For any series 5+7+9+.....the sum of how many terms will be 480 ?
use your formula
a = 5, d = 2 , n = ? , sum(n) = 480
(n/2)(10 + 2(n-1)) = 480
n(8 + 2n) = 960
8n + 2n^2 = 960
n^2 + 4n - 480 = 0
(n+24)(n-20) = 0
n = -24 or n = 20 , but n must be a whole number
the sum of 20 terms would be 480
check:
sum(20) = (20/2)(10 + 2(19)) = 480
or, since the sum of the first n odd numbers is n^2,
n^2 - 2^2 = 480
n^2 = 484
n = 22
The sum of the first 22 odd numbers is 484. Subtract the first 2, leaving 20, as above.
To find the sum of terms in the series 5+7+9+..., we need to determine the common difference (d) and find the number of terms (n) that will result in a sum of 480.
In this arithmetic series, the common difference (d) is 7 - 5 = 2.
To find the number of terms (n) that will sum up to 480, we can use the formula for the sum of an arithmetic series:
Sn = (n/2) * [2a + (n - 1)d],
where Sn is the sum of the series, a is the first term, n is the number of terms, and d is the common difference.
Plugging in the values we have:
480 = (n/2) * [2*5 + (n-1)*2].
Simplifying further:
480 = (n/2) * (10 + 2n - 2).
Multiplying through:
960 = n * (10 + 2n - 2).
960 = n * (8 + 2n).
960 = 8n + 2n^2.
Rearranging:
2n^2 + 8n - 960 = 0.
Dividing through by 2:
n^2 + 4n - 480 = 0.
Factoring the equation:
(n - 16)(n + 30) = 0.
So, n = 16 or n = -30.
Since the number of terms (n) cannot be negative, we discard n = -30.
Therefore, the sum of 480 can be achieved with 16 terms.
To find the sum of terms in a series, we can use the formula for the sum of an arithmetic series:
Sn = n/2 * (a + L)
Where:
Sn is the sum of the series,
n is the number of terms,
a is the first term,
L is the last term.
In this case, we are given the first term a = 5, and the common difference between terms is d = 7 - 5 = 2. We need to find the value of n when the sum Sn is equal to 480.
Let's first find the last term (L) of the series. We can use the formula for the nth term of an arithmetic series:
L = a + (n-1) * d
Substituting the given values, we can find L:
L = 5 + (n-1) * 2
L = 5 + 2n - 2
L = 2n + 3
Now, we can substitute the values of a, L, and Sn into the sum formula:
Sn = n/2 * (a + L)
480 = n/2 * (5 + 2n + 3)
480 = n/2 * (2n + 8)
Multiplying both sides by 2 to remove the fraction, we get:
960 = n * (2n + 8)
960 = 2n^2 + 8n
Rearranging the equation, we get a quadratic equation in standard form:
2n^2 + 8n - 960 = 0
Now, we can solve this quadratic equation using various methods such as factoring, completing the square, or applying the quadratic formula. The most efficient way in this case is factoring:
2n^2 + 8n - 960 = 0
n^2 + 4n - 480 = 0
We can rewrite this equation as:
(n - 16)(n + 20) = 0
Setting each factor equal to zero, we get:
n - 16 = 0, which gives n = 16
n + 20 = 0, which gives n = -20
Since the number of terms (n) cannot be negative in this context, we discard n = -20.
Therefore, the sum of 480 can be obtained with 16 terms in the series.