Sequences and the nth Term: Question 4

Syllabus C2.7, E2.7

Structured Core 4 marks

A linear sequence begins:

58,51,44,37,58, \quad 51, \quad 44, \quad 37, \quad \dots

(a) Find an expression, in terms of nn, for the nnth term of the sequence. [2]

(b) Find the value of nn for which the nnth term of the sequence is 19-19. [2]

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Worked solution

Part (a): find the nnth term

Step 1: Identify the common difference.

5158=7,4451=7,3744=751 - 58 = -7, \qquad 44 - 51 = -7, \qquad 37 - 44 = -7

The terms decrease by the same amount each time, so this is a linear sequence with common difference d=7d = -7.

Step 2: Start with the multiplier.

Since the common difference is 7-7, the nnth term contains 7n-7n. Compare 7n-7n with the actual sequence:

nn11223344
7n-7n7-714-1421-2128-28
term5858515144443737

Step 3: Find the constant to add.

Each actual term is 6565 more than 7n-7n (since 58(7)=6558-(-7)=65). So we add 6565:

nth term=7n+65n\text{th term} = -7n + 65

Quick check: for n=4n=4, 7(4)+65=28+65=37-7(4)+65 = -28+65 = 37

nth term=7n+65\boxed{n\text{th term} = -7n + 65}

Part (b): find nn when the term is 19-19

Set the expression from part (a) equal to 19-19:

7n+65=19-7n + 65 = -19

Subtract 6565 from both sides:

7n=1965=84-7n = -19 - 65 = -84

Divide both sides by 7-7:

n=847=12n = \frac{-84}{-7} = 12

Check: 7(12)+65=84+65=19-7(12) + 65 = -84 + 65 = -19

n=12\boxed{n = 12}

Final answers

  • (a) nnth term =7n+65= -7n + 65
  • (b) n=12n = 12