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, $\dfrac{5}{2}, 3, \dfrac{7}{2}, ……$

(iii) -1.2, -3.2, -5.2, -7.2, ……..

(iv) -10, -6, -2, 2, ………

(v) 3, 3 + $\sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$, ……….

(vi) 0.2, 0.22, 0.222, 0.2222, ……….

(vii) 0, -4, -8, -12, ……..

(viii) $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, ………..$

(ix) 1, 3, 9, 27, ……..

(x) a, 2a, 3a, 4a, ……….

(xi) a, a², a³, a⁴, ………..

(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},$ ………..

(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},$ ………..

(xiv) 1², 3², 5², 7², ……

(xv) 1², 5², 7², 73, ……

(i) Given,

2, 4, 8, 16, …

⇒ a₂ - a₁ = 4 - 2 = 2,

⇒ a₃ - a₂ = 8 - 4 = 4,

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(ii) Given,

2, $\dfrac{5}{2}, 3, \dfrac{7}{2}, ……$

⇒ a₂ - a₁ = $\dfrac{5}{2} - 2 = \dfrac{5 - 4}{2} = \dfrac{1}{2}$,

⇒ a₃ - a₂ = $3 - \dfrac{5}{2} = \dfrac{6 - 5}{2} = \dfrac{1}{2}$,

⇒ a₄ - a₃ = $\dfrac{7}{2} - 3 = \dfrac{7 - 6}{2} = \dfrac{1}{2}$.

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

So, the given list of numbers forms A.P. with common difference = $\dfrac{1}{2}$.

Next three terms are :

⇒ $\dfrac{7}{2} + \dfrac{1}{2} = \dfrac{8}{2}$ = 4,

⇒ 4 + $\dfrac{1}{2} = \dfrac{8 + 1}{2} = \dfrac{9}{2}$.

⇒ $\dfrac{9}{2} + \dfrac{1}{2} = \dfrac{10}{2}$ = 5.

Hence, for the given A.P. common difference = $\dfrac{1}{2}$ and next three terms are 4, $\dfrac{9}{2}$ and 5.

(iii) Given,

-1.2, -3.2, -5.2, -7.2, ….

⇒ 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 same every time.

So, the given list of numbers forms A.P. with common difference = -2.

Next three terms are :

⇒ -7.2 + (-2) = -7.2 - 2 = -9.2,

⇒ -9.2 + (-2) = -9.2 - 2 = -11.2

⇒ -11.2 + (-2) = -11.2 - 2 = -13.2

Hence, for the given A.P. common difference = -2 and next three terms are -9.2, -11.2 and -13.2

(iv) Given,

-10, -6, -2, 2, ………

⇒ 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 same every time.

So, the given list of numbers forms A.P. with common difference = 4.

Next three terms are :

⇒ 2 + 4 = 6,

⇒ 6 + 4 = 10,

⇒ 10 + 4 = 14.

Hence, for the given A.P. common difference = 4 and next three terms are 6, 10 and 14

(v) Given,

3, 3 + $\sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$, ……….

⇒ a₂ - a₁ = $3 + \sqrt{2} - 3 = \sqrt{2}$,

⇒ a₃ - a₂ = $3 + 2\sqrt{2} - (3 + \sqrt{2}) = \sqrt{2}$,

⇒ a₄ - a₃ = $3 + 3\sqrt{2} - (3 + 2\sqrt{2}) = \sqrt{2}$.

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

So, the given list of numbers forms A.P. with common difference = $\sqrt{2}$.

Next three terms are :

⇒ $3 + 3\sqrt{2} + \sqrt{2} = 3 + 4\sqrt{2}$,

⇒ $3 + 4\sqrt{2} + \sqrt{2} = 3 + 5\sqrt{2}$,

⇒ $3 + 5\sqrt{2} + \sqrt{2} = 3 + 6\sqrt{2}$.

Hence, for the given A.P. common difference = $\sqrt{2}$ and next three terms are $3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}$.

