Question 8 - A-Level Maths

OCR C2 2008 January — Question 8 11 marks

Exam Board	OCR
Module	C2 (Core Mathematics 2)
Year	2008
Session	January
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Geometric Sequences and Series
Type	Find N for S_∞ - S_N condition
Difficulty	Standard +0.3 This is a straightforward multi-part geometric series question requiring standard formula application (finding terms, finite sum, sum to infinity) and basic logarithm manipulation to solve an inequality. While part (iii) involves algebraic manipulation and logarithms, the steps are routine for C2 level with no novel insight required—slightly easier than average due to the scaffolded structure and standard techniques.
Spec	1.04i Geometric sequences: nth term and finite series sum 1.04j Sum to infinity: convergent geometric series \|r\|<1 1.06g Equations with exponentials: solve a^x = b

8 The first term of a geometric progression is 10 and the common ratio is 0.8.

Find the fourth term.
Find the sum of the first 20 terms, giving your answer correct to 3 significant figures.
The sum of the first $N$ terms is denoted by $S _ { N }$, and the sum to infinity is denoted by $S _ { \infty }$. Show that the inequality $S _ { \infty } - S _ { N } < 0.01$ can be written as $$0.8 ^ { N } < 0.0002 ,$$ and use logarithms to find the smallest possible value of $N$.

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Question 8:

Part (i)

Answer	Marks	Guidance
$u_4 = 10 \times 0.8^3 = 5.12$	M1, A1 [2]	Attempt $u_4$ using $ar^{n-1}$; Obtain 5.12 aef

Part (ii)

Answer	Marks	Guidance
$S_{20} = \frac{10(1-0.8^{20})}{1-0.8} = 49.4$	M1, A1 [2]	Attempt use of correct sum formula for a GP; Obtain 49.4

Part (iii)

Answer	Marks	Guidance
$\frac{10}{1-0.8} - \frac{10(1-0.8^N)}{(1-0.8)} < 0.01$	M1	Attempt $S_\infty$ using $\frac{a}{1-r}$
$50 - 50(1-0.8^N) < 0.01$	A1	Obtain $S_\infty = 50$, or unsimplified equiv
$0.8^N < 0.0002$ A.G.	M1	Link $S_\infty - S_N$ to 0.01 and attempt to rearrange
Show given inequality convincingly	A1	Show given inequality convincingly
$\log 0.8^N < \log 0.0002$	M1	Introduce logarithms on both sides
$N \log 0.8 < \log 0.0002$	M1	Use $\log a^b = b \log a$, and attempt to find $N$
$N > 38.169$, hence $N = 39$	A1 [7]	Obtain $N = 39$ only

# Question 8:

## Part (i)
| $u_4 = 10 \times 0.8^3 = 5.12$ | M1, A1 **[2]** | Attempt $u_4$ using $ar^{n-1}$; Obtain 5.12 aef |

## Part (ii)
| $S_{20} = \frac{10(1-0.8^{20})}{1-0.8} = 49.4$ | M1, A1 **[2]** | Attempt use of correct sum formula for a GP; Obtain 49.4 |

## Part (iii)
| $\frac{10}{1-0.8} - \frac{10(1-0.8^N)}{(1-0.8)} < 0.01$ | M1 | Attempt $S_\infty$ using $\frac{a}{1-r}$ |
| $50 - 50(1-0.8^N) < 0.01$ | A1 | Obtain $S_\infty = 50$, or unsimplified equiv |
| $0.8^N < 0.0002$ **A.G.** | M1 | Link $S_\infty - S_N$ to 0.01 and attempt to rearrange |
| Show given inequality convincingly | A1 | Show given inequality convincingly |
| $\log 0.8^N < \log 0.0002$ | M1 | Introduce logarithms on both sides |
| $N \log 0.8 < \log 0.0002$ | M1 | Use $\log a^b = b \log a$, and attempt to find $N$ |
| $N > 38.169$, hence $N = 39$ | A1 **[7]** | Obtain $N = 39$ only |

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8 The first term of a geometric progression is 10 and the common ratio is 0.8.\\
(i) Find the fourth term.\\
(ii) Find the sum of the first 20 terms, giving your answer correct to 3 significant figures.\\
(iii) The sum of the first $N$ terms is denoted by $S _ { N }$, and the sum to infinity is denoted by $S _ { \infty }$. Show that the inequality $S _ { \infty } - S _ { N } < 0.01$ can be written as

$$0.8 ^ { N } < 0.0002 ,$$

and use logarithms to find the smallest possible value of $N$.

\hfill \mbox{\textit{OCR C2 2008 Q8 [11]}}

This paper (10 questions)

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Q1 4 Q2 4 Q3 4 Q4 5 Q5 6 Q6 8 Q7 8 Q8 11 Q9 9 Q10 12

Answer	Marks	Guidance
\(\frac{10}{1-0.8} - \frac{10(1-0.8^N)}{(1-0.8)} < 0.01\)	M1	Attempt \(S_\infty\) using \(\frac{a}{1-r}\)
\(50 - 50(1-0.8^N) < 0.01\)	A1	Obtain \(S_\infty = 50\), or unsimplified equiv
\(0.8^N < 0.0002\) A.G.	M1	Link \(S_\infty - S_N\) to 0.01 and attempt to rearrange
Show given inequality convincingly	A1	Show given inequality convincingly
\(\log 0.8^N < \log 0.0002\)	M1	Introduce logarithms on both sides
\(N \log 0.8 < \log 0.0002\)	M1	Use \(\log a^b = b \log a\), and attempt to find \(N\)
\(N > 38.169\), hence \(N = 39\)	A1 [7]	Obtain \(N = 39\) only