Algebraic Expressions

EXERCISE-12.1

P1. 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 number m and n. (vii) Product of numbers y and z subtracted from 10.

(viii) Sum of numbers and b subtracted from their product.

Sol. (i) y − z

(ii)12x+y

(iii)  z2

(iv)14pq

(v) x2 + y2

(vi) 5 + 3 (mn)

(vii) 10 − yz

(viii)  ab − (a + b)

P2. (i)Identify the terms and their factors in the following expressions Show the terms and factors by tree diagrams.

(a) x − 3

(b) 1 + x + x2

(c) − y3

(d) 5xy2+7x2y

(e) − ab + 2b2 − 3a2

(ii) Identify terms and factors in the expressions given below:

(a) − 4x + 5

(b) − 4x + 5y 

(c) 5+ 3y2

(d) xy + 2x2y

(e) pq + q

(f) 1.2 ab − 2.4 b + 3.6 

(g) 34x+14

(h) 0.1p2 + 0.2 q2

Sol. (i) (a)

(b)

(c)

(d)

(e)

(ii)

Row

Expression

Terms

Factors

(a)

− 4x + 5

− 4x

5

− 4, x

5

(b)

− 4x + 5y

− 4x

5y

− 4, x

5, y

(c)

5y + 3y2

5y

3y2

5, y

3, yy

(d)

xy + 2x2y2

xy

2x2y2

xy

2, xxyy

(e)

pq q

pq

q

pq

q

(f)

1.2ab − 2.4b + 3.6a

1.2ab

− 2.4b

3.6a

1.2, ab

− 2.4, b

3.6, a

(g)

34x+14 34x14

34, x, 14

(h)

0.1p2 + 0.2q2

0.1p2

0.2q2

0.1, pp

0.2, qq

 P3. Identify the numerical coefficients of terms (other than constants) in the following expressions:

(i) 5 − 3t2

(ii) 1 + t2 + t3 

(iii) x + 2xy+ 3y

(iv) 100m + 1000n 

(v) − p2q2 + 7pq

(vi) 1.2a + 0.8b

(vii) 3.14 r2

(viii) 2 (b)

(ix) 0.1y + 0.01 y2

Sol.

Row

Expression

Terms

Coefficients

(i)

5 − 3t2

− 3t2

− 3

(ii)

1 + t + t2 + t3

t

t2

t3

1

1

1

(iii)

x + 2xy + 3y

x

2xy

3y

1

2

3

(iv)

100m + 1000n

100m

1000n

100

1000

(v)

− p2q2 + 7pq

− p2q2

7pq

− 1

7

(vi)

1.2a +0.8b

1.2a

0.8b

1.2

0.8

(vii)

3.14 r2

3.14 r2

3.14

(viii)

2(l + b)

2l

2b

2

2

(ix)

0.1+ 0.01y2

0.1y

0.01y2

0.1

0.01

P4. (a) Identify terms which contain x and give the coefficient of x.

(i) y2x + y

(ii) 13y2− 8yx 

(iii) x + y + 2

(iv) 5 + 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

Sol. (a)

Row

Expression

Terms with 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 + xy

x

xy

1

y

(vi)

12xy2 + 25

12xy2

12y2

(vii)

7xxy2

7x

xy2

7

y2

(b)

Row

Expression

Terms with y2

Coefficient of y2

(i)

8 − xy2

xy2

− x

(ii)

5y2 + 7x

5y2

5

(iii)

2x2y + 7y2

−15xy2

7y2

−15xy2

7

−15x

 P5. Classify into monomials, binomials and trinomials.

(i) 4y − 7z

(ii) y2

(iii) x + y − xy

(iv) 100

(v) ab − a − b

(vi) 5 − 3t

(vii) 4p2− 4pq2

(viii) 7mn

(ix) z2 − 3z + 8

(x) a2 + b2

(xi) z2 + z

(xii) 1 + x + x2

Sol. The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.

(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

P6. State whether a given pair of terms is of like or unlike terms.

