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

Answer 1

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.