If the sum of the first n terms of the series 4+7+10+... is 209, find n
a = 4
d = 3
n/2 (2a+(n-1)d) = 209
Now just plug in your numbers and solve for n.
To solve this problem, we need to find the value of n such that the sum of the first n terms of the given series is equal to 209.
The given series is an arithmetic progression with a common difference of 3.
The formula to find the sum of an arithmetic series is:
Sn = (n/2) * (2a + (n-1)d),
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
In this case, a = 4 and d = 3.
Therefore, the formula becomes:
Sn = (n/2) * (2*4 + (n-1)*3).
Simplifying further:
209 = (n/2) * (8 + 3n - 3).
Now, let's solve the equation to find n.
First, distribute the n/2 factor to the terms inside the parentheses:
209 = (4n + 3n^2 - 3n) / 2.
Multiply both sides of the equation by 2 to eliminate the fraction:
418 = 4n + 3n^2 - 3n.
Rearrange the equation to form a quadratic equation:
3n^2 + n - 418 = 0.
Now, solve this quadratic equation either by factoring, completing the square, or using the quadratic formula.
By factoring or using a calculator, we find that the roots of this equation are approximately -20.31 and 6.97.
Since the number of terms, n, must be a positive integer, we disregard the negative root.
Therefore, the value of n is approximately 7 (rounded to the nearest whole number).
Hence, the sum of the first 7 terms of the series is 209.