(i) 1, 100

(ii) 7x,52x 

(iii) − 29x, − 29y

(iv) 14xy, 42yx

(v) 4m2p, 4mp2

(vi) 12xz, 12 x2z2

Sol. The terms which have the same algebraic factors are called like terms. However, when the terms have different algebraic factors, these are called unlike terms.

(i) 1, 100 Like

(ii) − 7x52x  Like

(iii) −29x, −29y Unlike

(iv) 14xy, 42yx Like

(v) 4m2p, 4mp2 Unlike

(vi) 12xz, 12x2z2 Unlike

P7. Identify like terms in the following:

(a) −xy2, − 4yx2, 8x2, 2xy2, 7y, − 11x2, − 100x, −11yx, 20x2y, −6x2y, 2xy,3x

(b) 10pq, 7p, 8q, − p2q2, − 7qp, − 100q, − 23, 12q2p2, − 5p2, 41, 2405p, 78qp, 13p2qqp2, 701p2

Sol. (a) −xy2, 2xy2

−4yx2, 20x2y

8x2, −11x2, −6x2

7yy

−100x, 3x

−11xy, 2xy

(b) 10pq, −7qp, 78qp

7p, 2405p

8q, −100q

p2q2, 12p2q2

−23, 41

−5p2, 701p2

13p2qqp2

EXERCISE-12.2

P1. Simplify combining like terms:

(i) 21b − 32 + 7b − 20b

(ii) − z2 + 13z2 − 5z + 7z3 − 15z

(iii) p − (p − q) − q − (− p)

(iv) 3a−2bab − (a − b + ab) + 3ab + b − a

(v) 5x2y − 5x2 + 3y x2 − 3y2 + x− y+ 8xy2 −3y2

(vi) (3 y+ 5y − 4) − (8y − y2 − 4)

Sol. (i) 21b − 32 + 7− 20b = 21b + 7− 20b − 32

b (21 + 7 − 20) −32

= 8b − 32

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

= 7z3 + z2 (−1 + 13) + z (−5 − 15)

= 7z3 + 12z2 − 20z

(iii)  p − (p − q) − q − (q − p)

p − p + q − q − q + p

− q

(iv) 3a − 2b − ab − (a − b + ab) + 3ba + − a

= 3a − 2b − ab − a + b − ab + 3ab+b− a

= 3a − a − a − 2b + b + b ab ab+ 3ab

a (3 − 1 − 1) + b (− 2 + 1 + 1) + ab (−1 −1 + 3)

a + ab

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

= 5x2y + 3yx− 5x2 + x2 − 3y2 − y2 − 3y+ 8xy2

x2(5 + 3) + x2 (−5 + 1)+ y2(−3−1− 3) + 8xy2

= 8x2y − 4x2 − 7y2 + 8xy2

(vi) (3y+ 5y − 4) − (8y − y2 − 4)

= 3y2 + 5y − 4 − 8y + y2 + 4

= 3y2 + y2 + 5y − 8y − 4 + 4

y2 (3 + 1) + y (5 − 8) + 4 (1 − 1)

= 4y2 − 3y

P2. Add:

(i) 3mn, − 5mn, 8mn, −4mn

(ii) t − 8tz, 3tz − zz − 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 − 10+ 8xy, 4xy

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

(vii) 4x2y, − 3xy2, − 5xy2, 5x2y

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

(ix) ab − 4a, 4b − ab, 4a − 4b

(x) x− y2 − 1 , y2 − 1 − x2, 1− x2 − y2

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

(ii) (t − 8tz) + (3tz − z) + (z − t)

t − 8tz + 3tz − z + z − t

− t − 8tz + 3tz − z + z

t (1 − 1) + tz (− 8 + 3) + z (− 1 + 1)

= −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

= mn (− 7 + 12 + 9 − 2) + (5 + 2 − 8 − 3)

