Question
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Calculate the first term and the \( 12^{\text {th }} \) term of an arithmetic sequence if the \( 6^{\text {th }} \) term is 38 and the constant difference is 3 .

Ask by Floyd Watson. in South Africa
Mar 10,2025

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Answer

The first term is 23 and the 12th term is 56.

Solution

Let the first term be \( a \) and the constant difference be \( d = 3 \). 1. The \( 6^{th} \) term is given by: \[ a_6 = a + (6-1)d = a + 5 \times 3 = a + 15 \] Given \( a_6 = 38 \), we have: \[ a + 15 = 38 \] Solving for \( a \): \[ a = 38 - 15 = 23 \] 2. The \( 12^{th} \) term is: \[ a_{12} = a + (12-1)d = a + 11 \times 3 = a + 33 \] Substitute \( a = 23 \): \[ a_{12} = 23 + 33 = 56 \] Thus, the first term is \( 23 \) and the \( 12^{th} \) term is \( 56 \).

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Bonus Knowledge

To find the first term of the arithmetic sequence, we can use the formula for the nth term of an arithmetic sequence, which is given by: \[ a_n = a_1 + (n - 1) \cdot d \] where \( a_n \) is the nth term, \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the term number. Given that the \( 6^{th} \) term \( a_6 = 38 \) and \( d = 3 \), we can plug these values into the formula: \[ 38 = a_1 + (6 - 1) \cdot 3 \] \[ 38 = a_1 + 5 \cdot 3 \] \[ 38 = a_1 + 15 \] \[ a_1 = 38 - 15 \] \[ a_1 = 23 \] Now, to find the \( 12^{th} \) term \( a_{12} \): \[ a_{12} = a_1 + (12 - 1) \cdot d \] \[ a_{12} = 23 + (11) \cdot 3 \] \[ a_{12} = 23 + 33 \] \[ a_{12} = 56 \] So, the first term is 23 and the \( 12^{th} \) term is 56.

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