If the sum of the first n terms of the series,4+7+10+... is 209, find n
Looks like an AS, with a = 4 and d = 3
(n/2)(2a + (n-1)d) = sum(n)
(n/2)(8 + 3(n-1)) = 209
n(8 + 3n - 3) = 418
5n + 3n^2 - 418 = 0
3n^2 + 5n - 418 = 0
solve for n, reject the negative answer
difference = 3 = d
first term = 4 = a
https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
209 = (n/2) [ 4 + (n-1)3 ]
209 = (n/2) [ 1+3n ]
418 = n + 3 n^2
3 n^2 + 1 n - 418 = 0
about 11.64
hmmm, typo or my mistake ?
209 = (n/2) [ 4 + (n-1)3 ]
4 should be 8, 2 n not n
209 = (n/2) [ 8 + (n-1)3 ]
418 = 5n + 3 n^2
3 n^2 + 5 n - 418 = 0
Yh true
To find the value of n, we need to determine how many terms are in the series that have a sum of 209.
The given series is an arithmetic series with a common difference of 3. To find the sum of an arithmetic series, we can use the formula:
Sum = (n/2)(2a + (n-1)d)
Where:
- Sum is the sum of the series,
- n is the number of terms in the series,
- a is the first term of the series,
- d is the common difference of the series.
In our case:
- Sum = 209,
- a = 4,
- d = 3.
Now, let's substitute these values into the formula:
209 = (n/2)(2(4) + (n-1)(3))
Let's simplify the equation:
209 = (n/2)(8 + 3n - 3)
209 = (n/2)(3n + 5)
Distribute n/2:
209 = (3n^2 + 5n) / 2
Multiply both sides of the equation by 2:
418 = 3n^2 + 5n
Rearrange the equation to set it equal to zero:
3n^2 + 5n - 418 = 0
At this point, we need to solve this quadratic equation to find the value(s) of n. We can either factor the equation or use the quadratic formula.
Let's use the quadratic formula:
n = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation:
a = 3, b = 5, c = -418.
Calculate the discriminant:
discriminant = b^2 - 4ac = (5^2) - 4(3)(-418)
discriminant = 25 + 5016 = 5038
Now, substitute the values in the quadratic formula:
n = (-5 ± sqrt(5038)) / (2(3))
n = (-5 ± sqrt(5038)) / 6
Calculate both possible solutions:
n₁ = (-5 + sqrt(5038)) / 6
n₂ = (-5 - sqrt(5038)) / 6
Therefore, the value of n can be either n₁ or n₂, depending on the context of the problem.