Let the first term of the arithmetic progression (AP) be \( a \) and the common difference be \( d \).
Step 1: Set up equations
The sum of the first \( n \) terms of an AP is given by the formula: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] From the problem statement:
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The sum of the first 3 terms is 21: \[ S_3 = \frac{3}{2} \times (2a + 2d) = 21 \] Simplifying this, we get: \[ 3(2a + 2d) = 42 \] \[ 2a + 2d = 14 \quad \text{(Equation 1)} \] \[ a + d = 7 \quad \text{(Equation 2)} \]
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The sum of the first 5 terms is 55: \[ S_5 = \frac{5}{2} \times (2a + 4d) = 55 \] Simplifying this, we get: \[ 5(2a + 4d) = 110 \] \[ 2a + 4d = 22 \quad \text{(Equation 3)} \]
Step 2: Solve the equations
Now we have two equations:
- \( 2a + 2d = 14 \) (Equation 1)
- \( 2a + 4d = 22 \) (Equation 3)
We can subtract Equation 1 from Equation 3: \[ (2a + 4d) - (2a + 2d) = 22 - 14 \] This simplifies to: \[ 2d = 8 \] Thus, we find: \[ d = 4 \]
Now substitute \( d = 4 \) into Equation 2 to find \( a \): \[ a + 4 = 7 \] \[ a = 3 \]
Step 3: Find the seventh term
The \( n \)-th term of an AP is given by: \[ a_n = a + (n-1)d \] For the seventh term: \[ a_7 = a + 6d = 3 + 6 \times 4 = 3 + 24 = 27 \]
Step 4: Calculate the sum of the first ten terms
Using the sum formula: \[ S_{10} = \frac{10}{2} \times (2a + 9d) \] Calculating: \[ S_{10} = 5 \times (2 \times 3 + 9 \times 4) = 5 \times (6 + 36) = 5 \times 42 = 210 \]
Summary of results
- First term \( a = 3 \)
- Common difference \( d = 4 \)
- Seventh term \( a_7 = 27 \)
- Sum of the first ten terms \( S_{10} = 210 \)
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