Can any term of the $A.P.$ $242, 236, 230, \ldots$ be $0$? If yes,which term?
(N/A) The given $A.P.$ is $242, 236, 230, \ldots$
Here,the first term $a = 242$ and the common difference $d = 236 - 242 = -6$.
Let the $n^{th}$ term of the $A.P.$ be $0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $0 = 242 + (n - 1)(-6)$.
$-242 = -6(n - 1)$.
$n - 1 = \frac{242}{6} = \frac{121}{3} = 40.33$.
$n = 41.33$.
Since $n$ must be a positive integer,$0$ cannot be a term of this $A.P.$
Here,the first term $a = 242$ and the common difference $d = 236 - 242 = -6$.
Let the $n^{th}$ term of the $A.P.$ be $0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $0 = 242 + (n - 1)(-6)$.
$-242 = -6(n - 1)$.
$n - 1 = \frac{242}{6} = \frac{121}{3} = 40.33$.
$n = 41.33$.
Since $n$ must be a positive integer,$0$ cannot be a term of this $A.P.$
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