Determine the sum to 16 terms of the arithmetic series whose \( 7^{\text {th }} \) term is 19 and whose \( 10^{\text {th }} \) term is 25 . Find the arithmetic series
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To find the arithmetic series, we first need to determine the first term \( a \) and the common difference \( d \). Given that the \( 7^{\text{th}} \) term is 19 and the \( 10^{\text{th}} \) term is 25, we can use the formula for the \( n^{\text{th}} \) term of an arithmetic series: \[ T_n = a + (n - 1)d \] For \( T_7 = 19 \): \[ a + 6d = 19 \quad \text{(1)} \] For \( T_{10} = 25 \): \[ a + 9d = 25 \quad \text{(2)} \] Now, we can subtract equation (1) from equation (2): \[ (a + 9d) - (a + 6d) = 25 - 19 \] \[ 3d = 6 \quad \Rightarrow \quad d = 2 \] Substituting \( d = 2 \) back into equation (1): \[ a + 6(2) = 19 \] \[ a + 12 = 19 \quad \Rightarrow \quad a = 7 \] Now, we have \( a = 7 \) and \( d = 2 \). The series can be expressed as: \[ 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, \ldots \] To find the sum of the first 16 terms, we use the sum formula for an arithmetic series: \[ S_n = \frac{n}{2}(2a + (n-1)d) \] where \( n = 16 \). Substituting the values: \[ S_{16} = \frac{16}{2}(2(7) + (16-1)(2)) \] \[ S_{16} = 8(14 + 30) = 8 \times 44 = 352 \] The sum of the first 16 terms of the arithmetic series is \( 352 \), and the series itself is given by the ordered numbers starting from 7 and increasing by 2.
