To find the first, 28th, and 45th terms of an arithmetic sequence (AP), we first need to know the formula for the \( n \)-th term of an AP and the sum of the first \( n \) terms.
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Formula for the \( n \)-th term: \[ a_n = a + (n - 1)d \] where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.
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Formula for the sum of the first \( n \) terms: \[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \] where \( S_n \) is the sum of the first \( n \) terms.
In this problem, we know:
- The sum of the first 11 terms \( S_{11} = 891 \).
- The common difference \( d = 15 \).
- We are trying to find \( a \), the first term.
Step 1: Using the sum formula
Substituting the values into the sum formula for the first 11 terms:
\[ S_{11} = \frac{11}{2} \times (2a + (11 - 1) \times 15) \]
This simplifies to:
\[ 891 = \frac{11}{2} \times (2a + 10 \times 15) \]
Calculating \( 10 \times 15 \):
\[ 10 \times 15 = 150 \]
So the equation becomes:
\[ 891 = \frac{11}{2} \times (2a + 150) \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 1782 = 11 \times (2a + 150) \]
Now, dividing both sides by 11:
\[ 162 = 2a + 150 \]
Subtracting 150 from both sides:
\[ 12 = 2a \]
Dividing by 2 gives:
\[ a = 6 \]
Step 2: Finding the first, 28th, and 45th terms
Now that we have \( a = 6 \) and \( d = 15 \), we can find the terms.
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First term ( \( a_1 \) ): \[ a_1 = a = 6 \]
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28th term ( \( a_{28} \) ): \[ a_{28} = a + (28 - 1)d = 6 + (27 \times 15) \] \[ = 6 + 405 = 411 \]
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45th term ( \( a_{45} \) ): \[ a_{45} = a + (45 - 1)d = 6 + (44 \times 15) \] \[ = 6 + 660 = 666 \]
Summary of Results
- First term \( a_1 = 6 \)
- 28th term \( a_{28} = 411 \)
- 45th term \( a_{45} = 666 \)