Sequences and nth term – Year 9 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. The terms go up by 5 each time, so keep adding 5: \(19+5=\)24 and \(24+5=\)29. (1 mark)
   2. The term-to-term rule is add 5, because the difference between consecutive terms is always 5. (1 mark)
   3. A common difference of 5 means the nth term starts with \(5n\) (1 mark). The sequence \(5n\) is 5, 10, 15, 20, and each of our terms is 1 less, so the nth term is \(5n-1\) (1 mark).

   Check: \(5\times1-1=4\) and \(5\times4-1=19\), the first and fourth terms. The term-to-term rule needs the previous term, but the nth term finds any term directly; for example the 100th term is \(5\times100-1=499\).

2. Answer to Question 2 (4 marks)

   1. Substitute \(n=1\), 2 and 3 into \(4n-3\): \(4-3=1\), \(8-3=5\), \(12-3=9\). The first three terms are 1, 5, 9. (1 mark)
   2. Substitute \(n=20\): \(4\times20-3=80-3=\)77. (1 mark)
   3. Set the formula equal to 99: \(4n-3=99\). Add 3: \(4n=102\). Divide by 4: \(n=25.5\) (1 mark). The position must be a whole number, so no, 99 is not a term (1 mark).

   The 25th term is 97 and the 26th is 101, so 99 falls between them. Another quick check: every term is 1 more than a multiple of 4, and 99 is 3 more than 96, so it cannot appear.

3. Answer to Question 3 (4 marks)

   1. The terms go down by 4, so the common difference is \(-4\) and the nth term contains \(-4n\) (1 mark). The term before the first, the zero term, is \(25+4=29\), so the nth term is \(29-4n\) (1 mark).
   2. A term is negative when \(29-4n\lt0\), which gives \(4n\gt29\) and \(n\gt7.25\) (1 mark). The first whole number above 7.25 is 8, and the 8th term is \(29-32=\)\(-3\) (1 mark).

   Check (a): \(29-4=25\) and \(29-16=13\), the 1st and 4th terms. The 7th term is \(29-28=1\), still positive, which confirms that the 8th term is the first negative one. Writing \(4n+29\) would be wrong, because the sequence decreases.

4. Answer to Question 4 (4 marks)

   1. Each pattern uses 4 more matchsticks than the one before, so the common difference is 4 and the nth term starts with \(4n\) (1 mark). Pattern 1 needs 5, which is \(4+1\), so the nth term is \(4n+1\) (1 mark).
   2. Substitute \(n=25\): \(4\times25+1=\)101 matchsticks. (1 mark)
   3. Set the formula equal to 201: \(4n+1=201\), so \(4n=200\) and \(n=\)50. Pattern 50 uses 201 matchsticks. (1 mark)

   A typical shape behind \(4n+1\) is a row of n squares sharing sides: one starting stick, then 4 new sticks for each square. Notice that doubling the pattern number from 25 to 50 does not double the sticks, because of the +1.

5. Answer to Question 5 (4 marks)

   1. Divide any term by the one before: \(12\div3=4\), \(48\div12=4\), \(192\div48=4\). The common ratio is 4. (1 mark)
   2. Multiply the last term by the ratio: \(192\times4=\)768. (1 mark)
   3. The nth term of this geometric sequence is \(3\times4^{n-1}\). For \(n=7\): \(3\times4^{6}=3\times4096=\)12 288. (1 mark)
   4. An arithmetic sequence has a constant difference. Here the differences are 9, 36, 144, which keep growing, so the sequence is not arithmetic. (1 mark)

   Check (c) by continuing the list: 768, \(768\times4=3072\), \(3072\times4=12288\) gives the 5th, 6th and 7th terms.

Total : 20 marks