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150 is a term of the AP : \( 11,8,5,2 \ldots \)8. An AP consists of 50 terms of which 3 rd term is 12 and the last tem is 736 . tem. zero? |
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Answer» (i) 11, 8, 5, 2, .... is an A.P. a = 11 d = 8 - 11 = -3 150th term = a150 = a + (150 - 1)d = 11 + 149 x -3 = 11 - 447 = -436 (ii) a3 = 12, an = a50 = 736 \(\therefore\) a + 2d = 12---(1) a + 49d = 736----(2) ⇒ 47d = 724 (By (2) - (1)) ⇒ d = 724/47 ⇒ a = 12 - 2d = 12 - \(\frac{724\times2}{47}\) = \(\frac{564-1448}{47}\) = \(\frac{-448}{47}\) Let nth term be zero \(\therefore\) a +(n -1)d = 0 ⇒ \(\frac{-884}{47}+(n-1)\frac{724}{47}=0\) ⇒ 724(n - 1) = 884 ⇒ n -1 = 884/724 ⇒ n = 1608/724 which is not a whole number \(\therefore\) No term of A.P. will be zero. |
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