Answer :
Sure, let's work through this step-by-step to find the sum of the first 16 terms of the arithmetic series given that its third term is -15 and its eighth term is 10.
### Step 1: Set up the equations
We start with the general formula for the nth term of an arithmetic series:
[tex]\[ a_n = a + (n-1)d \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.
Given:
- The third term ([tex]\(a_3\)[/tex]) is -15.
- The eighth term ([tex]\(a_8\)[/tex]) is 10.
From these, we can set up two equations:
[tex]\[ a + 2d = -15 \quad \text{(1)} \][/tex]
[tex]\[ a + 7d = 10 \quad \text{(2)} \][/tex]
### Step 2: Solve for the common difference [tex]\( d \)[/tex]
To eliminate [tex]\(a\)[/tex], subtract Equation (1) from Equation (2):
[tex]\[ (a + 7d) - (a + 2d) = 10 - (-15) \][/tex]
[tex]\[ 5d = 25 \][/tex]
[tex]\[ d = 5 \][/tex]
### Step 3: Solve for the first term [tex]\( a \)[/tex]
Now substitute [tex]\(d = 5\)[/tex] back into Equation (1):
[tex]\[ a + 2(5) = -15 \][/tex]
[tex]\[ a + 10 = -15 \][/tex]
[tex]\[ a = -25 \][/tex]
### Step 4: Find the sum of the first 16 terms
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic series is given by the formula:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
For [tex]\( n = 16 \)[/tex], [tex]\( a = -25 \)[/tex], and [tex]\( d = 5 \)[/tex]:
[tex]\[ S_{16} = \frac{16}{2} \left(2(-25) + (16-1)(5)\right) \][/tex]
[tex]\[ S_{16} = 8 \left(-50 + 75\right) \][/tex]
[tex]\[ S_{16} = 8 \times 25 \][/tex]
[tex]\[ S_{16} = 200 \][/tex]
### Final Answer:
The sum of the first 16 terms of the arithmetic series is 200.
### Step 1: Set up the equations
We start with the general formula for the nth term of an arithmetic series:
[tex]\[ a_n = a + (n-1)d \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.
Given:
- The third term ([tex]\(a_3\)[/tex]) is -15.
- The eighth term ([tex]\(a_8\)[/tex]) is 10.
From these, we can set up two equations:
[tex]\[ a + 2d = -15 \quad \text{(1)} \][/tex]
[tex]\[ a + 7d = 10 \quad \text{(2)} \][/tex]
### Step 2: Solve for the common difference [tex]\( d \)[/tex]
To eliminate [tex]\(a\)[/tex], subtract Equation (1) from Equation (2):
[tex]\[ (a + 7d) - (a + 2d) = 10 - (-15) \][/tex]
[tex]\[ 5d = 25 \][/tex]
[tex]\[ d = 5 \][/tex]
### Step 3: Solve for the first term [tex]\( a \)[/tex]
Now substitute [tex]\(d = 5\)[/tex] back into Equation (1):
[tex]\[ a + 2(5) = -15 \][/tex]
[tex]\[ a + 10 = -15 \][/tex]
[tex]\[ a = -25 \][/tex]
### Step 4: Find the sum of the first 16 terms
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic series is given by the formula:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
For [tex]\( n = 16 \)[/tex], [tex]\( a = -25 \)[/tex], and [tex]\( d = 5 \)[/tex]:
[tex]\[ S_{16} = \frac{16}{2} \left(2(-25) + (16-1)(5)\right) \][/tex]
[tex]\[ S_{16} = 8 \left(-50 + 75\right) \][/tex]
[tex]\[ S_{16} = 8 \times 25 \][/tex]
[tex]\[ S_{16} = 200 \][/tex]
### Final Answer:
The sum of the first 16 terms of the arithmetic series is 200.
