n/2 (2*4 + (n-1)*3) = 209
1/2 (3n^2+5n) = 209
3n^2+5n-418 = 0
(3n+38)(n-11) = 0
n=11
1/2 (3n^2+5n) = 209
3n^2+5n-418 = 0
(3n+38)(n-11) = 0
n=11
209 = (n/2)(2(4) + (n-1)(3))
Now let's simplify this expression:
209 = (n/2)(8 + 3n - 3)
209 = (n/2)(5 + 3n)
Okay, let's distribute the n/2 to the terms inside the parenthesis:
209 = (5n/2) + (3n²/2)
Now we have a quadratic equation. Let's multiply everything by 2 to make the equation cleaner:
418 = 5n + 3n²
Rearranging the equation:
3n² + 5n - 418 = 0
After solving the quadratic equation, we find that n is approximately equal to 12. So, the sum of the first 12 terms in the series 4 + 7 + 10 + ... is 209.
The given series is an arithmetic progression with a common difference of 3. The general formula for the nth term of an arithmetic progression is given by:
an = a1 + (n - 1) * d
where:
an represents the nth term,
a1 represents the first term, and
d represents the common difference.
In this case, the first term a1 is 4 and the common difference d is 3.
Now we can use the formula for the sum of an arithmetic series to find n:
Sn = (n/2) * (a1 + an)
Given that the sum of the first n terms (Sn) is 209, we can substitute the known values into the formula:
209 = (n/2) * (4 + a1 + (n - 1) * d)
Simplifying further:
209 = (n/2) * (4 + 4 + 3n - 3)
Now, we can distribute and combine like terms:
209 = (n/2) * (7 + 3n)
Multiply both sides of the equation by 2 to eliminate the fraction:
418 = n*(7 + 3n)
Rearranging the equation to form a quadratic equation:
3n^2 + 7n - 418 = 0
Now we can solve this quadratic equation to find the possible values of n.