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) 5 7 2, , 3, ,2 2 . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√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, . . .

Solution:

(i) 2, 4, 8, 16 …

Here, the common difference is;

a₂ – a₁ = 4 – 2 = 2

a₃ – a₂ = 8 – 4 = 4

a₄ – a₃ = 16 – 8 = 8

Since a_n ₊₁ – a_nor the common difference is not the same every time.

Therefore, the given series does not form an A.P.

(ii) 2, 5/2, 3, 7/2 ….

Here,

a₂ – a₁ = 5/2 – 2 = 1/2

a₃ – a₂ = 3 – 5/2 = 1/2

a₄ – a₃ = 7/2 – 3 = 1/2

Since a_n ₊₁ – a_n or the common difference is the same every time.

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

The next three terms are;

a₅ = 7/2 + 1/2 = 4

a₆ = 4 + 1/2 = 9/2

a₇ = 9/2 + 1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …

Here,

a₂ – a₁ = (-3.2) – (-1.2) = -2

a₃ – a₂ = (-5.2) – (-3.2) = -2

a₄ – a₃ = (-7.2) – (-5.2) = -2

Since a_n ₊₁ – a_n or common difference is the same every time.

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

Hence, the next three 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 …

Here, the terms and their difference are;

a₂ – a₁ = (-6) – (-10) = 4

a₃ – a₂ = (-2) – (-6) = 4

a₄ – a₃ = 2 – (-2) = 4

Since a_n ₊₁ – a_n or the common difference is the same every time.

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

Hence, the next three terms are;

a₅ = 2 + 4 = 6

a₆ = 6 + 4 = 10

a₇ = 10 + 4 = 14

(v) Given, 3, 3 + √2, 3 + 2√2, 3 + 3√2

Here,

a₂ – a₁ = 3 + √2 – 3 = √2

a₃ – a₂ = (3 + 2√2) – (3 + √2) = √2

a₄ – a₃ = (3 + 3√2) – (3 + 2√2) = √2

Since a_n ₊₁ – a_n or the common difference is the same every time.

Therefore, d = √2, and the given series forms an A.P.

Hence, the next three terms are;

a₅ = (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 ….

Here,

a₂ – a₁ = 0.22 – 0.2 = 0.02

a₃ – a₂ = 0.222 – 0.22 = 0.002

a₄ – a₃ = 0.2222 – 0.222 = 0.0002

Since a_n ₊₁ – a_n or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

(vii) 0, -4, -8, -12 …

Here,

a₂ – a₁ = (-4) – 0 = -4

a₃ – a₂ = (-8) – (-4) = -4

a₄ – a₃ = (-12) – (-8) = -4

Since a_n ₊₁ – a_n or the common difference is the same every time.

Therefore, d = -4, and the given series forms an A.P.

Hence, the next three terms are;

a₅ = -12 – 4 = -16

a₆ = -16 – 4 = -20

a₇ = -20 – 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

Here,

a₂ – a₁ = (-1/2) – (-1/2) = 0

a₃ – a₂ = (-1/2) – (-1/2) = 0

a₄ – a₃ = (-1/2) – (-1/2) = 0

Since a_n ₊₁ – a_n or the common difference is the same every time.

Therefore, d = 0, and the given series forms an A.P.

Hence, the next three 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 …

Here,

a₂ – a₁ = 3 – 1 = 2

a₃ – a₂ = 9 – 3 = 6

a₄ – a₃ = 27 – 9 = 18

Since a_n ₊₁ – a_n or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

(x) a, 2 a, 3 a, 4 a …

Here,

a₂ – a₁ = 2a – a = a

a₃ – a₂ = 3a – 2a = a

a₄ – a₃ = 4a – 3a = a

Since a_n ₊₁ – a_n or the common difference is the same every time.

Therefore, d = a, and the given series forms an A.P.

Hence, the next three terms are;

a₅ = 4a + a = 5a

a₆ = 5a + a = 6a

a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴ …

Here,

a₂ – a₁ = a²– a = a( a – 1)

a₃ – a₂ = a³– a²= a²( a – 1)

a₄ – a₃ = a⁴ – a³= a³( a – 1)

Since a_n ₊₁ – a_n or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

(xii) √2, √8, √18, √32 …

Here,

a₂ – a₁ = √8 – √2 = 2√2 – √2 = √2

a₃ – a₂ = √18 – √8 = 3√2 – 2√2 = √2

a₄ – a₃ = 4√2 – 3√2 = √2

Since a_n ₊₁ – a_n or the common difference is the same every time.

Therefore, d = √2, and the given series forms an A.P.

Hence, the next three 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 …

Here,

a₂ – a₁ = √6 – √3 = √3 × √2 – √3 = √3(√2 – 1)

a₃ – a₂ = √9 – √6 = 3 – √6 = √3(√3 – √2)

a₄ – a₃ = √12 – √9 = 2√3 – √3 × √3 = √3(2 – √3)

Since a_n ₊₁ – a_n or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

(xiv) 1², 3², 5², 7² …

Or, 1, 9, 25, 49 …..

Here,

a₂ − a₁ = 9 − 1 = 8

a₃ − a₂= 25 − 9 = 16

a₄ − a₃ = 49 − 25 = 24

Since a_n ₊₁ – a_n or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

(xv) 1², 5², 7², 73 …

Or 1, 25, 49, 73 …

Here,

a₂ − a₁ = 25 − 1 = 24

a₃ − a₂= 49 − 25 = 24

a₄ − a₃ = 73 − 49 = 24

Since a_n ₊₁ – a_n or the common difference is the same every time.

Therefore, d = 24, and the given series forms an A.P.

Hence, the next three terms are;

a₅ = 73 + 24 = 97

a₆ = 97 + 24 = 121

a₇= 121 + 24 = 145