Class – 7 CH-12 ALGEBRAIC EXPRESSIONS MATHS NCERT SOLUTIONS

 

Class – 7 CH-12 ALGEBRAIC EXPRESSIONS  

MATHS NCERT SOLUTIONS

 Exercise 12.1

Question 1:   

Get the algebraic expressions in the following cases using variables, constants and arithmetic operations: 

(i) Subtraction of z from y.  

(ii) One-half of the sum of numbers x and y.  

(iii) The number z multiplied by itself. 

(iv) One-fourth of the product of numbers p and q.  

(v) Numbers x and y both squared and added. 

(vi) Number 5 added to three times the product of m and n.  (vii) Product of numbers y and z subtracted from 10. (viii) Sum of numbers a and b subtracted from their product. 

  SOLUTION 1:  

(i) y z    (ii) 

  

2

(iii) z2  (iv) pq

   

4

(v) x2 y2  (vi) 3mn5  

x y

(vii) 10 yz  (viii) ab a b 

 

 

 

Question 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 y 3  

(d) 5xy2 7x y2  (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2x y2 2    (e) pq q  (f) 1.2ab2.4b3.6a  

3 1 2 0.2q2 

(g) x  (h) 0.1p 

4 4

 

 

 

 SOLUTION 2:

(i) (a) x3  

Expression 

 

Terms  

 

Factors 

 

(b) 1 x x2  

Expression 

 

Terms 

 

Factors 

 

(c) y y 3  

  Expression 

 

  Terms 

 

  Factors 

 

(d) 5xy2 7x y2  

  Expression 

 

  Terms 

 

  Factors 

 

 

(e)  ab 2b2 3a2  

  Expression 

   

Terms 

   

Factors       

 

(ii) (a)  4x 5  (b)  4x 5y  

  Terms: 4x,5  Terms: 4x,5y  

  

  Factors: 4, ; x 5  Factors: 4, ; 5,x y  

  (c) 5y3y2  (d) xy2x y2 2  

  Terms: 5y y,3 2  Terms: xy x y,2 2 2  

  

  Factors: 5, ; 3, ,y y y    Factors: x y, ; 2 , , ,x x y y  

  (e) pq q  (f) 1.2ab2.4b3.6a  

  Terms: pq q,    Terms: 1.2ab, 2.4 ,3.6 b a  

  

  Factors: p q q, ;    Factors: 1.2, , ; a b 2.4, ; 3.6,b a  

  3 1

(g) x   

4 4 (h) 0.1p2 0.2q2  

  3 1

Terms: x,    

4 4 Terms: 0.1p2,0.2q2  

  3 1

Factors: , ; x   Factors: 0.1, p p, ; 0.2, ,q q 

4 4

 

 

Question 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) p q2 2 7pq    (vi) 1.2a0.8b  

(vii) 3.14r2  (viii) 2l b  (ix) 0.1y0.01y2 

 

  SOLUTION 3: 

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) p q2 2 7pq  p q2 2  1  

7 pq  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 2 2l b  2l  

2b  2 

2 

(ix) 0.1y0.01y2  0.1y  0.1 

0.01y2  0.01 

 

Question 4:  

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

(i) y x y2   (ii) 13y2 8yx  (iii) x y 2  

(iv) 5 z zx  (v) 1 x xy  (vi) 12xy2 25  

(vii) 7x xy 2  

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

  (i) 8xy2  (ii) 5y2 7x  (iii) 2x y2 15xy2 7y2 

  SOLUTION 4:  

(a)  

S.No. Expression Term with factor x  Coefficient of x  

(i) y x y2   y x2  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 xy 2  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) 2x y2 15xy2 7y2  15xy2  15x  

7y2  7 

 

Question 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) 4p q2 4pq2    (viii) 7mn    (ix) z2  3z 8  

(x) a2 b2  (xi) z2 z  (xii) 1 x x2 

  SOLUTION 5:  

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 3 t  Binomial 

(vii) 4p q2 4pq2  Binomial 

(viii) 7mn  Monomial 

(ix) z2  3z 8  Trinomial 

(x) a2 b2 Binomial 

(xi) z2 z Binomial 

(xii) 1 x x2  Trinomial 

 

