Arithmetic series starts k, 3k/4, k/2, k/4 with 15th term 90 + 2k; find the sum of the first 30 terms.

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First term = k Second term = 3k/4 Common difference d = (3k/4) − k = −k/4 15th term = k + 14d = k + 14(−k/4) k + 14(−k/4) = k − (14k)/4 = k − 3.5k = −2.5k But 15th term is given as 90 + 2k So: −2.5k = 90 + 2k −2.5k − 2k = 90 −4.5k = 90 k = 90 / (−4.5) = −20 Now first term a = k = −20 Common difference d = −k/4 = −(−20)/4 = 20/4 = 5 Sum of first 30 terms: S₃₀ = (30/2)[2a + (30-1)d] S₃₀ = 15[2(−20) + 29×5] S₃₀ = 15[-40 + 145] S₃₀ = 15[105] S₃₀ = 1575 Answer: 1575

4MA1/2H Nov 2021, Question 20: Sum of an Arithmetic Series

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