Sequences

1. 1

   The nth term of a sequence is $3n - 5$. What is the 10th term?

   • A

     $25$

   • B

     $30$

   • C

     $20$

   • D

     $35$

   Check answer

   Answer & explanation

   Correct: option A

   Substitute $n = 10$: $T_{10} = 3(10) - 5 = 30 - 5 = 25$.

2. 2

   What is the nth term of the arithmetic sequence $4, 7, 10, 13, \ldots$?

   • A

     $3n + 1$

   • B

     $3n - 1$

   • C

     $4n - 3$

   • D

     $n + 3$

   Check answer

   Answer & explanation

   Correct: option A

   Common difference $d = 3$; zeroth term $= 4 - 3 = 1$, so $T_n = 3n + 1$. Check: $T_1 = 3(1) + 1 = 4$ ✓.

3. 3

   A geometric sequence has first term 5 and common ratio 3. What is the 4th term?

   • A

     $45$

   • B

     $60$

   • C

     $135$

   • D

     $15$

   Check answer

   Answer & explanation

   Correct: option C

   $T_4 = 5 \times 3^{4-1} = 5 \times 27 = 135$.

4. 4

   The second differences of a sequence are all equal to 6. What is the coefficient of $n^2$ in the nth term?

   • A

     $6$

   • B

     $3$

   • C

     $12$

   • D

     $2$

   Check answer

   Answer & explanation

   Correct: option B

   The coefficient of $n^2$ is always half the second difference: $6 \div 2 = 3$.

5. 5

   Is 85 a term in the sequence with nth term $4n + 1$?

   • A

     Yes, it is the 21st term

   • B

     Yes, it is the 22nd term

   • C

     No, because the solution is not an integer

   • D

     No, because 85 is odd

   Check answer

   Answer & explanation

   Correct: option A

   Solve $4n + 1 = 85 \implies 4n = 84 \implies n = 21$. Since 21 is a positive integer, 85 is the 21st term.

6. 6

   Which of the following is a Fibonacci-type sequence?

   • A

     $2, 4, 8, 16, 32, \ldots$

   • B

     $1, 4, 9, 16, 25, \ldots$

   • C

     $3, 5, 8, 13, 21, \ldots$

   • D

     $2, 5, 8, 11, 14, \ldots$

   Check answer

   Answer & explanation

   Correct: option C

   In a Fibonacci-type sequence, each term is the sum of the two preceding terms: $3 + 5 = 8$, $5 + 8 = 13$, $8 + 13 = 21$ ✓. Option A is geometric; B is square numbers; D is arithmetic.

7. 7

   The sequence $2, 8, 18, 32, 50, \ldots$ has second differences of 4. What is the nth term?

   • A

     $2n^2$

   • B

     $n^2 + n$

   • C

     $4n^2 - 2$

   • D

     $2n^2 + 1$

   Check answer

   Answer & explanation

   Correct: option A

   $a = 4 \div 2 = 2$. Subtract $2n^2$: $2 - 2 = 0$, $8 - 8 = 0$, $18 - 18 = 0$ — remainder is 0, so $T_n = 2n^2$. Check: $T_3 = 2(9) = 18$ ✓.

8. 8

   What are the next two terms of the sequence $1, 3, 6, 10, 15, \ldots$?

   • A

     $20$ and $25$

   • B

     $21$ and $28$

   • C

     $18$ and $21$

   • D

     $20$ and $26$

   Check answer

   Answer & explanation

   Correct: option B

   These are triangular numbers with nth term $\frac{n(n+1)}{2}$. The 6th term $= \frac{6 \times 7}{2} = 21$; the 7th term $= \frac{7 \times 8}{2} = 28$.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q17 marks total

   Show all answers

   Scenario

   A sequence of patterns is made from small squares.

   1. a

      The number of squares in patterns 1, 2, 3 and 4 are 1, 5, 11 and 19. Find the nth term of this sequence.