  

Question 6:   

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

(i) 1,   100  (ii) 7x x,   (iii) 29x,29y  

(iv) 14xy,42yx  (v) 4m p mp2 ,4 2    (vi) 12xz,12x z2 2 

 

  SOLUTION 6: 

S.No. Pair of terms Like / Unlike terms 

(i) 1,  100 Like terms 

(ii) 7x x,   Like terms 

(iii) 29x,29y  Unlike terms 

(iv) 14xy,42yx  Like terms 

(v) 4m p mp2 ,4 2  Unlike terms 

(vi) 12xz,12x z2 2  Unlike terms 

 

 

Question 7:   

Identify like terms in the following: 

(a) xy2, 4 yx x2,8 2,2xy2,7 , 11y  x2 100 , 11x  yx,20x y2 , 6 x y xy x2, ,2 ,3  

(b) 10pq p q p q,7 ,8 , 2 2, 7 qp, 100 , 23,12 q  q p2 2, 5 p2,41,2405 ,78p qp,13p q qp2 , 2,701p2

 

  SOLUTION 7:  

(a) Like terms are:

(i) xy xy2,2 2  (ii) 4yx2,20x y2    (iii) 8x2, 11 x2, 6 x2  

(iv) 7y y,      

 

(b) Like terms are: (v) 100 ,3x x  (vi) 11yx xy,2   

(i) 10pq, 7 pq,78pq  (ii) 7 ,2405p p  (iii) 8 , 100q  q  

(iv) p q2 2,12p q2 2    (v) 12,41  (vi) 5p2,701p2  

(vii) 13p q qp2 , 2 

 

 Exercise 12.2  

Question 1:   

Simplify combining like terms: 

(i) 21b  32 7b 20b  

(ii)  z2 13z2  5x 7z3 15z  

(iii) p     p q q q p  

(iv) 3 2a    b ab a b ab3ab b a  

(v) 5x y2 5x2 3yx2 3y2   x2 y2 8xy2 3y2  (vi) 3y2   5 4y  8y y 2 4   SOLUTION 1:  

(i) 21 32 7 20 21 7 20 32b  b b  b b b  

       = 28 20 32b b = 8 32b  

(ii)  z2 13z2 5 7z z3 15 7z z3    z2 13z25 15z z  

       = 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)

 

5x y2 5x2 3yx2 3y2   x2 y2 8xy2 3y2 5x y2 3yx2 8xy2 5x2  x2 3y2  y2 3y2  

  = 5x y2 3x y2 8xy2 5x2 x23y2 y2 3y2  

  = 8x y2 8xy2 4x2 7y2  

(vi) 3y2 5 4y 8y y 2 4 3 y2 5 4 8y  y y 2 4  

           = 3y2 y25 8y y4 4   

           = 4y2  3y 0 = 4y2 3y  

 

 

Question 2:   

Add: 

(i) 3mn, 5 mn,8mn4mn  

(ii) t 8tz,3tz  z z, t  

(iii) 7mn5,12mn 2,9mn 8, 2mn3  

(iv) a   b 3,b a 3,a b 3  

(v) 14x10y 12xy 13,18 7x 10y8xy,4xy  

(vi) 5m7n n,3  4m 2,2m3mn5  

(vii) 4x y2 , 3 xy2, 5 xy x y2,5 2  

(viii) 3p q2 2 4pq 5, 10p q2 2,159pq7p q2 2  

(ix) ab4a,4bab,4a4b  

(x) x2  y2 1, y2  1 x2,1 x2 y2   SOLUTION 2:  

(i) 3mn, 5 mn mn,8 , 4 mn 3mn  5mn8mn  4mn  

= 3  5 8 4mn = 2mn  

 

(ii) t 8tz tz,3        z z, t t 8tz 3tz z z t  

  = t     t 8 3tz tz z z  

  = 1 1   t  8 3tz    1 1z  

  = 0 5 0 tz = 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 xy,4 14x10y12xy   13 18 7x 10y8xy 4xy  

