What are the first three terms of an arithmatic series in which, a1=9, an=105, and Sn=741?
741 / (9 + 105) = 6.5
... so there are 13 terms
(105 - 9) / (13 - 1) = 8
... the common difference is 8
741/(9+105)=6.5 (105-9)/(13-1)=8
To find the first three terms of an arithmetic series, we need to use the given information.
The first term of the arithmetic series, a₁, is given as 9.
The nth term of the arithmetic series, aₙ, is given as 105.
The sum of the first n terms of the arithmetic series, Sₙ, is given as 741.
We can use the formula for the nth term of an arithmetic series to find the common difference, d:
aₙ = a₁ + (n - 1) * d
Substituting the given values:
105 = 9 + (n - 1) * d
To find n, we can use the formula for the sum of the first n terms of an arithmetic series:
Sₙ = (n / 2) * (a₁ + aₙ)
Substituting the given values:
741 = (n / 2) * (9 + 105)
Simplifying the equation:
741 = (n / 2) * 114
Dividing both sides by 114:
6.5n = 741
n ≈ 114
Now, we can substitute the value of n into the equation for the nth term to find the common difference, d:
105 = 9 + (114 - 1) * d
105 = 9 + 113d
96 = 113d
d ≈ 0.8496
Now that we have the common difference, we can find the first three terms of the arithmetic series:
a₁ = 9
a₂ = a₁ + d = 9 + 0.8496 ≈ 9.8496
a₃ = a₂ + d = 9.8496 + 0.8496 ≈ 10.6992
Therefore, the first three terms of the arithmetic series are approximately 9, 9.8496, and 10.6992.
To find the first three terms of an arithmetic series, we need to determine the common difference (d) and the first term (a₁).
Given information:
a₁ = 9 (First term)
aₙ = 105 (Nth term)
Sₙ = 741 (Sum of n terms)
First, let's find the common difference (d):
The formula to calculate the sum of an arithmetic series is:
Sₙ = (n/2)(2a₁ + (n-1)d)
We know that Sₙ = 741 and a₁ = 9.
Substituting these values into the formula, we get:
741 = (n/2)(2 * 9 + (n-1)d)
Now, let's find aₙ and substitute it into the equation (n=?
Since aₙ = 105, we can solve for n:
105 = a₁ + (n-1)d
105 = 9 + (n-1)d
Now, we have a system of equations:
Equation 1: 741 = (n/2)(2 * 9 + (n-1)d)
Equation 2: 105 = 9 + (n-1)d
We can solve this system of equations to find the values of n and d.
For simplicity, let's solve equation 2 for d:
96 = (n-1)d
d = 96 / (n-1)
Now, substitute the value of d into equation 1:
741 = (n/2)(18 + 96 / (n-1))
Now, we can solve this equation to find the value of n.
Simplify the equation:
741 = 9n + 48 / (n-1)
Multiply both sides by (n-1) to eliminate the fraction:
741(n-1) = (9n(n-1)) + 48
Expand and simplify:
741n - 741 = 9n² - 9n + 48
Rearrange to form a quadratic equation:
9n² - 750n + 789 = 0
Now, we can solve this quadratic equation to find the values of n.
In this case, we can use the quadratic formula:
n = (-b ± √(b² - 4ac)) / (2a)
Here, a = 9, b = -750, and c = 789.
Using the quadratic formula, we get:
n₁ ≈ 86.15
n₂ ≈ -8.58
Since we are dealing with arithmetic series, we can discard the negative value for n. Thus, n = 86.
Now, substitute the value of n into equation 2 to find the common difference (d):
105 = 9 + (86-1)d
96 = 85d
d ≈ 1.13 (approximately)
Now, we know that the common difference (d) is approximately 1.13.
To find the first term (a₁), we can substitute the values of n and d into equation 2:
a₁ = 9 + (n-1)d
a₁ = 9 + (86-1)(1.13)
a₁ = 9 + 85(1.13)
a₁ ≈ 99.05 (approximately)
Therefore, the first three terms of the arithmetic series are approximately:
a₁ ≈ 99.05
a₂ ≈ 100.18
a₃ ≈ 101.31