(vi) Given,

0.2, 0.22, 0.222, 0.2222, ……..

⇒ a₂ - a₁ = 0.22 - 0.2 = 0.02,

⇒ a₃ - a₂ = 0.222 - 0.22 = 0.002.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(vii) Given,

0, -4, -8, -12, ……..

⇒ 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 same every time.

So, the given list of numbers forms A.P. with common difference = -4.

Next three terms are :

⇒ -12 + (-4) = -12 - 4 = -16,

⇒ -16 + (-4) = -16 - 4 = -20,

⇒ -20 + (-4) = -20 - 4 = -24.

Hence, for the given A.P. common difference = -4 and next three terms are -16, -20 and -24.

(viii) Given,

$-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, ………..$

⇒ a₂ - a₁ = $-\dfrac{1}{2} - \Big(-\dfrac{1}{2}\Big)$ = 0,

⇒ a₃ - a₂ = $-\dfrac{1}{2} - \Big(-\dfrac{1}{2}\Big)$ = 0,

⇒ a₄ - a₃ = $-\dfrac{1}{2} - \Big(-\dfrac{1}{2}\Big)$ = 0,

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

So, the given list of numbers forms A.P. with common difference = 0.

Next three terms are :

$\Rightarrow -\dfrac{1}{2} + 0 = -\dfrac{1}{2}$,

$\Rightarrow -\dfrac{1}{2} + 0 = -\dfrac{1}{2}$,

$\Rightarrow -\dfrac{1}{2} + 0 = -\dfrac{1}{2}$.

Hence, for the given A.P. common difference = 0 and next three terms are $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}$.

(ix) Given,

1, 3, 9, 27, ………

⇒ a₂ - a₁ = 3 - 1 = 2,

⇒ a₃ - a₂ = 9 - 3 = 6.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(x) Given,

a, 2a, 3a, 4a, …..

⇒ 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.

So, the given list of numbers forms A.P. with common difference = a.

Next three terms are :

⇒ 4a + a = 5a,

⇒ 5a + a = 6a,

⇒ 6a + a = 7a.

Hence, for the given A.P. common difference = a and next three terms are 5a, 6a and 7a.

(xi) Given,

a, a², a³, a⁴, ………

⇒ a₂ - a₁ = a² - a = a(a - 1),

⇒ a₃ - a₂ = a³ - a² = a²(a - 1).

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(xii) Given,

⇒ $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},$ ………..

⇒ $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$, ………..

⇒ a₂ - a₁ = $2\sqrt{2} - \sqrt{2} = \sqrt{2}$,

⇒ a₃ - a₂ = $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$,

⇒ a₄ - a₃ = $4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$.

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

So, the given list of numbers forms A.P. with common difference = $\sqrt{2}$.

Next three terms are :

⇒ $4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$,

⇒ $5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$,

⇒ $6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$.

Hence, for the given A.P. common difference = $\sqrt{2}$ and next three terms are $\sqrt{50}, \sqrt{72}, \sqrt{98}$.

(xiii) Given,

$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},$ ………..

⇒ a₂ - a₁ = $\sqrt{6} - \sqrt{3}$,

⇒ a₃ - a₂ = $\sqrt{9} - \sqrt{6}$.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(xiv) Given,

⇒ 1², 3², 5², 7², ……

⇒ 1, 9, 25, 49, ………

⇒ a₂ - a₁ = 9 - 1 = 8,

⇒ a₃ - a₂ = 25 - 9 = 16.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(xv) Given,

⇒ 1², 5², 7², 73, ……

⇒ 1, 25, 49, 73, ………

⇒ 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.

So, the given list of numbers forms A.P. with common difference = 24.

Next three terms are :

⇒ 73 + 24 = 97,

⇒ 97 + 24 = 121,

⇒ 121 + 24 = 145.

Hence, for the given A.P. common difference = 24 and next three terms are 97, 121 and 145.