−3mn−5
7. 4x2y,−3xy2,−5xy2,5x2y
8. 3p2q2−4pq+5,−10p2q2,15+9pq+7p2q2
9. ab−4a,4b−ab,4a−4b
10. x2−y2−1,y2−1−x2,1−x2−y2

Answer

(i) 3mn,−5mn,8mn,−4mn
=3mn+(−5mn)+8mn+(−4mn)
= (3−5+8−4)mn = 2mn

(ii) t−8tz,3tz−z,z−t=t−8tz+3tz−z+z−t
= t−t−8tz+3tz−z+z
= (1−1)t+(−8+3)tz+(−1+1)z
= 0−5tz+0 = −5tz

(iii) −7mn+5,12mn+2,9mn−8,−2mn−3
=−7mn+5+12mn+2+9mn−8+(−2mn)−3
= −7mn+12mn+9mn−2mn+5+2−8−3
= (−7+12+9−2)mn+7−11
= 12mn−4

(iv) a+b−3,b−a+3,a−b+3
= a + b - 3 + b - a + 3 + a - b + 3
= (a−a+a)+(b+b−b)−3+3+3
= a+b+3

(v) 14x+10y−12xy−13,18−7x−10y+8xy,4xy=14x+10y−12xy−13+18−7x−10y+8xy+4xy
= 14x−7x+10y−10y−12xy+8xy+4xy−13+18
= 7x+0y+0xy+5 = 7x+5

(vi) 5m−7n,3n−4m+2,2m−3mn−5=5m−7n+3n−4m+2+2m−3mn−5
= 5m−4m+2m−7n+3n−3mn+2−5
= (5−4+2)m+(−7+3)n−3mn−3
= 3m−4n−3mn−3

(vii) 4x2y,−3xy2,−5xy2,5x2y=4x2y+(−3xy2)+(−5xy2)+5x2y
= 4x2y+5x2y−3xy2−5xy2
= 9x2y−8xy2

(viii) 3p2q2−4pq+5,−10p2q2,15+9pq+7p2q2 = 3p2q2−4pq+5+(−10p2q2)+15+9pq+7p2q2
= 3p2q2−10p2q2+7p2q2−4pq+9pq+5+15
= (3−10+7)p2q2+(−4+9)pq+20
= 0p2q2+5pq+20= 5pq+20

(ix) ab−4a,4b−ab,4a−ab=ab−4a+4b−ab+4a−ab
= −4a+4a+4b−4b+ab−ab
= 0+0+0=0

(x) x2−y2−1,y2−1−x2,1−x2−y2 = x2−y2−1+y2−1−x2+1−x2−y2
= x2−x2−x2−y2+y2−y2−1−1+1
= (1−1−1)x2+(−1+1−1)y2−1−1+1
= −x2−y2−1
Exercise 12.2 Page Number 239

3

Subtract:
1. −5y2 from y2
2. 6xy from −12xy
3. (a−b) from (a+b)
4. a(b−5) from b(5−a)
5. −m2+5mn from 4m2−3mn+8
6. −x2+10x−5 from 5x−10
7. 5a2−7ab+5b2 from 3ab−2a2−2b2
8. 4pq−5q2−3p2 from 5p2+3q2−pq

Answer

(i) y2−(−5y2) = y2+5y2 = 6y2

(ii) −12xy−(6xy) = −12xy−6xy = −18xy

(iii) (a+b)−(a−b) = a+b−a+b = a−a+b+b = 2b

(iv) b(5−a) −a(b−5) = 5b−ab−ab+5a = 5b−2ab+5a = 5a+5b−2ab

(v) 4m2−3mn+8−(−m2+5mn) = 4m2−3mn+8+m2−5mn
= 4m2+m2−3mn−5mn+8
= 5m2−8mn+8

(vi) 5x−10−(−x2+10x−5) = 5x−10+x2−10x+5
= x2+5x−10x−10+5 = x2−5x−5

(vii) 3ab−2a2−2b2−(5a2−7ab+5b2) = 3ab−2a2−2b2−5a2+7ab−5b2
= 3ab+7ab−2a2−5a2−2b2−5b2
= 10ab−7a2−7b2
= −7a2−7b2+10ab

(viii) 5p2+3q2−pq−(4pq−5q2−3p2) = 5p2+3q2−pq−4pq+5q2+3p2
= 5p2+3p2+3q2+5q2−pq−4pq
= 8p2+8q2−5pq
Exercise 12.2 Page Number 240

4

(a) What should be added to x2+xy+y2 to obtain 2x2+3xy?
(b) What should be subtracted from 2a+8b+10 to get −3a+7b+16 ?

Answer

(a) Let p should be added.
Then according to question,
x2+xy+y2+p=2x2+3xy⇒ p=2x2+3xy−(x2+xy+y2)
⇒ p=2x2+3xy−x2−xy−y2⇒ p=2x2−x2−y2+3xy−xy
⇒ p=x2−y2+2xy
Hence, x2−y2+2xy should be added.

(b) Let qq should be subtracted.
Then according to question,
2a+8b+10−q=−3a+7b+16
⇒ −q=−3a+7b+16−(2a+8b+10)
⇒ −q=−3a+7b+16−2a−8b−10
⇒ −q=−3a−2a+7b−8b+16−10
⇒ −q=−5a−b+6⇒q=−(−5a−b+6)
⇒ q=5a+b−6
Exercise 12.2 Page Number 240

5

What should be taken away from 3x2−4y2+5xy+20 to obtain −x2−y2+6xy+20 ?

6. (a) From the sum of 3x−y+11 and −y−11, subtract 3x−y−11
(b) subtract the sum of 3x2−5x and −x2+2x+5 from the sum of  4+3x and 5−4x+2x2

Answer

Let qq should be subtracted.
Then according to question,
3x2−4y2+5xy+20−q=−x2−y2+6xy+20
⇒ q=3x24y2+5xy+20−(−x2−y2+6xy+20)
⇒ q=3x2−4y2+5xy+20+x2+y2−6xy−20
⇒ q=3x2+x2−4y2+y2+5xy−6xy+20−20
⇒ q=4x2−3y2−xy+0
Hence, 4x2−3y2−xyshould be subtracted.

6. (a) According to question,
(3x−y+11)+(−y−11)−(3x−y−11)(3x−y+11)+(−y−11)−(3x−y−11)
= 3x−3x−y−y+y+11−11+11
= (3−3)x−(1+1−1)y+11+11−11
= 0x−y+11 = −y+11

(b) According to question,
[(4+3x)+(5−4x+2x2)]−[(3x2−5x)+(−x2+2x+5)]
= [4+3x+5−4x+2x2]−[3x2−5x−x2+2x+5]
= [2x2+3x−4x+5+4]−[3x2−x2+2x−5x+5]
= [2x2−x+9]−[2x2−3x+5]
= 2x2−x+9−2x2+3x−5
= 2x2−2x2−x+3x+9−5
= 2x+4
Exercise 12.2 Page Number 240

1

If m=2, find the value of:
(i) m−2
(ii) 3m−5
(iii) 9−5m
(iv) 3m2−2m−7
(v) 5m/2−4

Answer

(i) m−2 = 2−2[Putting m=2]
= 0

(ii) 3m−5 = 3×2−5 [Putting m=2]
= 6 – 5 = 1

(iii) 9−5m = 9 – 5 x 2 [Putting m=2]
= 9 – 10 = −1

(iv) 3m2−2m−7= 3(2)2−2(2)−7 [Putting m=2]
= 3 x 4 – 2 x 2 – 7 = 12 – 4 – 7
= 12 – 11 = 1

(v) 5m2−4 = (5×2)/2−4
[Putting m=2]
= 5 – 4 = 1
Exercise 12.3 Page Number 242

2

If p=−2, find the value of:
(i) 4p+7
(ii) −3p2+4p+7
(iii) −2p3−3p2+4p+7

Answer

(i) 4p+7 = 4(−2)+7 [Putting p=−2]
= −8+7 = −1

(ii) −3p2+4p+7= −3(−2)2+4(−2)+7 [Putting p=−2]
= −3×4−8+7 = −12−8+7
= −20+7 = −13

(iii) −2p3−3p2+4p+7 = −2(−2)3−3(−2)2+4(−2)+7 [Putting p=−2]
= −2×(−8)−3×4−8+7= 16−12−8+7
= −20+23 = 3
Exercise 12.3 Page Number 242

3

Find the value of the following expressions, when x=−1:x=−1:
(i) 2x−7
(ii) −x+2
(iii) x2+2x+1
(iv) 2x2−x−2

