The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is

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In the given problem, let us take the first term as a and the common difference as d. Here, we are given that, a_9 = 449 ...............(1) a_449 = 9 .................(2) We need to find n Also, we know, a_n = a + ( n- 1) d For the 9th term (n = 9), a_9 = a + ( 9 -1) d 449 = a + 8d (Using 1 ) a = 449 - 8 d .................(3) Similarly, for the 449th term (n = 449), a_449 = a + ( 449 - 1 )d 9 = a + 448d (Using 2 ) a = 9 - 448 d .............(4) Subtracting (3) from (4), we get, a -a = ( 9 - 448d) - ( 449 - 8d) 0 = 9 - 448d - 449 + 8d 0 = -440 - 440d 440d = - 440 d = - 1 Now, to find a, we substitute the value of d in (3), a = 449 - 8 (-1) a = 449 + 8 a = 457 So, for the given A.P d = - 1 and a = 457 So, let us take the term equal to zero as the nth term. So, a_n = 457 + ( n- 1) ( -1 ) 0 = 457 - n + 1 n = 458 So, n = 458

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