In an arithmetic series the 3rd term is -8 and 12th term is -35. Determine the number of terms in the series if its sum is -100
a+2d = -8
a+11d = -35
so, 9d = -27
Now you can find a and d, and then solve
n/2 (2a+(n-1)d) = -100
n = 8
To solve this problem, we need to use the formula for the nth term of an arithmetic series and the formula for the sum of an arithmetic series.
Let's denote the first term of the series as 'a' and the common difference between terms as 'd'.
The formula for the nth term of an arithmetic series is given by:
an = a + (n - 1)d
We are given that the 3rd term is -8, so we can plug in these values to get:
a + (3 - 1)d = -8
Simplifying this equation, we have:
a + 2d = -8 .......(Equation 1)
Similarly, we are given that the 12th term is -35, so we can write:
a + (12 - 1)d = -35
Simplifying this equation, we have:
a + 11d = -35 .......(Equation 2)
Now, we have a system of two equations with two variables (a and d). We can solve these equations simultaneously to find the values of 'a' and 'd'.
Let's start by subtracting Equation 1 from Equation 2 to eliminate 'a' and solve for 'd':
(a + 11d) - (a + 2d) = -35 - (-8)
9d = -27
d = -27/9
d = -3
Now, we can substitute the value of 'd' back into Equation 1 to solve for 'a':
a + 2(-3) = -8
a - 6 = -8
a = -8 + 6
a = -2
So, we have found that the first term (a) is -2 and the common difference (d) is -3.
Next, we need to determine how many terms produce a sum of -100. To find this, we will use the formula for the sum of an arithmetic series:
Sn = (n/2)(a + L)
Where n is the number of terms, a is the first term, and L is the last term.
In this case, we don't know the value of L, but we can find it using the formula for the nth term of an arithmetic series:
Ln = a + (n - 1)d
We need to find the value of n for which the sum Sn equals -100, so we can set up the equation as:
-100 = (n/2)(a + Ln)
Substituting the known values of a, L, and d, we get:
-100 = (n/2)(-2 + (-2 + (n - 1)(-3)))
Simplifying this equation, we have:
-100 = (n/2)(-4 - 3n + 3)
Expanding, we get:
-100 = (-4n - 3n^2 + 3n)/2
Multiplying both sides by 2 to eliminate the fraction, we get:
-200 = -4n^2 - 3n^2 + 3n
Combining like terms, we have:
0 = -7n^2 + 3n - 200
This is a quadratic equation in the form of an^2 + bn + c = 0.
To solve, we can use factoring, completing the square, or the quadratic formula.
Factoring may not be easy in this case, so let's solve it using the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / 2a
For our quadratic equation, the coefficients are:
a = -7, b = 3, c = -200
Substituting these values into the quadratic formula, we have:
n = (-3 ± √(3^2 - 4(-7)(-200))) / 2(-7)
Simplifying under the square root, we have:
n = (-3 ± √(9 - 560)) / -14
Simplifying further, we have:
n = (-3 ± √(-551)) / -14
Since we cannot take the square root of a negative number in the real number system, this equation has no real solutions. Thus, there is no value of n that produces a sum of -100.
Therefore, we cannot determine the number of terms in the series that would result in a sum of -100.