= 12mn − 4

(iv) (a + − 3) + (b − a + 3) + (a − b + 3)

a + − 3 + b − a + 3 + a − b + 3

a − a + a + − − 3 + 3 + 3

a (1 − 1 + 1) + b (1 + 1 − 1) + 3 (− 1 + 1 + 1)

a + b + 3

(v) (14+ 10y − 12xy − 13) + (18 − 7x − 10y + 8yx) + 4xy

= 14+ 10y − 12xy − 13 + 18 − 7x − 10y + 8yx + 4xy

= 14− 7x + 10y − 10y − 12xy + 8yx + 4xy − 13 + 18

= x (14 − 7) + (10 − 10) + xy (− 12 + 8 + 4) − 13 + 18

= 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

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

= 3m − 4n − 3mn − 3

(vii) 4x2 − 3xy2 − 5xy2 + 5x2y = 4x2 + 5x2y − 3xy2 − 5xy2

x2 y (4 + 5) + xy2 (− 3 − 5)

= 9x2y − 8xy2

(viii) (3p2q2 − 4pq + 5) + (−10 p2q2) + (15 + 9pq + 7p2q2)

= 3p2q2 − 4pq + 5 − 10 p2q2 + 15 + 9pq + 7p2q2

= 3p2q2 − 10 p2q+ 7p2q− 4pq + 9pq + 5 + 15

p2q2 (3 − 10 + 7) + pq (− 4 + 9) + 5 + 15

= 5pq + 20

(ix) (ab − 4a) + (4− ab) + (4a − 4b)

ab − 4a + 4− ab + 4a − 4b

ab − ab − 4+ 4a + 4− 4b

ab (1 − 1) + a (− 4 + 4) + b(4 − 4)

= 0

(x) (x2 − y2 − 1) + (y2 − 1 − x2)+(1 − x2 − y2)

x2 − y2 − 1 + y2 − 1 − x2 + 1 − x2 − y2

x2 − x− x− y2 + y− y− 1 − 1 + 1

x2(1 − 1 − 1) + y2 (−1 + 1 − 1) + (− 1 − 1 + 1)

= − x2 − y2 − 1

P3. Subtract:

(i) − 5yfrom y2

(ii) 6xy from − 12xy

(iii) (a − b) from (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 + 3q− pq

Sol. (i) y2 − (−5y2) = y2 + 5y2 = 6y2

(ii) − 12xy − (6xy) = −18xy

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

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

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

= 4m2 + m2 − 3mn − 5 mn + 8

= 5m2 − 8mn + 8

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

x2 + 5− 10x − 10 + 5

x2 − 5− 5

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

= 3ab − 2a2 − 2b2 − 5a2 + 7ab − 5 b2

= 3ab + 7ab − 2a− 5a2 − 2b2 − 5 b2

= 10ab − 7a2 − 7b2

(viii) 4pq − 5q2 − 3p2 from 5p2 + 3q2 − pq

(5p2 + 3q2 − pq) − (4pq − 5q2− 3p2)

= 5p2 + 3q2 − pq − 4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2 + 5q2 − pq − 4pq

= 8p2 + 8q2 − 5pq

P4. (a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?

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

Sol. (a) Let a be the required term.

a + (x2 + y2 + xy) = 2x+ 3xy
a = 2x2 + 3xy − (x2 + y2 + xy)

a = 2x2 + 3xy − x2 − y2 − xy

a = 2x2 − x2 − y2 + 3xy − xy x2 − y2 + 2xy

(b) Let p be the required term.

(2a + 8b + 10) − = − 3a + 7b + 16

p = 2a + 8b + 10 − (− 3a + 7b + 16)

= 2a + 8b + 10 + 3a − 7− 16

= 2a + 3a + 8b − 7b + 10− 16

= 5− 6

P5. What should be taken away from 3x2 − 4y2 + 5xy + 20 to obtain − x2 − y+ 6xy + 20?

Sol. Let p be the required term.

(3x2 − 4y2 + 5xy + 20) − p = − x2 − y2 + 6xy + 20

p = (3x2 − 4y2 + 5xy + 20) − (− x2 − y2 + 6xy + 20)

= 3x2 − 4y2 + 5xy +20+x2y2 − 6xy − 20

= 3x2 + x− 4y2 + y2 + 5xy−6xy+20 − 20

= 4x2 − 3y2 − xy

P6. (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.