  = 14x 7x 10y10y12xy8xy 4xy 13 18  

  = 7x 0y 0xy5 = 7 5x  

(vi) 5m7n n,3  4m 2,2m3mn       5 5m 7n 3n 4m 2 2m 3mn5  

  = 5 4m    m 2 7 3 3m n n mn 2 5  

  = 5 4 2m   7 3n3mn3  

  = 3 4 3m n mn3  

(vii) 4x y2 , 3 xy2, 5 xy x y2,5 2 4x y2  ( 3xy2) ( 5xy2)5x y2  

  = 4x y2 5x y2 3xy2 5xy2  

  = 9x y2 8xy2  

(viii) 3p q2 2 4pq 5, 10p q2 2,159pq7p q2 2  

= 3p q2 2 4pq  5  10p q2 215 9 pq7p q2 2 

  = 3p q2 2 10p q2 2 7p q2 2 4pq9pq 5 15  

    = 3 10 7 p q2 2    4 9 pq 20      = 0p q2 2 5pq20 = 5pq 20  

(ix) ab4a b ab a ab ab,4  ,4       4a 4b ab 4a ab  

  =      4 4 4 4a a b b 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 

  = x x x2        2 2 y2 y2 y2 1 1 1  

  = 1 1 1  x2     1 1 1y2   1 1 1  

  =   x2 y2 1 

 

Question 3:   

Subtract: 

(i) 5y2 from y2  

(ii) 6xy from 12xy  

(iii) a b  from a b   

(iv) a b 5 from b5a  

(v)  m2 5mn from 4m2 3mn8  

(vi)  x2 10x5 from 5x10  

(vii) 5a2 7ab5b2 from 3ab2a2 2b2  

(viii) 4pq5q2 3p2 from 5p2 3q2 pq 

 

  SOLUTION 3:  

(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  

= 5 2b ab5a  

= 5 5 2a b ab  (v) 4m2 3mn  8  m2 5mn  

= 4m2 3mn  8 m2 5mn  

  = 4m2  m2 3mn5mn8  

  = 5m2 8mn8  

(vi) 5 10x    x2 10 5x   

= 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 

 

 

Question 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 ? 

  SOLUTION 4:  

(a) Let p should be added. Then according to question, 

x2    xy y2 p 2x2 3xy     

  p2x2 3xyx2 xy y 2  

 p2x2 3xy x  2 xy y2   

 p 2x2   x2 y2 3xy xy  

 p x  2 y2 2xy  

Hence, x2  y2 2xy should be added. 

(b) Let q 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 

 

Question 5:   What should be taken away from 3x2 4y2 5xy20 to obtain   x2 y2 6xy20 ? 

  SOLUTION 5:  

Let q should be subtracted. 

Then according to question, 

  3x2 4y2 5xy    20 q x2 y2 6xy20  

 q3x2 4y2 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 3y xy2  0  

Hence, 4x2 3y xy2  should be subtracted. 

 

Question 6:   

(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11. 

(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and    –x2 + 2x + 5. 

  SOLUTION 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 3 x5 4 2 x x2   3x2 5x   x2 2 5x   

 = 4 3 5 4 2 x  x x2   3x2 5x x 2 2 5x    = 2x2 3 4 5 4x x     3x2 x2 2 5 5x x   

 = 2x2  x 9   2x2 3 5x   

= 2x2   x 9 2x2  3x 5  

= 2x2 2x2    x 3x 9 5  

 = 2 4x 

 

 

 

 Exercise 12.3  

If m  2, find the value of:

(i) m2    (ii) 3m5  (iii) 95m   

(iv) 3m2  2m 7  (v) 5m 4 

2

Question 1:   

 SOLUTION 1:  

(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 

5m 5 2

(v) 4 = 4  [Putting m 2 ] 

 

2 2

  = 5 – 4 = 1 

 

 

Question 2:   

If p2, find the value of: 

(i) 4p7  (ii) 

 SOLUTION 2:  

 

