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In an A.P., the sum of the first $n$ terms is given by $S_n = 6n - n^2$. Find its $30^{th}$ term.
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Here $S_n = 6n - n^2$
For $n = 1$, $S_1 = a_1 = 6(1) - (1)^2 = 6 - 1 = 5$
So, the first term $a = 5$.
For $n = 2$, $S_2 = a_1 + a_2 = 6(2) - (2)^2 = 12 - 4 = 8$
We know $S_2 = a_1 + a_2 = a + (a+d) = 2a+d$.
So, $2a+d = 8$.
Substitute $a=5$: $2(5) + d = 8$
$10 + d = 8$
$d = 8 - 10 = -2$
Now, find the $30^{th}$ term $a_{30}$:
$a_{30} = a + (30-1)d$
$a_{30} = 5 + 29(-2)$
$a_{30} = 5 - 58$
$a_{30} = -53$
For $n = 1$, $S_1 = a_1 = 6(1) - (1)^2 = 6 - 1 = 5$
So, the first term $a = 5$.
For $n = 2$, $S_2 = a_1 + a_2 = 6(2) - (2)^2 = 12 - 4 = 8$
We know $S_2 = a_1 + a_2 = a + (a+d) = 2a+d$.
So, $2a+d = 8$.
Substitute $a=5$: $2(5) + d = 8$
$10 + d = 8$
$d = 8 - 10 = -2$
Now, find the $30^{th}$ term $a_{30}$:
$a_{30} = a + (30-1)d$
$a_{30} = 5 + 29(-2)$
$a_{30} = 5 - 58$
$a_{30} = -53$