Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Question

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?

Answer 1

(i) 11, 8, 5, 2, .... is an A.P.

a = 11 d = 8 - 11  = -3

150^th term = a₁₅₀ = a + (150 - 1)d

 = 11 + 149 x -3

 = 11 - 447 = -436

(ii) a₃ = 12, a_n = 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 n^th 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.