WMA12/01 Jan 2020, Question 8: Arithmetic Series: Proofs and Conditions

Part (i) · 3 marks

Show that the sum to n terms of an arithmetic series with first term a and common difference d equals n/2 {2a + (n − 1)d}.

1. 1S = a + (a + d) + (a + 2d) + … + {a + (n − 2)d} + {a + (n − 1)d}
2. 2S = {a + (n − 1)d} + {a + (n − 2)d} + … + (a + d) + a
3. 3Adding term by term, each pair sums to 2a + (n − 1)d
4. 42S = n{2a + (n − 1)d}
5. 5S = n/2 {2a + (n − 1)d}

AnswerShown: 2S = n{2a + (n − 1)d} ⇒ S = n/2 {2a + (n − 1)d}

Part (iia) · 1 mark

For the sequence uₙ = 5n + 3(−1)ⁿ, calculate the value of the fifth term u₅.

1. 1u₅ = 5 × 5 + 3(−1)⁵
2. 2= 25 − 3 = 22

Answeru₅ = 22

Part (iib) · 3 marks

For uₙ = 5n + 3(−1)ⁿ, find the sum of the terms from n = 1 to n = 59.

1. 1Split: ∑ 5n + ∑ 3(−1)ⁿ
2. 2∑ₙ₌₁⁵⁹ 5n is an AP with a = 5, d = 5: (59/2)(10 + 58 × 5) = 8850
3. 3∑ₙ₌₁⁵⁹ 3(−1)ⁿ = −3 + 3 − 3 + … (59 terms) = −3
4. 4Total = 8850 − 3 = 8847

Answer8847

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