Certainly! Let's solve this step by step.
We are given the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series as [tex]\( 209 \)[/tex]. The first term ([tex]\( a \)[/tex]) is [tex]\( 4 \)[/tex], and the common difference ([tex]\( d \)[/tex]) is [tex]\( 3 \)[/tex]. We need to find the value of [tex]\( n \)[/tex].
First, let's recall the formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \][/tex]
Plugging in the values we know:
[tex]\[ 209 = \frac{n}{2} \left(2 \cdot 4 + (n - 1) \cdot 3\right) \][/tex]
Simplify inside the parentheses:
[tex]\[ 209 = \frac{n}{2} \left(8 + 3(n - 1)\right) \][/tex]
[tex]\[ 209 = \frac{n}{2} \left(8 + 3n - 3\right) \][/tex]
[tex]\[ 209 = \frac{n}{2} \left(3n + 5\right) \][/tex]
Next, eliminate the fraction by multiplying both sides by [tex]\( 2 \)[/tex]:
[tex]\[ 418 = n(3n + 5) \][/tex]
Rearrange the equation into standard quadratic form:
[tex]\[ 3n^2 + 5n - 418 = 0 \][/tex]
Now, solve this quadratic equation using the quadratic formula [tex]\( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -418 \)[/tex].
First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 5^2 - 4 \cdot 3 \cdot (-418) \][/tex]
[tex]\[ \Delta = 25 + 5024 \][/tex]
[tex]\[ \Delta = 5049 \][/tex]
Now, apply the quadratic formula:
[tex]\[ n = \frac{-5 \pm \sqrt{5049}}{2 \cdot 3} \][/tex]
Calculate the roots:
[tex]\[ n_1 = \frac{-5 + \sqrt{5049}}{6} \][/tex]
[tex]\[ n_2 = \frac{-5 - \sqrt{5049}}{6} \][/tex]
Evaluating these roots:
[tex]\[ n_1 \approx \frac{-5 + 71.05}{6} \approx \frac{66.05}{6} \approx 11.01 \][/tex]
[tex]\[ n_2 \approx \frac{-5 - 71.05}{6} \approx \frac{-76.05}{6} \approx -12.68 \][/tex]
Since [tex]\( n \)[/tex] must be a positive integer, we discard the negative root:
[tex]\[ n = 11 \][/tex]
Therefore, the number of terms [tex]\( n \)[/tex] is:
[tex]\[ \boxed{11} \][/tex]
