Higher November 2023 Paper 1 Q22
22 The 2nd term of a geometric sequence is \(3 + 2\sqrt{2}\)
The 3rd term of the sequence is \(13 + 9\sqrt{2}\)
Find the value of the common ratio of the sequence.
Give your answer in the form \(a + \sqrt{b}\) where \(a\) and \(b\) are integers.
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(3 + \sqrt{2}\) | P1 | for start of process to find the common ratio, eg writes \(\dfrac{13 + 9\sqrt{2}}{3 + 2\sqrt{2}}\) |
| P1 | for process to rationalise the denominator, eg \(\dfrac{13 + 9\sqrt{2}}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}\) | |
| P1 | (dep P2) for expanding terms, condone one error in expansion of numerator or denominator, eg \(\dfrac{39 + 27\sqrt{2} - 18\sqrt{2}\sqrt{2} - 26\sqrt{2}}{9 + 6\sqrt{2} - 6\sqrt{2} - 4\sqrt{2}\sqrt{2}}\) or \(\dfrac{39 + 27\sqrt{2} - 26\sqrt{2} - 36}{9 - 8}\) oe | |
| A1 | cao |
Additional guidance
Award P1 for process to rationalise the denominator of \(\dfrac{3 + 2\sqrt{2}}{13 + 9\sqrt{2}}\)
A correct answer with no supportive working gets 0 marks.