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The first term of an arithmetic series is 51 and the eighth term is 100 . a) Find the twentieth term of the series. b) Determine the sum of the first twenty terms of the series.

Ask by Turner Ellis. in Ghana
Feb 15,2025

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a) The twentieth term is 184. b) The sum of the first twenty terms is 2350.

Solution

We are given an arithmetic sequence where the first term (a₁) is 51 and the eighth term (a₈) is 100. Step 1: Determine the common difference (d). For an arithmetic sequence, the nth term is given by:   aₙ = a₁ + (n - 1)d For the eighth term:   100 = 51 + 7d Subtract 51 from both sides:   7d = 49 Divide by 7:   d = 7 a) Find the twentieth term. Using the formula for the nth term:   a₂₀ = a₁ + 19d = 51 + 19(7)   Calculate 19 × 7 = 133,   Thus, a₂₀ = 51 + 133 = 184 b) Determine the sum of the first twenty terms. The sum of the first n terms of an arithmetic sequence is:   Sₙ = (n/2)(a₁ + aₙ) For n = 20:   S₂₀ = (20/2)(51 + 184) = 10(235) = 2350 Final Answers: a) The twentieth term is 184. b) The sum of the first twenty terms is 2350.

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To find the common difference \( d \) of the arithmetic series, use the information provided. The first term \( a_1 = 51 \) and the eighth term \( a_8 = 100 \). The eighth term can be expressed as: \[ a_8 = a_1 + 7d \] Substituting the known values: \[ 100 = 51 + 7d \] Solving for \( d \): \[ 49 = 7d \] \[ d = 7 \] Now, we can find the twentieth term \( a_{20} \): \[ a_{20} = a_1 + 19d \] \[ a_{20} = 51 + 19 \times 7 \] \[ a_{20} = 51 + 133 = 184 \] Next, to find the sum of the first twenty terms \( S_{20} \): \[ S_n = \frac{n}{2}(a_1 + a_n) \] Where \( n = 20 \) and \( a_{20} = 184 \): \[ S_{20} = \frac{20}{2}(51 + 184) \] \[ S_{20} = 10(235) = 2350 \] So, the answers are: a) The twentieth term of the series is \( 184 \). b) The sum of the first twenty terms of the series is \( 2350 \).

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