Monday, April 25, 2011

12. Algebraic Expressions

Exercise 12.1

Question 1: 
Get the algebraicexpressions 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 a and b subtracted from their product.

Answer:
(i) yz
(ii)
(iii) z2
(iv)
(v) x2 + y2
(vi) 5 + 3 (mn)
(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 diagrams.
(a) x − 3 (b) 1 + x + x2 (c) y y3
(d) (e) − ab + 2b2 − 3a2
(ii) Identify terms and factors in the expressions given below:
(a) − 4x + 5 (b) − 4x + 5y (c) 5y + 3y2
(d) (e) pq + q
(f) 1.2 ab − 2.4 b + 3.6 a (g)
(h) 0.1p2 + 0.2 q2

Answer:
(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, y, y
(d)
xy + 2x2y2
xy
2x2y2
x, y
2, x, x, y, y
(e)
pq + q
pq
q
p, q
q
(f)
1.2ab − 2.4b + 3.6a
1.2ab
− 2.4b
3.6a
1.2, a, b
− 2.4, b
3.6, a
(g)
(h)
0.1p2 + 0.2q2
0.1p2
0.2q2
0.1, p, p
0.2, q, q

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) − p2q2 + 7pq (vi) 1.2a + 0.8b
(vii) 3.14 r2 (viii) 2 (l + b) (ix) 0.1y + 0.01 y2

Answer:

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)
0.1y + 0.01y2
0.1y
0.01y2
0.1
0.01

Question 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)
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 + x + xy
x
xy
1
y
(vi)
12xy2 + 25
12xy2
12y2
(vii)
7 + xy2
xy2
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

Question 5:
Classify into monomials, binomials and trinomials.
(i) 4y7z (ii) y2 (iii) x + yxy
(iv) 100 (v) abab (vi) 5 − 3t
(vii) 4p2q − 4pq2 (viii) 7mn (ix) z2 − 3z + 8
(x) a2 + b2 (xi) z2 + z (xii) 1 + x + x2

Answer:
The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.
(i) 4y − 7z
Binomial
(ii) y2
Monomial
(iii) x + yxy
Trinomial
(iv) 100
Monomial
(v) abab
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

Question 6:

State whether a given pair of terms is of like or unlike terms.
(i) 1, 100 (ii) (iii) − 29x, − 29y
(iv) 14xy, 42yx (v) 4m2p, 4mp2 (vi) 12xz, 12 x2z2

Answer:
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) − 7x,
Like
(iii) −29x, −29y
Unlike
(iv) 14xy, 42yx
Like
(v) 4m2p, 4mp2
Unlike
(vi) 12xz, 12x2z2
Unlike

Question 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) −xy2, 2xy2
−4yx2, 20x2y
8x2, −11x2, −6x2
7y, y
−100x, 3x
−11xy, 2xy
(b) 10pq, −7qp, 78qp
7p, 2405p
8q, −100q
p2q2, 12p2q2
−23, 41
−5p2, 701p2
13p2q, qp2



Exercise 12.2  


Question 1:
Simplify combining like terms:
(i) 21b − 32 + 7b − 20b
(ii) − z2 + 13z2 − 5z + 7z3 − 15z
(iii) p − (pq) − q − (q p)
(iv) 3a − 2bab − (ab + ab) + 3ab + b − a
(v) 5x2y − 5x2 + 3y x2 − 3y2 + x2 y2 + 8xy2 −3y2
(vi) (3 y2 + 5y − 4) − (8yy2 − 4)

Answer:
(i) 21b − 32 + 7b − 20b = 21b + 7b − 20b − 32
= b (21 + 7 − 20) −32
= 8b − 32
(ii) − z2 + 13z2 − 5z + 7z3 − 15z = 7z3z2 + 13z2 − 5z − 15z
= 7z3 + z2 (−1 + 13) + z (−5 − 15)
= 7z3 + 12z2 − 20z
(iii) p − (pq) − q − (qp) = pp + qqq + p
= p q
(iv) 3a − 2bab − (ab + ab) + 3ba + b a
= 3a − 2bab a + bab + 3ab + b a
= 3aaa − 2b + b + bab ab + 3ab
= a (3 − 1 − 1) + b (− 2 + 1 + 1) + ab (−1 −1 + 3)
= a + ab
(v) 5x2y − 5x2 + 3yx2 − 3y2 + x2y2 + 8xy2 − 3y2
= 5x2y + 3yx2 − 5x2 + x2 − 3y2y2 − 3y2 + 8xy2
= x2y (5 + 3) + x2 (−5 + 1) + y2(−3 − 1 − 3) + 8xy2
= 8x2y − 4x2 − 7y2 + 8xy2
(vi) (3y2 + 5y − 4) − (8yy2 − 4)
= 3y2 + 5y − 4 − 8y + y2 + 4
= 3y2 + y2 + 5y − 8y − 4 + 4
= y2 (3 + 1) + y (5 − 8) + 4 (1 − 1)
= 4y2 − 3y

