(i) Given A.P. 21, 18, 15, ….

First term (a) = 21

Common difference

(d) = 18 – 21 = -3

∵ a_n = a + (n – 1)d

According to question

-81 = 21 + (n – 1)(-3)

⇒ -81-21 = (n – 1) × -3

⇒ -102 = (n – 1) × -3

⇒ (n – 1) = -102/3

⇒ n – 1 = 34

⇒ n = 34 + 1 = 35

Hence, 35^th term of given series is -81.

(ii) Given A.P. 84, 80, 76…..

First term (a) = 84

Common difference (d) = 80 – 84 = -4

∵ a_n = a + (n – 1)d

According to question,

0 = 84 + (n – 1)(-4)

⇒ -84 = (n – 1) × -4

⇒ (n – 1) = -84/-4

⇒ n – 1 = 21

⇒ n = 21 + 1 = 22

Hence. 22^nd term of given A.P. is zero.

(iii) Given A.P. 5, 11, 17, 23 …..

First term (a) = 5

Common difference

(d) = 11 – 5 = 6

∵ a_n = a + (n – 1)d

According to question.,

⇒ 301 = 5 + (n – 1)(6)

⇒ 301 – 5 = 6(n—1)

⇒ 6(n – 1) = 296

⇒ (n – 1) = 296/6

⇒ n – 1 = 49.33

⇒ n = 49.33 + 1 = 50.33

∴ Value of n cannot be fraction, it means n is not a whole number. Thus no term can be 301 in given series A.P.

(iv) Given A.P. : 11, 8, 5, 2 …

First term (a) = 11 and common difference (d) = 8 – 11 = – 3

Let n^th term, a_n = -150

⇒ a + (n – 1)d = -150

⇒ 11 + (n – 1) × (-3) = -150

⇒ -3(n – 1) = -150 – 11 = -161

⇒ (n – 1) = -161/-3

= 53.6 (approx)

∴ n = 53.6 + 1 = 54.6

⇒ n is not a whole no.

Hence, -150 is no  term is given A.P.