An arithmetic sequence has first term a and common difference d. The sum of the first 16 terms of the sequence is 260 5 (a) Show that 4a + 30d = 65 5 (b) Given that the sum of the first 60 terms is 315, find the sum of the first 41 terms. 5 (c) S_n is the sum of the first n terms of the sequence. Explain why the value you found in part (b) is the maximum value of S_n.

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An arithmetic sequence has first term a and common difference d - AQA - A-Level Maths Mechanics - Question 5 - 2019 - Paper 1

Question 5

An arithmetic sequence has first term a and common difference d. The sum of the first 16 terms of the sequence is 260 5 (a) Show that 4a + 30d = 65 5 (b) Given th... show full transcript

Worked Solution & Example Answer:An arithmetic sequence has first term a and common difference d - AQA - A-Level Maths Mechanics - Question 5 - 2019 - Paper 1

Step 1

Explain why the value you found in part (b) is the maximum value of S_n.

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Answer

The maximum value of the sum $S_n$ occurs when all terms are positive. In this case:

For terms before the 41st term, the sequence terms are increasing and positive,
After the 41st term, we begin to include negative terms due to the negative common difference $d = -0.5$ .

Therefore, terms after the 41st term will reduce the overall sum. Hence, the sum $S_{41}$ gives the maximum value before the 41st term starts to affect the total negatively.

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