OCR C2 2016 June — Question 6 11 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2016
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeCompare two growth models
DifficultyStandard +0.8 This question combines arithmetic and geometric progressions with a comparison requiring algebraic manipulation. Part (i) is routine, part (ii) is standard GP sum formula application, but part (iii) requires finding when a finite AP sum exceeds an infinite GP sum, involving quadratic solving and interpretation—this elevates it above typical C2 questions which rarely combine sequences in this way.
Spec1.04h 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

6 An arithmetic progression \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by \(u _ { 1 } = 5\) and \(u _ { n + 1 } = u _ { n } + 1.5\) for \(n \geqslant 1\).
  1. Given that \(u _ { k } = 140\), find the value of \(k\). A geometric progression \(w _ { 1 } , w _ { 2 } , w _ { 3 } , \ldots\) is defined by \(w _ { n } = 120 \times ( 0.9 ) ^ { n - 1 }\) for \(n \geqslant 1\).
  2. Find the sum of the first 16 terms of this geometric progression, giving your answer correct to 3 significant figures.
  3. Use an algebraic method to find the smallest value of \(N\) such that \(\sum _ { n = 1 } ^ { N } u _ { n } > \sum _ { n = 1 } ^ { \infty } w _ { n }\).

Question 6:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
\(u_k = 5 + 1.5(k-1)\)M1* Attempt \(n\)th term of AP using \(a = 5\) and \(d = 1.5\); M0 for \(5 + 1.5k\)
\(5 + 1.5(k-1) = 140\)M1d* Equate to 140 and attempt to solve for \(k\); must be valid solution attempt
\(k = 91\)A1 [3] Obtain 91; answer only gains full credit
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(S_{16} = \frac{120(1 - 0.9^{16})}{1 - 0.9} = 978\)M1 Attempt to find sum of 16 terms of GP with \(a = 120\), \(r = 0.9\); must use correct formula
A1 [2]Obtain 978 or better; if \(> 3\)sf allow rounding to 977.6 with no errors seen
Part (iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{N}{2}(10 + (N-1) \times 1.5) > \frac{120}{1-0.9}\)B1 Correct sum to infinity stated; could be 1200 or unsimplified
\(N(1.5N + 8.5) > 2400\)B1 Correct \(S_N\) stated; any correct expression including unsimplified
\(3N^2 + 17N - 4800 > 0\)M1* Link \(S_N\) of AP to \(S_\infty\) of GP and attempt to rearrange; must rearrange to three term quadratic
A1Obtain correct 3 term quadratic
M1d*Attempt to solve quadratic
\(N = 38\)A1 [6] Obtain \(N = 38\) (must be equality); A0 for \(N \geq 38\); A0 if second value of \(N\) given in final answer
# Question 6:

## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_k = 5 + 1.5(k-1)$ | M1* | Attempt $n$th term of AP using $a = 5$ and $d = 1.5$; M0 for $5 + 1.5k$ |
| $5 + 1.5(k-1) = 140$ | M1d* | Equate to 140 and attempt to solve for $k$; must be valid solution attempt |
| $k = 91$ | A1 [3] | Obtain 91; answer only gains full credit |

## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{16} = \frac{120(1 - 0.9^{16})}{1 - 0.9} = 978$ | M1 | Attempt to find sum of 16 terms of GP with $a = 120$, $r = 0.9$; must use correct formula |
| | A1 [2] | Obtain 978 or better; if $> 3$sf allow rounding to 977.6 with no errors seen |

## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{N}{2}(10 + (N-1) \times 1.5) > \frac{120}{1-0.9}$ | B1 | Correct sum to infinity stated; could be 1200 or unsimplified |
| $N(1.5N + 8.5) > 2400$ | B1 | Correct $S_N$ stated; any correct expression including unsimplified |
| $3N^2 + 17N - 4800 > 0$ | M1* | Link $S_N$ of AP to $S_\infty$ of GP and attempt to rearrange; must rearrange to three term quadratic |
| | A1 | Obtain correct 3 term quadratic |
| | M1d* | Attempt to solve quadratic |
| $N = 38$ | A1 [6] | Obtain $N = 38$ (must be equality); A0 for $N \geq 38$; A0 if second value of $N$ given in final answer |

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6 An arithmetic progression $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by $u _ { 1 } = 5$ and $u _ { n + 1 } = u _ { n } + 1.5$ for $n \geqslant 1$.\\
(i) Given that $u _ { k } = 140$, find the value of $k$.

A geometric progression $w _ { 1 } , w _ { 2 } , w _ { 3 } , \ldots$ is defined by $w _ { n } = 120 \times ( 0.9 ) ^ { n - 1 }$ for $n \geqslant 1$.\\
(ii) Find the sum of the first 16 terms of this geometric progression, giving your answer correct to 3 significant figures.\\
(iii) Use an algebraic method to find the smallest value of $N$ such that $\sum _ { n = 1 } ^ { N } u _ { n } > \sum _ { n = 1 } ^ { \infty } w _ { n }$.

\hfill \mbox{\textit{OCR C2 2016 Q6 [11]}}