3p2  4p 7  (iii) 2p3 3p2  4p 7 

 

(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     

  =       2  8 3 4 8 7   

= 16 12  8 7  

  =  20 23 = 3 

 

Question 3:   

Find the value of the following expressions, when x1:  

(i) 2x7  (ii)  x 2  (iii) 

(iv) 2x2  x 2 

  SOLUTION 3:  

(i) 2x7 = 2 1 7     

=  2 7 = 9  

(ii)  x 2 =    1 2      

  = 1 + 2 = 3 

(iii) x2  2x 1 =    12 2 1 1     

  = 1 – 2 + 1  

= 2 – 2 = 0 

(iv) 2x2  x 2 = 2 1   2  1 2    

  = 2 x 1 + 1 – 2  

= 2 + 1 – 2  

= 3 – 2 = 1 

 

Question 4:   

If a  2,b 2, find the value of: 

(i) a2 b2  (ii) a2  ab b2   

  SOLUTION 4:  

(i) a2 b2 = 22   22   

= 4 + 4 = 8 

(ii) a2  ab b2  

= 22 2 2    22       = 4 – 4 + 4 = 4 

 

  [Putting p 2] x2  2x 1   

[Putting x1] 

[Putting x 1] 

[Putting x1] 

[Putting x1] 

(iii) a2 b2 

[Putting a  2,b 2 ] 

[Putting a  2,b 2 ] 

(iii) a2 b2 = 22   22  [Putting a  2,b 2 ] 

= 4 – 4 = 0 

 

 

Question 5:   

When a  0,b 1, find the value of the given expressions: 

(i) 2a2b  (ii) 2a2  b2 1  (iii) 2a b2  2ab2 ab  (iv) a2  ab 2 

  SOLUTION 5:  

(i) 2a2b  = 2 0 2 1     [Putting a  0,b 1] 

   = 0 – 2 = 2  

(ii) 2a2  b2 1 = 2 0 2    1 12   [Putting a  0,b 1] 

           = 2 x 0 + 1 + 1 = 0 + 2 = 2 

(iii) 2a b2  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 

 

 

Question 6:   

Simplify the expressions and find the value if x is equal to 2: 

(i) x 7 4x5  (ii) 3x  2 5 7 x   

(iii) 6x5x 2    (iv) 4 2 x  1 3x 11 

  SOLUTION 6:  

(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 5 7 x = 3x  6 5x 7 = 3x  5x 6 7  

  = 8x1 = 8 x 2 – 1  [Putting x1] 

  = 16 – 1 = 15 

(iii) 6 5x x2 = 6x 5x 10 = 11x10  

  = 11 x 2 – 10  [Putting x1] 

  = 22 – 10 = 12 

 

(iv) 4 2 1 3 11 x   x = 8x  4 3x 11 = 8x  3x 4 11  

  = 11x7 = 11 x 2 + 7 [Putting x1] 

  = 22 + 7 = 29 

 

Question 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 

  SOLUTION 7:  

(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 12  

(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 

 

Question 8:   

(i) If z 10, find the value of z3 3z10 .  

(ii) If p 10, find the value of p2  2p 100. 

  SOLUTION 8:  

(i) z3 3z 10 = 103 310 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 

 

Question 9:   

What should be the value of a if the value of 2x2  x a equals to 5, when x 0 ? 

  SOLUTION 9:  

Given: 2x2   x a 5  

2

  2 0    0 a 5  [Putting x  0] 

  0  0 a 5     

  a5  

Hence, the value of a is 5. 

 

Question 10:   

Simplify the expression and find its value when a 5 and b3: 2a ab2   3 ab   SOLUTION 10:  

Given: 2a ab2   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 12.4  

Question 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)       …    … 

6 11            16       21…   5n1 ...  

 

(b)       … 

      4  

        7             10       13…   3n1 ...  

(c)        … 

7 12               17        22…   5n 2 ...  

If the number of digits formed is taken to be n, the number of segments required to form n 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    

 

  SOLUTION 1:  

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 

(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 

 

 

Question 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 --- 

  SOLUTION 2:  

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  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 ---