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# If the 3rd and the 9th term of an AP are $4$ and $-8$ respectively, which term of this AP is zero?

Given:

The 3rd and the 9th term of an AP are $4$ and $-8$ respectively.

To do:

We have to find which term of the AP is zero.

Solution:

Let $a$ be the first term and $d$ be the common difference.

3rd term $a_3=a+2d=4$........(i)

9th term $a_9=a+8d=-8$......(ii)

Subtracting (i) from (ii), we get,

$a+8d-a-2d=-8-4$

$6d=-12$

$d=\frac{-12}{6}$

$d=-2$

This implies,

$a+2(-2)=4$

$a=4+4=8$

Let $a_n$ be zero.

Therefore,

$a_{n}=a+(n-1)d=0$

$8+(n-1)(-2)=0$

$-2n+2=-8$

$2n=2+8$

$2n=10$

$n=5$

The 5th term is 0.

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