(i) 3, 6, 12, 24, ......
Common difference,
d1 = 6 – 3 = 3
Common difference,
d2= 12 – 6 = 6
Since,
d1 ≠ d2
Therefore, it’s not an A.P.
(ii) 0, - 4, -8, -12, ......
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) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), \(\frac{1}{8}\), ......
Common difference,
Common difference,
Since,
d1 ≠ d2
Therefore, it’s not an A.P.
(iv) 12, 2, -8, -18, ......
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