Solution
An arithmetic series has:
$$ \text{First term: } a = 16, \quad \text{Common difference: } d = -5. $$
The sum to $n$ terms of an arithmetic series is given by:
$$ S_n = \frac{n}{2} \left( 2a + (n-1)d \right). $$
We are tasked to find the least value of $n$ such that:
$$ S_n < -450. $$
Step 1: Substituting the known values
Substitute $a = 16$ and $d = -5$ into the formula for $S_n$:
$$ S_n = \frac{n}{2} \left( 2(16) + (n-1)(-5) \right). $$
Simplify:
$$ S_n = \frac{n}{2} \left( 32 - 5(n-1) \right). $$
Expand $32 - 5(n-1)$:
$$ 32 - 5(n-1) = 32 - 5n + 5 = 37 - 5n. $$
Thus:
$$ S_n = \frac{n}{2}(37 - 5n). $$
Simplify further:
$$ S_n = \frac{37n - 5n^2}{2}. $$
Step 2: Solve $S_n < -450$
We require:
$$ \frac{37n - 5n^2}{2} < -450. $$
Multiply through by $2$ to eliminate the fraction:
$$ 37n - 5n^2 < -900. $$
Rearrange into standard quadratic form:
$$ 5n^2 - 37n - 900 > 0. $$
Step 3: Solve the quadratic inequality
First, solve the quadratic equation
$$ 5n^2 - 37n - 900 = 0 $$
using the quadratic formula:
$$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$
where
$$ a = 5, \quad b = -37, \quad c = -900. $$
Substitute these values:
$$ n = \frac{-(-37) \pm \sqrt{(-37)^2 - 4(5)(-900)}}{2(5)}. $$
Simplify:
$$ n = \frac{37 \pm \sqrt{1369 + 18000}}{10}. $$
$$ n = \frac{37 \pm \sqrt{19369}}{10}. $$
Calculate:
$$ \sqrt{19369} \approx 139.2. $$
Thus:
$$ n = \frac{37 \pm 139.2}{10}. $$
Solve for the two roots:
$$ n = \frac{37 + 139.2}{10} = \frac{176.2}{10} = 17.62 $$
$$ n = \frac{37 - 139.2}{10} = \frac{-102.2}{10} = -10.22 $$
Since $n$ must be a positive integer, we take
$$ n > 17.62. $$
Therefore, the least integer value of $n$ is:
$$ n = 18 $$
Final Answer
The least value of $n$ is:
$$ \boxed{18} $$
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