Question 2:
(i) 3mn, − 5mn, 8mn, −4mn
(ii) t − 8tz, 3tzz, zt
(iii) − 7mn + 5, 12mn + 2, 9mn − 8, − 2mn − 3
(iv) a + b − 3, ba + 3, ab + 3
(v) 14x + 10y − 12xy − 13, 18 − 7x − 10y + 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, 4bab, 4a − 4b
(x) x2 y2 − 1 , y2 − 1 − x2, 1− x2 y2

Answer:
(i) 3mn + (−5mn) + 8mn + (−4mn) = mn (3 − 5 + 8 − 4)
= 2mn
(ii) (t − 8tz) + (3tzz) + (zt) = t − 8tz + 3tzz + zt
= t t − 8tz + 3tzz + 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 + b − 3) + (ba + 3) + (ab + 3)
= a + b − 3 + ba + 3 + ab + 3
= aa + a + b + b b − 3 + 3 + 3
= a (1 − 1 + 1) + b (1 + 1 − 1) + 3 (− 1 + 1 + 1)
= a + b + 3
(v) (14x + 10y − 12xy − 13) + (18 − 7x − 10y + 8yx) + 4xy
= 14x + 10y − 12xy − 13 + 18 − 7x − 10y + 8yx + 4xy
= 14x − 7x + 10y − 10y − 12xy + 8yx + 4xy − 13 + 18
= x (14 − 7) + y (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 y − 3xy2 − 5xy2 + 5x2y = 4x2 y + 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 p2q2 + 7p2q2 − 4pq + 9pq + 5 + 15
= p2q2 (3 − 10 + 7) + pq (− 4 + 9) + 5 + 15
= 5pq + 20
(ix) (ab − 4a) + (4b ab) + (4a − 4b)
= ab − 4a + 4b ab + 4a − 4b
= abab − 4a + 4a + 4b − 4b
= ab (1 − 1) + a (− 4 + 4) + b(4 − 4)
= 0
(x) (x2y2 − 1) + (y2 − 1 − x2) + (1 − x2y2)
= x2y2 − 1 + y2 − 1 − x2 + 1 − x2y2
= x2x2 x2 y2 + y2 y2 − 1 − 1 + 1
= x2(1 − 1 − 1) + y2 (−1 + 1 − 1) + (− 1 − 1 + 1)
= − x2y2 − 1

Question 3:
Subtract:
(i) − 5y2 from y2
(ii) 6xy from − 12xy
(iii) (ab) 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

Answer:
(i) y2 − (−5y2) = y2 + 5y2 = 6y2
(ii) − 12xy − (6xy) = −18xy
(iii) (a + b) − (ab) = a + b a + b = 2b
(iv) b (5 − a) − a (b − 5) = 5babab + 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) = 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 − 5 b2
= 3ab + 7ab − 2a2 − 5a2 − 2b2 − 5 b2
= 10ab − 7a2 − 7b2
(viii) 4pq − 5q2 − 3p2 from 5p2 + 3q2pq
(5p2 + 3q2pq) − (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?
Answer:
(a) Let a be the required term.
a + (x2 + y2 + xy) = 2x2 + 3xy
a = 2x2 + 3xy − (x2 + y2 + xy)
a = 2x2 + 3xyx2y2xy
a = 2x2x2y2 + 3xyxy
= x2y2 + 2xy
(b) Let p be the required term.
(2a + 8b + 10) − p = − 3a + 7b + 16
p = 2a + 8b + 10 − (− 3a + 7b + 16)
= 2a + 8b + 10 + 3a − 7b − 16
= 2a + 3a + 8b − 7b + 10− 16
= 5a + b − 6

