(i) if a_k+1 – a_k is same for different values of k then the series is an AP.

We have, a₁ = 2, a₂ = 4, a₃ = 8 and a₄ = 16

a₄ – a₃ = 16 – 8 = 8

a₃ – a₂

= 8 – 4

= 5

a₂ – a₁

= 4 – 2

= 2

Here, a_k+1 – a_k is not same for all values of k.

Hence, the given series is not an AP.

(ii) As per the question:

a1 = 2,

a2 = 5/2,

a3 = 3

And

a4 = 7/2

a₄ – a₃ = – 3= 1/2

a₃ – a₂ = 3 – = 1/2

a₂ – a₁ = – 2 = 1/2

Now, we can observe that a_k+1 – a_k is same for all values of k.

Hence, it is an AP.

And, the common difference = 1/2

Next three terms of this series are:

a₅ = a + 4d

= 2 + 4* 1/2

= 4

a₆ = a + 5d

= 2 + 5 * 1/2

=

a₇ = a + 6d

= 2 + 6 * 1/2

= 5

Hence,

The next three terms of the AP are: 4, 9/2 and 5

(iii) a₄ – a₃ = - 7.2 + 5.2 = - 2

a₃ – a₂ = - 5.2 + 3.2 = - 2

a₂ – a₁ = - 3.2 + 1.2 = - 2

In this, a_k+1 – a_k is same for all values of k.

Hence, the given series is an AP.

Common difference = - 2

Next three terms of the series are:

a₅ = a + 4d

= -1.2 + 4 × (-2)

= -1.2 - 8 = -9.2

a₆ = a + 5d

= -1.2 + 5 × (-2)

= -1.2 - 10 = -11.2

a₇ = a + 6d

= -1.2 + 6 × (-2)

= -1.2 - 12 = -13.2

Next three terms of AP are: - 9.2, - 11.2 and – 13.2

(iv)a₄ – a₃ = 2 + 2 = 4

a₃ – a₂ = - 2 + 6 = 4

a₂ – a₁ = - 6 + 10 = 5

Here, a_k+1 – a_k is same for all values of k

Hence, the given series is an AP.

Common difference = 4

Next three terms of the AP are:

a₅ = a + 4d

= -10 + 4 × 4

= -10 + 16 = 6

a₆ = a + 5d

= -10 + 5 × 4

= -10 + 20 = 10

a₇ = a + 6d

= -10 + 6 × 4

= -10 + 24 = 14

Next three terms of AP are: 6, 10 and 14.

(v) a₄ – a₃ = 3 + 3√2 – 3 - 2√2 = √2

a₃ – a₂ = 3 + 2√2 – 3 - √2 = √2

a₂ – a₁ = 3 + √2 – 3 = √2

Here, a_k+1 – a_k is same for all values of k

Hence, the given series is an AP.

Common difference = √2

Next three terms of the AP are

a₆ = a + 5d = 3 + 5√2

a₇ = a + 6d = 3 + 6√2

Next three terms of AP are: 3 + 4√2, 3 + 5√2 and 3 + 6√2

a₄ – a₃ = 0.2222 – 0.222 = 0.0002

a₃ – a₂ = 0.222 – 0.22 = 0.002

a₂ – a₁ = 0.22 – 0.2 = 0.02

Here, a_k+1 – a_k is not same for all values of k

Hence, the given series is not an AP.

(vi) a₄ – a₃ = 0.2222 – 0.222 = 0.0002

a₃ – a₂ = 0.222 – 0.22 = 0.002

a₂ – a₁ = 0.22 – 0.2 = 0.02

Here, a_k+1 – a_k is not same for all values of k

Hence, the given series is not an AP.

(vii) Here; a₄ – a₃ = - 12 + 8 = - 4

a₃ – a₂ = - 8 + 4 = - 4

a₂ – a₁ = - 4 – 0 = - 4

Since a_k+1 – a_k is same for all values of k.

Hence, this is an AP.

The next three terms can be calculated as follows:

a₅ = a + 4d = 0 + 4(- 4) = - 16

a₆ = a + 5d = 0 + 5(- 4) = - 20

a₇ = a + 6d = 0 + 6(- 4) = - 24

Thus, next three terms are; - 16, - 20 and – 24

(viii) Here, it is clear that d = 0

Since a_k+1 – a_k is same for all values of k.

Hence, it is an AP.

The next three terms will be same, i.e. – 1/2

(ix) a₄ – a₃ = 27 – 9 = 18

a₃ – a₂ = 9 – 3 = 6

a₂ – a₁ = 3 – 1 = 2

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP.

(x) a₄ – a₃ = 4a – 3a = a

a₃ – a₂ = 3a – 2a = a

a₂ – a₁ = 2a – a = a

Since a_k+1 – a_k is same for all values of k.

Hence, it is an AP.

Next three terms are:

a₅ = a + 4d = a + 4a = 5a

a₆ = a + 5d = a + 5a = 6a

a₇ = a + 6d = a + 6a = 7a

Next three terms are; 5a, 6a and 7a.

(xi) Here, exponent is increasing in each subsequent term.

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP.

(xii) Different terms of this AP can also be written as follows:

√2, 2√2, 3√2, 4√2, ………..

a4 – a3 = 4√2 - 3√2 = √2

a3 – a2 = 3√2 - 2√2 = √2

a2 – a1 = 2√2 - √2 = √2

Since ak+1 – ak is same for all values of k.

Hence, it is an AP.

Next three terms can be calculated as follows:

a5 = a + 4d = √2 + 4√2 = 5√2

a6 = a + 5d = √2 + 5√2 = 6√2

a7 = a + 6d = √2 + 6√2 = 7√2

Next three terms are; 5√2, 6√2 and 7√2

(xiii) a₄ – a₃ = √12 - √9 = 2√3 – 3

a₃ – a₂ = √9 - √6 = 3 - √6

a₂ – a₁ = √6 - √3

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP

(xiv) The given terms can be written as follows:

1, 9, 25, 49, …

Here, a₄ – a₃ = 49 – 25 = 24

a₃ – a₂ = 25 – 9 = 16

a₂ – a₁ = 9 – 1 = 8

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP