OCR H240/01 2018 September — Question 11 12 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2018
SessionSeptember
Marks12
TopicGeometric Sequences and Series
TypeRelationship between two GPs
DifficultyChallenging +1.2 This question requires setting up and solving simultaneous equations involving GP and AP formulas, then finding a sum to infinity. While it involves multiple steps and algebraic manipulation with surds, the techniques are standard A-level: writing general terms, substituting given conditions, and solving. The convergence condition provides a helpful constraint. The 'show that' format removes uncertainty about the answer, making it moderately above average difficulty but not requiring novel insight.
Spec1.01a Proof: structure of mathematical proof and logical steps1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1
11 In this question you must show detailed reasoning. The \(n\)th term of a geometric progression is denoted by \(g _ { n }\) and the \(n\)th term of an arithmetic progression is denoted by \(a _ { n }\). It is given that \(g _ { 1 } = a _ { 1 } = 1 + \sqrt { 5 } , g _ { 3 } = a _ { 2 }\) and \(g _ { 4 } + a _ { 3 } = 0\). Given also that the geometric progression is convergent, show that its sum to infinity is \(4 + 2 \sqrt { 5 }\).
AnswerMarks Guidance
DR\(a + d = ar^2\) B1
\(a + 2d + ar^3 = 0\)B1 Correct equation for \(a_3 + g_4 = 0\) (Allow \(a, a_i, g_1\) or \(1 + \sqrt{5}\) and may be different in each term eg \(a_i + d = e_i r^2\))
\(a + 2(ar^2 - a) + ar^3 = 0\)M1 Eliminate \(d\) (Could be \(a, a_1, g_1\) or \(1 + \sqrt{5}\) but must now be consistent throughout (soi))
\(r^3 + 2r^2 - 1 = 0\)A1 Obtain correct cubic
\(f(-1) = -1 + 2 - 1 = 0\) hence \((r+1)\) is a factorB1 Identify \((r + 1)\) as a factor, with justification
\((r + 1)(r^2 + r - 1) = 0\)M1 Attempt to find all 3 roots of cubic
\(r = -1, \frac{-1 \pm \sqrt{5}}{2}\)A1 For all three (1.1a)
GP is convergent so \(-1 < r < 1\), so \(r = \frac{-1 + \sqrt{5}}{2}\)B1 Identify correct value of \(r\), with reason
\(S_\infty = \frac{1 + \sqrt{5}}{1 - \frac{1}{2}(-1 + \sqrt{5})} = \frac{1 + \sqrt{5}}{1 - \frac{1}{2}(-1 + \sqrt{5})}\)M1 Attempt sum to infinity, using their \(r\) (Need \(-1 <\) their \(r < 1\))
\(= \frac{2(1 + \sqrt{5})}{2 - (-1 + \sqrt{5})} = \frac{2(1 + \sqrt{5})}{3 - \sqrt{5}}\)A1 Simplify to correct expression
\(= \frac{2(1 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{2(3 + 3\sqrt{5} + 3\sqrt{5} + 5)}{9 - 5}\)M1 Rationalise denominator (Must also attempt expansion)
\(= \frac{2(8 + 4\sqrt{5})}{4} = 4 + 2\sqrt{5}\)A1 Obtain given answer www

Total: [12]

DR | $a + d = ar^2$ | B1 | Correct equation for $a_2 = g_3$
| $a + 2d + ar^3 = 0$ | B1 | Correct equation for $a_3 + g_4 = 0$ (Allow $a, a_i, g_1$ or $1 + \sqrt{5}$ and may be different in each term eg $a_i + d = e_i r^2$)
| $a + 2(ar^2 - a) + ar^3 = 0$ | M1 | Eliminate $d$ (Could be $a, a_1, g_1$ or $1 + \sqrt{5}$ but must now be consistent throughout (soi))
| $r^3 + 2r^2 - 1 = 0$ | A1 | Obtain correct cubic
| $f(-1) = -1 + 2 - 1 = 0$ hence $(r+1)$ is a factor | B1 | Identify $(r + 1)$ as a factor, with justification
| $(r + 1)(r^2 + r - 1) = 0$ | M1 | Attempt to find all 3 roots of cubic
| $r = -1, \frac{-1 \pm \sqrt{5}}{2}$ | A1 | For all three (1.1a)
| GP is convergent so $-1 < r < 1$, so $r = \frac{-1 + \sqrt{5}}{2}$ | B1 | Identify correct value of $r$, with reason
| $S_\infty = \frac{1 + \sqrt{5}}{1 - \frac{1}{2}(-1 + \sqrt{5})} = \frac{1 + \sqrt{5}}{1 - \frac{1}{2}(-1 + \sqrt{5})}$ | M1 | Attempt sum to infinity, using their $r$ (Need $-1 <$ their $r < 1$)
| $= \frac{2(1 + \sqrt{5})}{2 - (-1 + \sqrt{5})} = \frac{2(1 + \sqrt{5})}{3 - \sqrt{5}}$ | A1 | Simplify to correct expression
| $= \frac{2(1 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{2(3 + 3\sqrt{5} + 3\sqrt{5} + 5)}{9 - 5}$ | M1 | Rationalise denominator (Must also attempt expansion)
| $= \frac{2(8 + 4\sqrt{5})}{4} = 4 + 2\sqrt{5}$ | A1 | Obtain given answer www
Total: [12]
11 In this question you must show detailed reasoning.
The $n$th term of a geometric progression is denoted by $g _ { n }$ and the $n$th term of an arithmetic progression is denoted by $a _ { n }$. It is given that $g _ { 1 } = a _ { 1 } = 1 + \sqrt { 5 } , g _ { 3 } = a _ { 2 }$ and $g _ { 4 } + a _ { 3 } = 0$.

Given also that the geometric progression is convergent, show that its sum to infinity is $4 + 2 \sqrt { 5 }$.

\hfill \mbox{\textit{OCR H240/01 2018 Q11 [12]}}
This paper (12 questions)
View full paper
Q1 5 Q2 6 Q3 6 Q4 7 Q5 6 Q6 7 Q7 11 Q8 9 Q9 8 Q10 13 Q11 12 Q12 10