The sum of an arithmetic series is1,9,5 of the first term is 5 and the difference is 8,find the number of term in series
To solve this problem, 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, n is the number of terms, a is the first term, and d is the common difference.
Given:
Sn = 1,9,5
a = 5
d = 8
Substituting these values into the formula, we get:
1 = (n/2)(2(5) + (n-1)(8))
1 = (n/2)(10 + 8n - 8)
1 = (n/2)(2n + 2)
1 = (n/2)(n + 1)
Multiplying both sides of the equation by 2 to eliminate the fraction, we get:
2 = n(n + 1)
Expanding the equation, we get:
2 = n^2 + n
Rearranging the equation, we get:
n^2 + n - 2 = 0
Using the quadratic formula to solve for n, we have:
n = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 1, and c = -2. Plugging these values into the quadratic formula, we have:
n = (-1 ± √(1^2 - 4(1)(-2))) / (2(1))
n = (-1 ± √(1 + 8)) / 2
n = (-1 ± √9) / 2
n = (-1 ± 3) / 2
We have two possible solutions for n:
1. n = (-1 + 3) / 2 = 2 / 2 = 1
2. n = (-1 - 3) / 2 = -4 / 2 = -2
Since the number of terms cannot be negative, we discard the second solution. Therefore, the number of terms in the series is 1.