Foundation June 2018 Paper 3 Q22
22 Here is a rule for a sequence.
After the first two terms, each term is half the sum of the previous two terms
(a) Here is a sequence that follows this rule.
2 10 6 ……. ……. …….
Show that the 6th term is the first one that is not a whole number. [3 marks]
(b) A different sequence follows the same rule.
The 1st term is 4
The 3rd term is 9.5
4 ……. 9.5
Work out the 2nd term. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((10 + 6) \div 2\) or 8 as fourth term | M1 | oe |
| (their fourth term + 6) \(\div\) 2 or 7 as fifth term | M1 | oe |
| 8 and 7 and 7.5 | A1 |
Additional guidance
| 8, 7, 7.5 with no working seen or on dotted lines | M1M1A1 |
| The fourth or fifth term must come from a correct method | |
| 14, 10, 12 | M0M1 |
| 14, 10, 18 without seeing correct method (14, 10, 18 is from using the pattern +8, –4) | M0M0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(9.5 \times 2\) or 19 or \(19 \div 2\) (= 9.5) | M1 | oe |
| their \(19 - 4\) | M1dep | |
| 15 | A1 | |
| Alternative method 2 | ||
| \(9.5 - 4\) or 5.5 | M1 | |
| their 5.5 + 9.5 | M1dep | |
| 15 | A1 | |
| Alternative method 3 | ||
| \(\dfrac{x + 4}{2} = 9.5\) | M1 | oe |
| \(x + 4 = 19\) | M1dep | |
| 15 | A1 | |
| Alternative method 4 | ||
| \(9.5 - 4 \div 2\) or 7.5 or \(4 \div 2 + 7.5 = 9.5\) | M1 | |
| their \(7.5 \times 2\) | M1dep | |
| 15 | A1 | |
Additional guidance
| If answer line blank look for clear indication of second term on dotted line | |
| \(4 + 15 = 19\), \(19 \div 2 = 9.5\) with incorrect answer or blank answer line | M1M1A0 |
| \(2 + 7.5 = 9.5\) followed by \(7.5 + 7.5\) | M1M1 |