To solve the problem, we need to find the value(s) of [tex]\( k \)[/tex] such that the sum of the sequence [tex]\( a_n = 48 - 3n \)[/tex] from [tex]\( n = 1 \)[/tex] to [tex]\( n = k \)[/tex] equals 330.
First, let’s identify the first term [tex]\( a_1 \)[/tex] and the [tex]\( k \)[/tex]-th term [tex]\( a_k \)[/tex]:
[tex]\[
a_1 = 48 - 3 \cdot 1 = 45
\][/tex]
[tex]\[
a_k = 48 - 3k
\][/tex]
The sum of the first [tex]\( k \)[/tex] terms of an arithmetic sequence can be found using the formula:
[tex]\[
S_k = \frac{k}{2} (a_1 + a_k)
\][/tex]
Given [tex]\( S_k = 330 \)[/tex], we can substitute [tex]\( a_1 \)[/tex] and [tex]\( a_k \)[/tex]:
[tex]\[
330 = \frac{k}{2} \left(45 + (48 - 3k)\right)
\][/tex]
Simplify the expression inside the parentheses:
[tex]\[
330 = \frac{k}{2} (93 - 3k)
\][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[
660 = k (93 - 3k)
\][/tex]
This expands to a quadratic equation:
[tex]\[
660 = 93k - 3k^2
\][/tex]
Rearrange it to standard form:
[tex]\[
3k^2 - 93k + 660 = 0
\][/tex]
To solve this quadratic equation, we use the quadratic formula [tex]\( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -93 \)[/tex], and [tex]\( c = 660 \)[/tex].
First, calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = (-93)^2 - 4 \cdot 3 \cdot 660 = 8649 - 7920 = 729
\][/tex]
Next, take the square root of the discriminant:
[tex]\[
\sqrt{\text{Discriminant}} = \sqrt{729} = 27
\][/tex]
Now, apply the quadratic formula:
[tex]\[
k = \frac{-(-93) \pm 27}{2 \cdot 3} = \frac{93 \pm 27}{6}
\][/tex]
This gives us two solutions:
[tex]\[
k_1 = \frac{93 + 27}{6} = \frac{120}{6} = 20
\][/tex]
[tex]\[
k_2 = \frac{93 - 27}{6} = \frac{66}{6} = 11
\][/tex]
Since both [tex]\( k_1 = 20 \)[/tex] and [tex]\( k_2 = 11 \)[/tex] are positive integers, they are valid solutions. Therefore, the possible values of [tex]\( k \)[/tex] are:
[tex]\[
\boxed{20, 11}
\][/tex]
