Which term of the AP


Which term of the AP $64,60,56,52,48, \ldots .$ is $0 ?$



To Find: we need to find $n$ when $a_{n}=0$

Given: The series is $64,60,56,52,48, \ldots$ and $a_{n}=0$

$a_{1}=64, a_{2}=60$ and $d=60-64=-4$

(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ )

Formula Used: $a_{n}=a+(n-1) d$


$0-64=(n-1)(-4)[$ subtract 64 on both sides $]$


$64=(n-1) 4$ [Divide both side by '-']

$16=(n-1)$ [Divide both side by 4 ]

$\mathrm{n}=17^{\text {th }}$ [add 1 on both sides]

The $17^{\text {th }}$ term of this AP is equal to 0 .


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