Which term of the $AP: 21, 18, 15, \ldots$ is $-81$? Also,is any term $0$? Give reason for your answer.
(N/A) For the given $AP$,the first term $a = 21$ and the common difference $d = 18 - 21 = -3$.
We use the formula for the $n$-th term: $a_n = a + (n - 1)d$.
To find which term is $-81$,we set $a_n = -81$:
$-81 = 21 + (n - 1)(-3)$
$-81 = 21 - 3n + 3$
$-81 = 24 - 3n$
$3n = 24 + 81$
$3n = 105$
$n = 35$.
Thus,the $35$th term of the $AP$ is $-81$.
To check if any term is $0$,we set $a_n = 0$:
$0 = 21 + (n - 1)(-3)$
$0 = 21 - 3n + 3$
$3n = 24$
$n = 8$.
Since $n = 8$ is a positive integer,the $8$th term of the $AP$ is $0$.
We use the formula for the $n$-th term: $a_n = a + (n - 1)d$.
To find which term is $-81$,we set $a_n = -81$:
$-81 = 21 + (n - 1)(-3)$
$-81 = 21 - 3n + 3$
$-81 = 24 - 3n$
$3n = 24 + 81$
$3n = 105$
$n = 35$.
Thus,the $35$th term of the $AP$ is $-81$.
To check if any term is $0$,we set $a_n = 0$:
$0 = 21 + (n - 1)(-3)$
$0 = 21 - 3n + 3$
$3n = 24$
$n = 8$.
Since $n = 8$ is a positive integer,the $8$th term of the $AP$ is $0$.
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