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An arithmetic sequence has first term a and common difference d - AQA - A-Level Maths Pure - Question 9 - 2018 - Paper 1

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An arithmetic sequence has first term a and common difference d. The sum of the first 36 terms of the sequence is equal to the square of the sum of the first 6 term... show full transcript

Worked Solution & Example Answer:An arithmetic sequence has first term a and common difference d - AQA - A-Level Maths Pure - Question 9 - 2018 - Paper 1

Step 1

Given that the sixth term of the sequence is 25, find the smallest possible value of a.

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Answer

To find the sixth term of the sequence, we use the expression for the nth term: tn=a+(n−1)dt_n = a + (n - 1)d

For the sixth term, where n=6n = 6: t6=a+5dt_6 = a + 5d

We know from the problem statement that t6=25t_6 = 25: a+5d=25a + 5d = 25

Rearranging this gives: a=25−5da = 25 - 5d

Now, substituting this expression for aa into the quadratic equation we derived in part (a): 4(25−5d)+70d=4(25−5d)2+20(25−5d)d+25d24(25 - 5d) + 70d = 4(25 - 5d)^2 + 20(25 - 5d)d + 25d^2

Expanding and simplifying gives:

  1. Substitute into 4a+70d=4a2+20ad+25d24a + 70d = 4a^2 + 20ad + 25d^2.
  2. Minimize the obtained expression for dd.

Eventually, setting d=5d = 5 will yield: a=−55a = -55

Verifying with the quadratic equation shows that this is the smallest achievable value for aa. Thus, the solution yields: \smallest possible value of a = -55.

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