Which of the following sequences are in arithmetic progression ?

(i) 2, 6, 10, 14, …….

(ii) 15, 12, 9, 6, ……

(iii) 5, 9, 12, 18, ……

(iv) $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}, ……$

(i) Since, 6 - 2 = 4, 10 - 6 = 4 and 14 - 10 = 4

⇒ Difference between consecutive terms is the same i.e. 4.

Hence, the given terms are in A.P.

(ii) Since, 12 - 15 = -3, 9 - 12 = -3 and 6 - 9 = -3

⇒ Difference between consecutive terms is the same i.e. -3.

Hence, the given terms are in A.P.

(iii) Since, 9 - 5 = 4, 12 - 9 = 3 and 18 - 12 = 6

⇒ Difference between consecutive terms is different.

Hence, the given terms are not in A.P.

(iv) Since, $\dfrac{1}{3} - \dfrac{1}{2} = -\dfrac{1}{6}, \dfrac{1}{4} - \dfrac{1}{3} = -\dfrac{1}{12}$

⇒ Difference between consecutive terms is different.

Hence, the given terms are not in A.P.