Question 5 :

What should be taken away from 3x2 − 4y2 + 5xy + 20 to obtain
x2y2 + 6xy + 20?
Answer:
Let p be the required term.
(3x2 − 4y2 + 5xy + 20) − p = − x2y2 + 6xy + 20
p = (3x2 − 4y2 + 5xy + 20) − (− x2y2 + 6xy + 20)
= 3x2 − 4y2 + 5xy + 20 + x2 + y2 − 6xy − 20
= 3x2 + x2 − 4y2 + y2 + 5xy − 6xy + 20 − 20
= 4x2 − 3y2xy

Question 6:

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

Answer:
(a) (3xy + 11) + (− y − 11)
= 3xy + 11 − y − 11
= 3xy y + 11 − 11
= 3x − 2y
(3x − 2y) − (3xy − 11)
= 3x − 2y − 3x + y + 11
= 3x − 3x − 2y + y + 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 − 5xx2 + 2x + 5
= 3x2x2 − 5x + 2x + 5
= 2x2 − 3x + 5
(− x + 2x2 + 9) − (2x2 − 3x + 5)
= − x + 2x2 + 9 − 2x2 + 3x − 5
= − x + 3x + 2x2 − 2x2 + 9 − 5
= 2x + 4


Exercise 12.3

Question 1:
If m = 2, find the value of:
(i) m − 2 (ii) 3m − 5 (iii) 9 − 5m
(iv) 3m2 − 2m − 7 (v)
 Answer:
(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)
 Question 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 = − 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

Question 3:

Find the value of the following expressions, when x = − 1:
(i) 2x − 7 (ii) − x + 2 (iii) x2 + 2x + 1
(iv) 2x2x − 2

Answer:

(i) 2x − 7
= 2 × (−1) − 7 = −9
(ii) − x + 2 = − (−1) + 2 = 1 + 2 = 3
(iii) x2 + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1
= 1 − 2 + 1 = 0
(iv) 2x2x − 2 = 2 (−1) × (−1) − (−1) − 2
= 2 + 1 − 2 = 1

Question 4:

If a = 2, b = − 2, find the value of:
(i) a2 + b2 (ii) a2 + ab + b2 (iii) a2b2

Answer:
(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) a2b2
= (2)2 − (−2)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) 2a2 b + 2ab2 + ab (iv) a2 + ab + 2

Answer:

(i) 2a + 2b = 2 × (0) + 2 × (−1) = 0 − 2 = −2
(ii) 2a2 + b2 + 1
= 2 × (0)2 + (−1) × (−1) + 1
= 0 + 1 + 1 = 2
(iii) 2a2b + 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

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) + 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
= 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

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

Answer:
(i) 3x − 5 − x + 9 = 3xx − 5 + 9
= 2x + 4 = (2 × 3) + 4 = 10
(ii) 2 − 8x + 4x + 4 = 2 + 4 − 8x + 4x
= 6 − 4x = 6 − (4 × 3) = 6 − 12 = −6
(iii) 3a + 5 − 8a + 1 = 3a − 8a + 5 + 1
= − 5a + 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 + a − 2b − 4 − 5
= 3a − 2b − 9s
= 3 × (−1) − 2 (−2) − 9
= − 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

Answer:
(i) z3 − 3 (z − 10) = z3 − 3z + 30
= (10 × 10 × 10) − (3 × 10) + 30
= 1000 − 30 + 30 = 1000
(ii) p2 − 2p − 100
= (−10) × (−10) − 2 (−10) − 100
= 100 + 20 − 100 = 20

Question 9:
What should be the value of a if the value of 2x2 + xa equals to 5, when x = 0?

Answer:

2x2 + xa = 5, when x = 0
(2 × 0) + 0 − a = 5
0 − a = 5
a = −5

Question 10:

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

Answer:

2 (a2 + ab) + 3 − ab = 2a2 + 2ab + 3 − ab
= 2a2 + 2abab + 3
= 2a2 + ab + 3
= 2 × (5 × 5) + 5 × (−3) + 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)

(b)

(c)

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 −
, , .

Answer:

(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

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
-
-
-
-
10, 001
-
Answer:
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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