# The $9^{th}$ term of an AP is $499$ and its $499^{th}$ term is $9$. Which of its term is equal to zero.

**Given: **The $9^{th}$ term of an AP is $499$ and its $499^{th}$ term is $9$.

**To do: **To find which of its term is equal to zero.

**Solution: **

Let $a$ and $d$ be the first term and common difference of the A.P. respectively.

As given, $9^{th}$ term of an A.P. $a_{9}=a+( 9-1)d=a+8d=499\ ....\ ( i)$

$499^{th}$ term $a_{499}=a+( 499-1)=a+498d=9\ ....\ ( ii)$.

Subtracting $( i)$ from $( ii)$-

$a+498d-a-8d=9-499$

$\Rightarrow 490d=-490$

$\Rightarrow d=-\frac{490}{490}$

$\Rightarrow d=-1$, On putting this value in $( i)$

$a+8( -1)=499$

$\Rightarrow a-8=499$

$\Rightarrow a=499+8$

$\Rightarrow a=507$

Let the $n^{th}$ term of the A.P. is $0$.

$\Rightarrow a_n=a+( n-1)d=0$

$\Rightarrow 507+( n-1)\times -1=0$

$\Rightarrow n-1=507$

$\Rightarrow n=507+1$

$\Rightarrow n=508$

Thus, $508^{th}$ term of the A.P. is $0$.

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