NCERT solution Class 10 arithmetic progressions

NCERT Solutions

Class 10 Maths

Arithmetic Progressions

NCERT Questions

Ex. 5.1 Q.1
Ex. 5.1 Q.2
Ex. 5.1 Q.3
Ex. 5.1 Q.4
Ex. 5.2 Q.1
Ex. 5.2 Q.2
Ex. 5.2 Q.3
Ex. 5.2 Q.4
Ex. 5.2 Q.5
Ex. 5.2 Q.6
Ex. 5.2 Q.7
Ex. 5.2 Q.8
Ex. 5.2 Q.9
Ex. 5.2 Q.10
Ex. 5.2 Q.11
Ex. 5.2 Q.12
Ex. 5.2 Q.13
Ex. 5.2 Q.14
Ex. 5.2 Q.15
Ex. 5.2 Q.16
Ex. 5.2 Q.17
Ex. 5.2 Q.18
Ex. 5.2 Q.19
Ex. 5.2 Q.20
Ex. 5.3 Q.1
Ex. 5.3 Q.2
Ex. 5.3 Q.3
Ex. 5.3 Q.4
Ex. 5.3 Q.5
Ex. 5.3 Q.6
Ex. 5.3 Q.7
Ex. 5.3 Q.8
Ex. 5.2 Q.9
Ex. 5.3 Q.10
Ex. 5.3 Q.11
Ex. 5.2 Q.12
Ex. 5.3 Q.13
Ex. 5.3 Q.14
Ex. 5.3 Q.15
Ex. 5.3 Q.16
Ex. 5.3 Q.17
Ex. 5.3 Q.18
Ex. 5.3 Q.19
Ex. 5.3 Q.20
Ex. 5.4 Q.1
Ex. 5.4 Q.2
Ex. 5.4 Q.3
Ex. 5.4 Q.4
Ex. 5.4 Q.5

Ex. 5.1 Q.4

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, . . . (ii) 2, 5/2, 3, 7/2, . . . (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

(iv) – 10, – 6, – 2, 2, . . . (v) 3, 3 + √2, 3 + 2√2, . . . (vi) 0.2, 0.22, 0.222, 0.2222, . . .

(vii) 0, – 4, – 8, –12, . . . (viii) -1/2, -1/2, -1/2, -1/2 …. (ix)1, 3, 9, 27, . . .

(x) a, 2a, 3a, 4a, . . . (xi) a, a², a³, a⁴, . . . (xii) √2, √8, √18 , √32, . . .

(xiii) √3, √6, √9 , √12 , . . . (xiv) 1², 3², 5², 7², . . . (xv) 1², 5², 7², 73, . . .

(i) 2, 4, 8, 16, . . .

It can be observed that a₂ – a₁ = 4 − 2 = 2 a₃ – a₂ = 8 − 4 = 4 a₄ – a₃ = 16 − 8 = 8

i.e., a_k+1 – a_k is not the same every time.

Therefore, the given numbers are not forming an A.P.

(ii) 2, 5/2, 3, 7/2, . . .

It can be observed that a₂ – a₁ = 5/2 − 2 = 1/2 a₃ – a₂ = 3 – 5/2 = 1/2 a₄ – a₃ = 7/2 − 3 = 1/2

i.e., a_k+1 – a_k is the same every time.

Therefore, d = 1/2 and the given numbers are in A.P.

Three more terms are: a₅ = 7/2 + 1/2 = 4 a₆ = 4 + 1/2 = 9/2 a₇ = 9/2 + 1/2 = 5

– 1.2, – 3.2, – 5.2, – 7.2, . . . It can be observed that a₂ – a₁ = -3.2 – (-1.2) = -3.2 + 1.2 = -2 a₃ – a₂ = -5.2 – (-3.2) = -5.2 + 3.2 = -2 a₄ – a₃ = -7.2 – (-5.2) = -7.2 + 5.2 = -2 i.e., a_k+1 – a_k is the same every time.

Therefore, d = -2 and the given numbers are in A.P.

Three more terms are:

a₅ = -7.2 - 2 = -9.2 a₆ = -9.2 - 2 = -11.2 a₇ = -11.2 - 2 = -13.2

– 10, – 6, – 2, 2, . . . It can be observed that a₂ – a₁ = -6 – (-10) = -6 + 10 = 4 a₃ – a₂ = -2 – (-6) = -2 + 6 = 4 a₄ – a₃ = 2 – (-2) = 2 + 2 = 4

i.e., a_k+1 – a_k is the same every time.

Therefore, d = 4 and the given numbers are in A.P.

Three more terms are:

a₅ = 2 + 4 = 6 a₆ = 6 + 4 = 10 a₇ = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, . . .

