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

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,

$a_{e}=a+(n-1) d$

For the $9^{\text {th }}$ term $(n=9)$,

$a_{9}=a+(9-1) d$

$449=a+8 d$ (Using 1)

$a=449-8 d$ ...........(3)

Similarly, for the $449^{\text {th }}$ term $(n=449)$,

$a_{49}=a+(449-1) d$

$9=a+448 d$(Using 2)

$a=9-448 d$.........(4)

Subtracting (3) from (4), we get,

$a-a=(9-448 d)-(449-8 d)$

$0=9-448 d-449+8 d$

$0=-440-440 d$

$440 d=-440$

$d=-1$

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, $n=458$