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