Let a be the first term and d be the common difference. We know that, nth term = an = a + (n − 1)d It is given that a = 40, d = −3 and an = 0 According to the question,⇒ 0 = 40 + (n − 1)(−3) ⇒ 0 = 40 − 3n + 3 ⇒ 3n = 43 ⇒ n = \[\frac{43}{3}\] .... (1) Here, n is the number of terms, so must be an integer. Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... . |