Sol. (a) (3x − y + 11) + (− y − 11)

= 3x − + 11 − y − 11

= 3x − − y + 11 − 11

= 3x − 2y

(3x − 2y) − (3x − − 11)

= 3x − 2y − 3x + + 11

= 3x − 3x − 2y + + 11

= − y + 11

(b) (4 + 3x) + (5 − 4x + 2x2) = 4 + 3x + 5 − 4x + 2x2

= 3x − 4x + 2x2 + 4 + 5

= − x + 2x2 + 9

(3x2 − 5x) + (− x2 + 2x + 5) = 3x2 − 5x − x2 + 2x + 5

= 3x2 − x− 5x + 2x + 5

= 2x2 − 3x + 5

(− x + 2x2 + 9) − (2x2 − 3x + 5)

= − x + 2x2 + 9 − 2x2 + 3x − 5

= − x + 3x + 2x2 − 2x+ 9 − 5

= 2x + 4

EXERCISE-12.3

P1. If m = 2, find the value of:

(i) m − 2

(ii) 3m − 5

(iii) 9 − 5m

(iv) 3m2 − 2m − 7

(v) 5m24

Sol. (i) m − 2 = 2 − 2 = 0

(ii) 3m − 5 = (3 × 2) − 5 = 6 − 5 = 1

(iii) 9 − 5m = 9 − (5 × 2) = 9 −10 = −1

(iv) 3m2 − 2m − 7 = 3 × (2 × 2) − (2 × 2) − 7 = 12 − 4 − 7 = 1

(v) 5m24=5×254=1

P2. If p = −2, find the value of:

(i) 4p + 7

(ii) −3p2 + 4p + 7

(iii) −2p3 − 3p2 + 4p + 7

Sol. (i) 4p + 7 = 4 × (−2) + 7 = − 8 + 7 = −1

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

(iii) −2p3 − 3p2 + 4p + 7

= −2 (−2) × (−2) × (−2) − 3 (−2) × (−2) + 4 × (−2) + 7

= 16 − 12 − 8 + 7 = 3

P3. Find the value of the following expressions, when x = − 1:

(i) 2x − 7

(ii) − x + 2

(iii) x2 + 2x + 1

(iv) 2x2 − x − 2

Sol. (i) 2− 7 = 2 × (−1) − 7 = −9

(ii) − + 2 = − (−1) + 2 = 1 + 2 = 3

(iii) x2 + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1 = 1 − 2 + 1 = 0

(iv) 2x2 − x − 2 = 2 (−1) × (−1) − (−1) − 2 = 2 + 1 − 2 = 1

P4. If a = 2, b = − 2, find the value of:

(i) ab2

(ii) a2 + ab + b2

(iii) a2 − b2

Sol. (i) a2 + b2 = (2)2 + (−2)2 = 4 + 4 = 8

(ii) a2 + ab + b2 = (2 × 2) + 2 × (−2) + (−2) × (−2) = 4 − 4 + 4 = 4

(iii) a2 − b2 = (2)2 − (−2)2 = 4 − 4 = 0

P5. When a = 0, b = − 1, find the value of the given expressions:

(i) 2a + 2b 

(ii) 2ab2 + 1

(iii) 2ab + 2ab2 + ab

(iv) a2 + ab + 2

Sol. (i) 2a + 2= 2 × (0) + 2 × (−1) = 0 − 2 = −2

(ii) 2a2 + b2 + 1 = 2 × (0)2 + (−1) × (−1) + 1 = 0 + 1 + 1 = 2

(iii) 2a2+ 2ab2 + ab = 2 × (0)2 × (−1) + 2 × (0) × (−1) × (−1) + 0 × (−1) = 0 + 0 + 0 = 0

(iv) a2 + ab + 2 = (0)2 + 0 × (−1) + 2 = 0 + 0 + 2 = 2

P6. 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

(iii) 6x + 5 (− 2)