It can be observed that a₂ – a₁ = 3 + √2 - 3 = √2 a₃ – a₂ = 3 + 2√2 - 3 - √2 = √2 a₄ – a₃ = 3 + 3√2 - 3 - 2√2 = √2

i.e., a_k+1 – a_k is the same every time.

Therefore, d = √2 and the given numbers are in A.P.

Three more terms are: a₅ = 3 + 3√2 + √2 = 3 + 4√2 a₆ = 3 + 4√2 + √2 = 3 + 5√2 a₇ = 3 + 5√2 + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222, . . . It can be observed that a₂ – a₁ = 0.22 – 0.2 = 0.02 a₃ – a₂ = 0.222 – 0.22 = 0.002 a₄ – a₃ = 0.2222 – 0.222 = 0.0002

i.e., a_k+1 – a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(vii) 0, – 4, – 8, –12, . . .

It can be observed that a₂ – a₁ = -4 – 0 = -4 a₃ – a₂ = -8 – (-4) = -8 + 4 = -4 a₄ – a₃ = -12 – (-8) = -12 + 8 = -4

i.e., a_k+1 – a_k is the same every time.

Therefore, d = -4 and the given numbers are in A.P.

Three more terms are:

a₅ = -12 - 4 = -16 a₆ = -16 - 4 = -20

a₇ = -20 - 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

It can be observed that a₂ – a₁ = (-1/2) – (-1/2) = -1/2 + 1/2 = 0 a₃ – a₂ = (-1/2) – (-1/2) = -1/2 + 1/2 = 0 a₄ – a₃ = (-1/2) – (-1/2) = -1/2 + 1/2 = 0

i.e., a_k+1 – a_k is the same every time.

Therefore, d = 0 and the given numbers are in A.P.

Three more terms are: a₅ = (1/2) + 0 = -1/2 a₆ = (1/2) + 0 = -1/2 a₇ = (1/2) + 0 = -1/2 (ix)1, 3, 9, 27, . . .

It can be observed that a₂ – a₁ = 3 − 1 = 2 a₃ – a₂ = 9 − 3 = 6 a₄ – a₃ = 27 − 9 = 18

i.e., a_k+1 – a_k is not the same every time.

Therefore, the given numbers are not forming an A.P.

(x) a, 2a, 3a, 4a, . . .

It can be observed that a₂ – a₁ = 2a - a = a a₃ – a₂ = 3a - 2a = a a₄ – a₃ = 4a - 3a = a

i.e., a_k+1 – a_k is the same every time.

Therefore, d = a and the given numbers are in A.P.

Three more terms are:

a₅ = 4a + a = 5a, a₆ = 5a + a = 6a, a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴, . . .

It can be observed that a₂ – a₁ = a² − a = a(a - 1) a₃ – a₂ = a³ – a² = a²(a - 1) a₄ – a₃ = a⁴ – a³ = a³(a – 1)

i.e., a_k+1 – a_k is not the same every time.

Therefore, the given numbers are not forming an A.P.

(xii) √2, √8, √18 , √32, . . . It can be observed that a₂ – a₁ = √8 - √2 = 2√2 - √2 = √2 a₃ – a₂ = √18 - √8 = 2√2 - 2√2 = √2 a₄ – a₃ = √32 - √18 = 4√2 - 3√2 = √2

i.e., a_k+1 – a_k is the same every time.

Therefore, d = √2 and the given numbers are in A.P.

Three more terms are: a₅ = √32 + √2 = 4√2 + √2 = 5√2 = √50 a₆ = 5√2 + √2 = 6√2 = √72 a₇ = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9 , √12 , . . .

It can be observed that

a₂ – a₁ = √6 - √2 = √(3 * 2) - √2 = √3(√2 - 1) a₃ – a₂ = √9 - √6 = 3 - √(3 * 2) = √3(√3 - √2) a₄ – a₃ = √12 - √9 = 2√3 - √(3 * 3) = √3(2 - √3)

i.e., a_k+1 – a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xiv) 1², 3², 5², 7², . . . = 1, 9, 25, 49, ……… It can be observed that a₂ – a₁ = 9 − 1 = 8 a₃ – a₂ = 25 − 9 = 16 a₄ – a₃ = 49 − 25 = 24

i.e., a_k+1 – a_k is not the same every time.

Therefore, the given numbers are not forming an A.P.

(xv) 1², 5², 7², 73, . . . = 1, 25, 49, 73 It can be observed that a₂ – a₁ = 25 − 1 = 24

a₃ – a₂ = 49 − 25 = 24 a₄ – a₃ = 73 − 49 = 24

i.e., a_k+1 – a_k is the same every time.

Therefore, d = 24 and the given numbers are in an A.P.

Three more terms are:

a₅ = 73 + 24 = 97, a₆ = 97 + 24 = 121, a₇ = 121 + 24 = 145

Video solution:

Arithmetic Progressions Class 10 Maths NCERT Chapter 5 NCERT Solutions

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