How many terms of the series 5 + 7 + 9 must be taken so that sum may be 480?
Let's first find the common difference of the arithmetic sequence.
d = 7 - 5 = 2
Now we can find the nth term of the sequence:
a_n = a_1 + (n-1)d
We want the sum of the first n terms to be 480, so we can use the formula for the sum of an arithmetic sequence:
S_n = (n/2)(a_1 + a_n)
Substituting in the appropriate values:
480 = (n/2)(5 + a_1 + 2(n-1))
Simplifying:
960 = n(2a_1 + 2n - 1)
We know a_1 = 5 and d = 2, so we can substitute that in to get:
960 = n(10 + 2n - 1)
960 = n(2n + 9)
2n^2 + 9n - 960 = 0
Using the quadratic formula:
n = (-9 ± sqrt(9^2 - 4(2)(-960))) / (2(2))
n = (-9 ± sqrt(9665)) / 4
We can disregard the negative root since n must be a positive integer. So:
n = (-9 + sqrt(9665)) / 4 ≈ 23.57
Since we can't take a partial term, we know we need to take at least 24 terms to get a sum of 480. We can check this:
5 + 7 + 9 + ... + 115 + 117 = 480
So the answer is 24 terms.
To find the number of terms required to obtain a sum of 480 in the series 5 + 7 + 9, you can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is:
Sn = (n/2) * (2a + (n-1)d)
Where:
Sn is the sum of the series,
n is the number of terms,
a is the first term, and
d is the common difference.
In this case:
a = 5
d = 7 - 5 = 2 (since the common difference between consecutive terms is 2)
Using the given sum (Sn = 480), the formula can be rearranged to solve for n:
480 = (n/2) * (2 * 5 + (n-1) * 2)
Multiplying out:
480 = (n/2) * (10 + 2n - 2)
Simplifying:
480 = (n/2) * (2n + 8)
Now, solving for n:
480 = n * (n + 4)
Rearranging:
n^2 + 4n - 480 = 0
This quadratic equation can be factored as:
(n + 24)(n - 20) = 0
Setting each factor to zero:
n + 24 = 0 or n - 20 = 0
Solving for n:
n = -24 or n = 20
Since the number of terms cannot be negative, the answer is:
n = 20
Therefore, you would need to take 20 terms of the series 5 + 7 + 9 to obtain a sum of 480.