Foundation November 2020 Paper 1 Q21
21
(a) All the terms of a geometric progression are positive.
The second and fourth terms are shown.
………. 4 ………. 16
Work out the first and third terms. [2 marks]
(b) The first two terms of an arithmetic progression are shown.
\(p\) \(5p\) …..
The sum of the first three terms is 90
Work out the value of \(p\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| First term 2 and Third term 8 | B2 | B1 one correct or First term \(2^1\) or Third term \(2^3\) or First term \(-2\) and Third term \(-8\) or \(4x^2 = 16\) (any letter) oe equation or \(ar = 4\) and \(ar^3 = 16\) |
Additional guidance
| If answer lines are blank, mark progression first and then working lines | |
| Correct answer for 1st term or 3rd term in the progression, but incorrect numerical term on answer line | B0 for that term |
| Correct answer for 1st term or 3rd term in the progression, with non-contradictory algebraic term on answer line | B1 for that term |
| Correct answers for 1st term and 3rd term in the progression, with non-contradictory algebraic terms on answer lines | B2 |
| First term 2 Third term \(2^3\) | B1 |
| First term \(-2\) Third term 10 | B0 |
| \(4x = \dfrac{16}{x}\) (any letter) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| 3rd term \(= 9p\) | M1 | oe implied by a total of \(15p\) |
| \(p + 5p +\) their 3rd term \(= 90\) or \(15p = 90\) | M1 | oe their 3rd term must be a linear expression in terms of \(p\) \(90 \div 15\) implies M1M1 |
| 6 | A1ft | ft their 3rd term, which must be a linear expression in \(p\), or their equation in the form sum of 3 linear terms in \(p = 90\) allow ft answers rounded to 1dp or better |
| Alternative method 2 | ||
| \(90 \div 3\) or 30 | M1 | oe |
| \(5p =\) their 30 | M1dep | oe |
| 6 | A1 | |
Additional guidance
| For A1ft, if not an integer, the answer must be given as a decimal, fully simplified fraction or fully simplified mixed number Once awarded, ignore further incorrect conversions eg \(p + 5p + 25p = 90\), \(31p = 90\), \(p = \dfrac{90}{31}\), \(p = 3\) (ignore conversion) | M0M1A1ft |
| Their 3rd term may first appear in their addition, eg \(p + 5p + 10p = 90\) implies that \(10p\) is their 3rd term | M0M1 |
| (3rd term \(5p + 4\)), \(p + 5p + 5p + 4 = 90\), \(p = 7.8\) | M0M1A1ft |
| (3rd term \(10p\)), \(p + 5p + 10p = 90\), \(p = 5.625\) | M0M1A1ft |
| Sum \(15p\) and/or answer 6 may come from incorrect 3rd term, eg eg1 (3rd term \(10p\)), \(p + 5p + 10p = 15p\), \((15p = 90)\), \(p = 6\) receives 2nd mark only; they have an incorrect 3rd term and an incorrect total for their 3 terms, but their answer is correct for their total, so equating to 90 is implied even if not seen eg2 (3rd term \(10p\)), \(p\), \(5p\), \(10p\), \(15p = 90\), \(p = 6\) | M0M1A0ft M0M0A0ft |
| If their 3rd term has an algebraic coefficient the 2nd mark can be awarded for a correct equation, but A1 cannot be awarded eg (3rd term \(np\)), \(p + 5p + np = 90\) | M0M1A0 |