Question

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

(i) 2, 4, 8, 16 …

(ii)

(iii) − 1.2, − 3.2, − 5.2, − 7.2 …

(iv) − 10, − 6, − 2, 2 …

(v)

(vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, − 4, − 8, − 12 …

(viii)

(ix) 1, 3, 9, 27 …

(x) a, 2a, 3a, 4a …

(xi) a, a², a³, a⁴ …
(xii)

(xiii)

(xiv) 1², 3², 5², 7² …

(xv) 1², 5², 7², 73 …

Answer 1

(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)

It can be observed that

i.e., a_k+1− a_k is same every time.

Therefore, and the given numbers are in A.P.

Three more terms are

(iii) −1.2, −3.2, −5.2, −7.2 …

It can be observed that

a₂ − a₁ = (−3.2) − (−1.2) = −2

a₃ − a₂ = (−5.2) − (−3.2) = −2

a₄ − a₃ = (−7.2) − (−5.2) = −2

i.e., a_k+1− a_k is same every time. Therefore, d = −2

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

(iv) −10, −6, −2, 2 …

It can be observed that

a₂ − a₁ = (−6) − (−10) = 4

a₃ − a₂ = (−2) − (−6) = 4

a₄ − a₃ = (2) − (−2) = 4

i.e., a_k+1 − a_k is same every time. Therefore, d = 4

The given numbers are in A.P.

Three more terms are

a₅ = 2 + 4 = 6

a₆ = 6 + 4 = 10

a₇ = 10 + 4 = 14

(v)

It can be observed that

i.e., a_{k+1 −} a_k is same every time. Therefore,

The given numbers are in A.P.

Three more terms are

(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) = −4

a₄ − a₃ = (−12) − (−8) = −4

i.e., a_k+1 − a_k is same every time. Therefore, d = −4

The given numbers are in A.P.

Three more terms are

a₅ = − 12 − 4 = −16

a₆ = − 16 − 4 = −20

a₇ = − 20 − 4 = −24

(viii)

It can be observed that

i.e., a_k+1 − a_k is same every time. Therefore, d = 0

The given numbers are in A.P.

Three more terms are

(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 in 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 same every time. Therefore, d = a

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 in A.P.

(xii)

It can be observed that

i.e., a_k+1 − a_k is same every time.

Therefore, the given numbers are in A.P.

And,

Three more terms are

(xiii)

It can be observed that

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 …

Or, 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 in A.P.

(xv) 1², 5², 7², 73 …

Or 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 same every time.

Therefore, the given numbers are in A.P.

And, d = 24

Three more terms are

a₅ = 73+ 24 = 97

a₆ = 97 + 24 = 121

a₇ = 121 + 24 = 145