# Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)  (xi) (xii) (i) Common difference, d1 = 6 – 3 = 3

Common difference, d2= 12 – 6 = 6

Since, d1 ≠ d2

Therefore, it’s not an A.P.

(ii) Common difference, d1 = -4 – 0 = -4

Common difference, d2= -8 – (-4) = - 4

Since, d1 = d2

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

(iii) Common difference, d1= - = Common difference, d2 = - = Since, d1 ≠ d2

Therefore, it’s not an A.P.

(iv) Common difference, d1= 2 – 12 = -10

Common difference, d2= -8 -2 = -10

Since,d1 = d2

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

(v) Common difference, d1= 3 – 3 = 0

Common difference, d2= 3 – 3 = 0

Since, d1=d2

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

(vi) Common difference, d1= p + 90 – p = 90

Common difference, d2= p + 180 – p – 90 = 90

Since, d1=d2

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

(vii) Common difference, d1= 1.7 – 1.0 = 0.7

Common difference, d2= 2.4 – 1.7 = 0.7

Since, d1=d2

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

(viii) Common difference, d1= -425 + 225 = -200

Common difference, d2= -625 + 425 = -200

Since, d1=d2

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

(ix) Common difference, d1= 10 + 26 – 10 = 26 = 64

Common difference, d2 = 10 + 27 – 10 – 26 = 26 (2 – 1) = 64

Since, d1=d2

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

(x)  Common difference, d1 = (a + 1) + b – a – b = 1

Common difference, d2 = (a + 1) + (b + 1) – (a + 1) – b = 1

Since, d1 = d2

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

(xi) Common difference, d1= 32 – 12 = 8

Common difference, d2 = 52 – 32 = 25 – 9 = 16

Since, d1≠d2

Therefore, it’s not an A.P.

(xii) Common difference, d1 = 52 – 12 = 24

Common difference, d2 = 72 – 52 = 24

Since, d1 = d2

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

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