(iv) 4 (2x −1) + 3+ 11

Sol. (i) x + 7 + 4 (x − 5) = x + 7 + 4x − 20

x + 4x + 7 − 20

= 5x − 13

= (5 × 2) − 13

= 10 − 13 = −3

(ii) 3 (x + 2) + 5x − 7 = 3x + 6 + 5x − 7

= 3x + 5x + 6 − 7 = 8x − 1

= (8 × 2) − 1 = 16 − 1 =15

(iii) 6x + 5 (x − 2) = 6x + 5x − 10

= 11x − 10

= (11 × 2) − 10 = 22 − 10 = 12

(iv) 4 (2x − 1) + 3x + 11 = 8x − 4 + 3x + 11

= 11x + 7

= (11 × 2) + 7

= 22 + 7 = 29

P7. 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

Sol. (i)3x − 5 − x + 9

= 3x − x − 5 + 9

= 2x + 4

= (2 × 3) + 4

= 10

(ii) 2 − 8+ 4x + 4

= 2 + 4 − 8+ 4x

= 6 − 4= 6 − (4 × 3)

= 6 − 12 = −6

(iii) 3a + 5 − 8+ 1

= 3a − 8a + 5 + 1

= − 5+ 6 = −5 × (−1) + 6

= 5 + 6 = 11

(iv) 10 − 3b − 4 − 5b 

= 10 − 4− 3b − 5b

= 6 − 8b = 6 − 8 × (−2)

= 6 + 16 = 22

(v) 2a − 2b − 4 − 5 + a

= 2a + − 2b − 4 − 5

= 3a − 2b − 9s

= 3 × (−1) − 2 (−2) − 9

= − 3 + 4 − 9 = −8

P7. 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

Sol. (i) 3x − 5 − x + 9

= 3x − x − 5 + 9

= 2x + 4

= (2 × 3) + 4

= 10

(ii) 2 − 8+ 4x + 4

= 2 + 4 − 8+ 4x

= 6 − 4

= 6 − (4 × 3)

= 6 − 12 = −6

(iii) 3a + 5 − 8+ 1

= 3a − 8a + 5 + 1

= − 5+ 6

= −5 × (−1) + 6

= 5 + 6 = 11

(iv) 10 − 3b − 4 − 5b 

= 10 − 4− 3b − 5b

= 6 − 8b 

= 6 − 8 × (−2)

= 6 + 16

= 22

(v) 2a − 2b − 4 − 5 + a

= 2a + − 2b − 4 − 5

= 3a − 2b − 9s

= 3 × (−1) − 2 (−2) − 9

= − 3 + 4 − 9 = −8

P9.What should be the value of a if the value of 2x2 + x − a equals to 5, when x = 0?

Sol. 2x2 + x − a = 5, when x = 0

(2 × 0) + 0 − a = 5

0 − a = 5

a = −5

P10. Simplify the expression and find its value when a = 5 and b = −3.

2 (aab) + 3 − ab

Sol. 2 (a2 + ab) + 3 − ab 

= 2a2 + 2ab + 3 − ab

= 2a2 + 2ab − ab + 3

= 2a2 + ab + 3

= 2 × (5 × 5) + 5 × (−3) + 3

= 50 − 15 + 3 = 38

EXERCISE-12.4

P1. 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.

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 −

Sol. (a) It is given that the number of segments required to form n digits of the kind  is (5n + 1).

Number of segments required to form 5 digits = (5 × 5 + 1)

= 25 + 1 = 26

Number of segments required to form 10 digits = (5 × 10 + 1)

= 50 + 1 = 51

Number of segments required to form 100 digits = (5 × 100 + 1)

= 500 + 1 = 501

(b) It is given that the number of segments required to form n digits of the kind  is (3n + 1).

Number of segments required to form 5 digits = (3 × 5 + 1)

= 15 + 1 = 16

Number of segments required to form 10 digits = (3 × 10 + 1)

= 30 + 1 = 31

Number of segments required to form 100 digits = (3 × 100 + 1)

= 300 + 1 = 301

(c) It is given that the number of segments required to form n digits of the kind  is (5n + 2).

Number of segments required to form 5 digits = (5 × 5 + 2)

= 25 + 2 = 27

Number of segments required to form 10 digits = (5 × 10 + 2)

= 50 + 2 = 52

Number of segments required to form 100 digits = (5 × 100 + 2)

= 500 + 2 = 502

P2. 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

10, 001

Sol.    The given table can be completed as follows.

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

10,001-

 

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