Answer :

(i)

Common difference, d_{1} = 6 – 3 = 3

Common difference, d_{2}= 12 – 6 = 6

Since, d_{1} ≠ d_{2}

Therefore, it’s not an A.P.

Common difference, d_{1} = -4 – 0 = -4

Common difference, d_{2}= -8 – (-4) = - 4

Since, d_{1} = d_{2}

Therefore, it’s an A.P. with common difference, d = -4

Common difference, d_{1}= - =

Common difference, d_{2} = - =

Since, d_{1} ≠ d_{2}

Therefore, it’s not an A.P.

Common difference, d_{1}= 2 – 12 = -10

Common difference, d_{2}= -8 -2 = -10

Since,d_{1} = d_{2}

Therefore, it’s an A.P. with common difference, d = -10

Common difference, d_{1}= 3 – 3 = 0

Common difference, d_{2}= 3 – 3 = 0

Since, d_{1}=d_{2}

Therefore, it’s an A.P. with common difference, d = 0

Common difference, d_{1}= p + 90 – p = 90

Common difference, d_{2}= p + 180 – p – 90 = 90

Since, d_{1}=d_{2}

Therefore, it’s an A.P. with common difference, d = 90

Common difference, d_{1}= 1.7 – 1.0 = 0.7

Common difference, d_{2}= 2.4 – 1.7 = 0.7

Since, d_{1}=d_{2}

Therefore, it’s an A.P. with common difference, d = 0.7

Common difference, d_{1}= -425 + 225 = -200

Common difference, d_{2}= -625 + 425 = -200

Since, d_{1}=d_{2}

Therefore, it’s an A.P. with common difference, d = -200

Common difference, d_{1}= 10 + 2^{6} – 10 = 2^{6} = 64

Common difference, d_{2} = 10 + 2^{7} – 10 – 2^{6} = 2^{6} (2 – 1) = 64

Since, d_{1}=d_{2}

Therefore, it’s an A.P. with common difference, d = 64

Common difference, d_{1} = (a + 1) + b – a – b = 1

Common difference, d_{2} = (a + 1) + (b + 1) – (a + 1) – b = 1

Since, d_{1} = d_{2}

Therefore, it’s an A.P. with common difference, d = 1

Common difference, d_{1}= 3^{2} – 1^{2} = 8

Common difference, d_{2} = 5^{2} – 3^{2} = 25 – 9 = 16

Since, d_{1}≠d_{2}

Therefore, it’s not an A.P.

Common difference, d_{1} = 5^{2} – 1^{2} = 24

Common difference, d_{2} = 7^{2} – 5^{2} = 24

Since, d_{1} = d_{2}

Therefore, it’s an A.P. with common difference, d = 24

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