      [3]
      Show answer

      Model answer · 3 marks

      First differences: 4, 6, 8 — not constant. Second differences: 2, 2 — constant, so quadratic. $a = 2 \div 2 = 1$. Subtract $n^2$: $1-1=0$, $5-4=1$, $11-9=2$, $19-16=3$ — linear sequence with nth term $n-1$. So $T_n = n^2 + (n-1) = n^2 + n - 1$. Check: $T_1 = 1+1-1 = 1$ ✓; $T_4 = 16+4-1 = 19$ ✓.

   2. b

      Is 199 a term in this sequence? Show your working.

      [2]
      Show answer

      Model answer · 2 marks

      $n^2 + n - 1 = 199 \implies n^2 + n - 200 = 0$. Discriminant $= 1 + 800 = 801$. $\sqrt{801} \approx 28.3$ — not an integer, so $n$ is not an integer. 199 is not in the sequence.

   3. c

      Which pattern number first has more than 300 squares?

      [2]
      Show answer

      Model answer · 2 marks

      $n^2 + n - 1 > 300 \implies n^2 + n - 301 > 0$. Roots: $n = \frac{-1 \pm \sqrt{1205}}{2} \approx \frac{-1 + 34.71}{2} \approx 16.85$. Since $n$ must be a positive integer, the first pattern with more than 300 squares is pattern 17. Check: $T_{17} = 289 + 17 - 1 = 305 > 300$ ✓; $T_{16} = 256 + 16 - 1 = 271 < 300$ ✓.

   Examiner note

   Part (a) requires the full differences table for method marks. Part (b) must include a conclusion about whether the result is an integer.

2. Q25 marks total

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   Scenario

   Here are the first five terms of a geometric sequence: $2, 6, 18, 54, 162, \ldots$

   1. a

      Write down the common ratio.

      [1]
      Show answer

      Model answer · 1 mark

      Common ratio $r = 6 \div 2 = 3$.

   2. b

      Find the nth term of the sequence.

      [2]
      Show answer

      Model answer · 2 marks

      First term $a = 2$, common ratio $r = 3$. $T_n = 2 \times 3^{n-1}$.

   3. c

      The $k$th term of the sequence is 1458. Find the value of $k$.

      [2]
      Show answer

      Model answer · 2 marks

      $2 \times 3^{k-1} = 1458 \implies 3^{k-1} = 729 = 3^6 \implies k - 1 = 6 \implies k = 7$. Check: $T_7 = 2 \times 3^6 = 2 \times 729 = 1458$ ✓.

   Examiner note

   For part (c), students may use trial and improvement or logarithms; either approach is acceptable provided it is clearly shown.

3. Q36 marks total

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   Scenario

   The first four terms of a sequence are $5, 14, 29, 50, \ldots$

   1. a

      Show that the sequence is quadratic by calculating second differences.

      [2]
      Show answer

      Model answer · 2 marks

      First differences: $14-5=9$, $29-14=15$, $50-29=21$. Second differences: $15-9=6$, $21-15=6$. The second differences are constant (both equal 6), so the sequence is quadratic.

   2. b

      Find the nth term of the sequence.

      [3]
      Show answer

      Model answer · 3 marks

      $a = 6 \div 2 = 3$. Subtract $3n^2$: $5-3=2$, $14-12=2$, $29-27=2$, $50-48=2$. The remainder is the constant sequence 2. So $T_n = 3n^2 + 2$. Check: $T_1 = 3 + 2 = 5$ ✓; $T_4 = 48 + 2 = 50$ ✓.

   3. c

      Find the value of the 20th term.

      [1]
      Show answer

      Model answer · 1 mark

      $T_{20} = 3(400) + 2 = 1200 + 2 = 1202$.

   Examiner note

   Students should present the differences table clearly in part (a). Part (b) must show the subtraction of $an^2$ and identification of the linear remainder.