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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