Which of the following lists of numbers form an A.P. ? If they form an A.P., find the common difference d and write the next three terms:

(i) 4, 10, 16, 22, ….

(ii) -2, 2, -2, 2, ….

(iii) 2, 4, 8, 16, ….

(iv) $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, ….$

(v) -10, -6, -2, 2, …

(vi) 1², 3², 5², 7², ….

(i) Given, 4, 10, 16, 22, ….

Here, a₂ - a₁ = 10 - 4 = 6,     a₃ - a₂ = 16 - 10 = 6,

          a₄ - a₃ = 22 - 16 = 6

i.e. any term - preceding term = 6, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = 6.

For the next three terms, we have:
    a₅ = a₄ + d = 22 + 6 = 28,
    a₆ = a₅ + d = 28 + 6 = 34,
    a₇ = a₆ + d = 34 + 6 = 40.

Hence, the given series is in A.P. with common difference d = 6 and the next three terms : 28, 34, 40.

(ii) Given, -2, 2, -2, 2, ….

Here, a₂ - a₁ = 2 - (-2) = 4,     a₃ - a₂ = -2 - 2 = -4,

          a₄ - a₃ = 2 - (-2) = 4

⇒ a₂ - a₁ = a₄ - a₃ ≠ a₃ - a₂.

Thus the difference of any term from its preceding term is not a fixed number.

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

(iii) Given, 2, 4, 8, 16, ….

Here, a₂ - a₁ = 4 - 2 = 2,     a₃ - a₂ = 8 - 4 = 4,

          a₄ - a₃ = 16 - 8 = 8

⇒ a₂ - a₁ ≠ a₃ - a₂ ≠ a₄ - a₃.

Thus the difference of any term from its preceding term is not a fixed number.

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

(iv) Given, $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, ….$

Here, a₂ - a₁ = $\dfrac{5}{2} - 2 = \dfrac{1}{2}$,     a₃ - a₂ = $3 - \dfrac{5}{2} = \dfrac{1}{2}$,

          a₄ - a₃ = $\dfrac{7}{2} - 3 = \dfrac{1}{2}$

i.e. any term - preceding term = $\dfrac{1}{2}$, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = $\dfrac{1}{2}$.

For the next three terms, we have:

    a₅ = a₄ + d = $\dfrac{7}{2} + \dfrac{1}{2} = \dfrac{8}{2} = 4$,

    a₆ = a₅ + d = $4 + \dfrac{1}{2} = \dfrac{9}{2}$,

    a₇ = a₆ + d = $\dfrac{9}{2} + \dfrac{1}{2} = \dfrac{10}{2} = 5$.

Hence, the given series is in A.P. with common difference d = $\dfrac{1}{2}$ and the next three terms : 4, $\dfrac{9}{2}$, 5.

(v) Given, -10, -6, -2, 2, …

Here, a₂ - a₁ = -6 - (-10) = 4,     a₃ - a₂ = -2 - (-6) = 4,

          a₄ - a₃ = 2 - (-2) = 4

i.e. any term - preceding term = 4, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = 4.

For the next three terms, we have:
    a₅ = a₄ + d = 2 + 4 = 6,
    a₆ = a₅ + d = 6 + 4 = 10,
    a₇ = a₆ + d = 10 + 4 = 14.

Hence, the given series is in A.P. with common difference d = 4 and the next three terms : 6, 10, 14.

(vi) Given, 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

⇒ a₂ - a₁ ≠ a₃ - a₂ ≠ a₄ - a₃.

Thus the difference of any term from its preceding term is not a fixed number.

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