Answer

(i) 2x−7 = 2(−1)−7
[Putting x=−1]
= −2−7 = −9

(ii) −x+2 = −(−1)+2
[Putting x=−1]
= 1 + 2 = 3

(iii) x2+2x+1= (−1)2+2(−1)+1 [Putting x=−1]
= 1 – 2 + 1 = 2 – 2 = 0

(iv) 2x2−x−2 = 2(−1)2−(−1)−2 [Putting x=−1]
= 2 x 1 + 1 – 2 = 2 + 1 – 2
= 3 – 2
= 1
Exercise 12.3 Page Number 242

4

If a=2,b=−2find the value of:
(i) a2+b2
(ii) a2+ab+b2
(iii) a2−b2

Answer

(i) a2+b2 = (2)2+(−2)2 [Putting a=2,b=−2]
= 4 + 4 = 8

(ii) a2+ab+b2 = (2)2+(2)(−2)+(−2)2 [Putting a=2,b=−2]
= 4 – 4 + 4 = 4

(iii) a2−b2 = (2)2−(−2)2 [Putting a=2,b=−2]
= 4 – 4 = 0
Exercise 12.3 Page Number 242

5

When a=0,b=−1,find the value of the given expressions:
(i) 2a+2b
(ii) 2a2+b2+1
(iii) 2a2b+2ab2+ab
(iv) a2+ab+2

Answer

(i) 2a+2b = 2(0)+2(−1)
[Putting a=0,b=−1]
= 0 – 2 = −2

(ii) 2a2+b2+1 = 2(0)2+(−1)2+1 [Putting a=0,b=−1]
= 2 x 0 + 1 + 1 = 0 + 2 = 2

(iii) 2a2b+2ab2+ab = 2(0)2(−1)+2(0)(−1)2+(0)(−1) [Putting a=0,b=−1]
= 0 + 0 + 0 = 0

(iv) a2+ab+2 = (0)2+(0)(−1)+2 [Putting a=0,b=−1]
= 0 + 0 + 2 = 2
Exercise 12.3 Page Number 242

6

Simplify the expressions and find the value if x = 2:
(i) x+7+4(x−5)
(ii) 3(x+2)+5x−7
(iii) 6x+5(x−2)
(iv) 4(2x−1)+3x+11

Answer

(i) x+7+4(x−5) = x+7+4x−20 = x+4x+7−20
= 5x−13 = 5×2−13 [Putting x=2]
= 10−13 = −3

(ii) 3(x+2)+5x−7 = 3x+6+5x−7 = 3x+5x+6−7
= 8x−1 = 8 x 2 – 1 [Putting x=2]
= 16 – 1 = 15

(iii) 6x+5(x−2) = 6x+5x−10 = 11x−10
= 11 x 2 – 10 [Putting x=2]
= 22 – 10 = 12

(iv) 4(2x−1)+3x+11 = 8x−4+3x+11 = 8x+3x−4+11
= 11x+7 = 11 x 2 + 7 [Putting x=−1]
= 22 + 7 = 29
Exercise 12.3 Page Number 242

7

Simplify these expressions and find their values if x=3,a=−1,b=−2:
(i) 3x−5−x+9
(ii) 2−8x+4x+4
(iii) 3a+5−8a+1
(iv) 10−3b−4−5b
(v) 2a−2b−4−5+a

Answer

(i) 3x−5−x+9 = 3x−x−5+9 = 2x+4
= 2×3+4 [Putting x=3]
= 6 + 4 = 10

(ii) 2−8x+4x+4 = −8x+4x+2+4 = −4x+6
= −4×3+6 [Putting x=3]
= −12+6=−6

(iii) 3a+5−8a+1 = 3a−8a+5+1 = −5a+6
= −5(−1)+6 [Putting a=−1]
= 5 + 6 = 11

(iv) 10−3b−4−5b = −3b−5b+10−4 = −8b+6
= −8(−2)+6 [Putting b=−2]
= 16 + 6 = 22

(v) 2a−2b−4−5+a = 2a+a−2b−4−5
= 3a−2b−9 = 3(−1)−2(−2)−9 [Putting a=−1, b=−2]
= −3+4−9 = −8
Exercise 12.3 Page Number 242

8

(i) If z=10, find the value of z3−3(z−10).
(ii) If p=−10, find the value of p2−2p−100.

Answer

(i) z3−3(z−10) = (10)3−3(10−10) [Putting z=10]
= 1000 – 3 x 0 = 1000 – 0 = 1000

(ii) p2−2p−100 = (−10)2−2(−10)−100 [Putting p=−10]
= 100+20−100 = 20
Exercise 12.3 Page Number 242

9

What should be the value of aa if the value of 2x2+x−a2x2+x−a equals to 5, when x=0x=0 ?

Answer

Given: 2x2+x−a=5
⇒ 2(0)2+0−a=5 [Putting x=0x=0]
⇒ 0+0−a=5⇒ a=−5
Hence, the value of a is −5.
Exercise 12.3 Page Number 242

10

Simplify the expression and find its value when a=5 and b=−3: 2(a2+ab)+3−ab

Answer

Given: 2(a2+ab)+3−ab
⇒ 2a2+2ab+3−ab ⇒ 2a2+2ab−ab+3
⇒ 2a2+ab+3
⇒ 2(5)2+(5)(−3)+3 [Putting a=5, b=−3]
⇒ 2 x 25 – 15 + 3
⇒ 50 – 15 + 3
⇒ 38
Exercise 12.3 Page Number 242

1

Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
(a)

11, 16, 21… (5n+1)...
(b) 

7, 10, 13… (3n+1)...
(c) 

7, 12, 17, 22…
If the number of digits formed is taken to be n,n, the number of segments required to form nn digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind
 

Answer

S. No.

Symbol

Digit’s number

Pattern’s Formulae

No. of Segments

(i)



5

5n+1

26

10

51

100

501

(ii)



5

3n+1

16

10

31

100

301

(iii)



5

5n+2

27

10

52

100

502

(i) 5n+1
Putting n=5,5 x 5 + 1 = 25 + 1 = 26
Putting n=10, 5 x 10 + 1 = 50 + 1 = 51
Putting n=100, 5 x 100 + 1 = 500 + 1 = 501

(ii) 3n+1
Putting n=5, 3 x 5 + 1 = 15 + 1 = 16
Putting n=10, 3 x 10 + 1 = 30 + 1 = 31
Putting n=100,3 x 100 + 1 = 300 + 1 = 301
(iii) 5n+2
Putting n=5, 5 x 5 + 2 = 25 + 2 = 27
Putting n=10, 5 x 10 + 2 = 50 + 2 = 52
Putting n=100, 5 x 100 + 2 = 500 + 2 = 502

Exercise 12.4 Page Number 246

2

Use the given algebraic expression to complete the table of number patterns:

S.No.

Expression

Terms

1st

2nd

3rd

4th

5th

…

10th

…

100th

…

(i)

2n−1

1

3

5

7

9

---

19

---

---

---

(ii)

3n+2

2

5

8

11

---

---

---

---

---

---

(iii)

4n+1

5

9

13

17

---

---

---

---

---

---

(iv)

7n+20

27

34

41

48

---

---

---

---

---

---

(v)

n2+1

2

5

10

17

---

---

---

---

10001

---

Answer

 

(i) 2n−1
Putting n=100, 2 x 100 – 1 = 200 – 1 = 199

(ii) 3n+2
Putting n=5, 3 x 5 + 2 = 15 + 2 = 17
Putting n=10, 3 x 10 + 2 = 30 + 2 = 32
Putting n=100, 3 x 100 + 2 = 300 + 2 = 302

(iii) 4n+1
Putting n=5,4 x 5 + 1 = 20 + 1 = 21
Putting n=10, 4 x 10 + 1 = 40 + 1 = 41
Putting n=100, 4 x 100 + 1 = 400 + 1 = 401

(iv) 7n+20
Putting n=5, 7 x 5 + 20 = 25 + 20 = 55
Putting n=10, 7 x 10 + 20 = 70 + 20 = 90
Putting n=100, 7 x 100 + 20 = 700 + 20 = 720

(v) n2+1
Putting n=5, 5 x 5 + 1 = 25 + 1 = 26
Putting n=10, 10 x 10 + 1 = 100 + 1 = 101
Putting n=100, 100 x 100 + 1 = 10000 + 1 = 10001
Now complete table is,

S.No.

Expression

Terms

1st

2nd

3rd

4th

5th

…

10th

…

100th

…

(i)

2n−1

1

3

5

7

9

---

19

---

199

---

(ii)

3n+2

2

5

8

11

17

---

32

---

302

---

(iii)

4n+1

5

9

13

17

21

---

41

---

401

---

(iv)

7n+20

27

34

41

48

55

---

90

---

720

---

(v)

n2+1

2

5

10

17

26

---

101

---

10001

---
Exercise 12.4 Page Number 146

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