| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| 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 geometric series question requiring standard formulas (r = 4.8/5 = 0.96, S_∞ = a/(1-r) = 125) and basic logarithm manipulation to solve 0.96^n < 0.008. The algebra is guided by 'show that' scaffolding, making it slightly easier than average but still requiring multiple connected steps. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{4.8}{5} = 0.96 \Rightarrow S_\infty = \frac{5}{0.04} = 125\) | B1* | For correct value of \(r\) used |
| B1 | For correct use of \(\frac{a}{1-r}\) to show given answer AG | |
| dep° |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_n = \frac{5(1 - 0.96^n)}{1 - 0.96}\) | B1 | For correct, unsimplified, \(S_n\) |
| M1 | For linking \(S_n\) to 124 (\(>\) or \(=\)) and multiplying through by 0.04, or equiv. | |
| A1 | For showing the given result correctly, with correct inequality throughout AG | |
| Hence \(1 - 0.96^n > 0.992 \Rightarrow 0.96^n < 0.008\) | B1 | For correct log statement seen or implied (ignore sign) |
| M1 | For dividing both sides by log 0.96 | |
| \(\text{Hence } n > \frac{\log 0.008}{\log 0.96} = 118.3\) | A1 | For correct (integer) value 119 |
**(i)**
$r = \frac{4.8}{5} = 0.96 \Rightarrow S_\infty = \frac{5}{0.04} = 125$ | B1* | For correct value of $r$ used
| B1 | For correct use of $\frac{a}{1-r}$ to show given answer AG
| | dep°
**(ii)**
$S_n = \frac{5(1 - 0.96^n)}{1 - 0.96}$ | B1 | For correct, unsimplified, $S_n$
| M1 | For linking $S_n$ to 124 ($>$ or $=$) and multiplying through by 0.04, or equiv.
| A1 | For showing the given result correctly, with correct inequality throughout AG
Hence $1 - 0.96^n > 0.992 \Rightarrow 0.96^n < 0.008$ | B1 | For correct log statement seen or implied (ignore sign)
| M1 | For dividing both sides by log 0.96
$\text{Hence } n > \frac{\log 0.008}{\log 0.96} = 118.3$ | A1 | For correct (integer) value 119
Least value of $n$ is 119
---5 In a geometric progression, the first term is 5 and the second term is 4.8 .\\
(i) Show that the sum to infinity is 125 .\\
(ii) The sum of the first $n$ terms is greater than 124 . Show that
$$0.96 ^ { n } < 0.008$$
and use logarithms to calculate the smallest possible value of $n$.
\hfill \mbox{\textit{OCR C2 2006 Q5 [8]}}