| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric sequences requiring standard formulas (nth term and sum). Part (i) is direct substitution, part (ii) requires solving an inequality with logarithms (standard C2 technique), and part (iii) uses the sum formula. While multi-part, each step follows textbook procedures without requiring problem-solving insight or novel approaches, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(100 \, 000 \times 0.9^3 = 72900\) | M1 | For relevant use of \(ar^n\) or equiv |
| (ii) \(100 \, 000 \times 0.9^5 = 5000\) | A1, B1 | For the correct answer 72900; For a correct equation or inequality |
| Hence \(\log 0.9 = \log 0.05\) | M1 | For complete solution method by logs or trial |
| So \(x = 28.4, 28\) or \(29\); or \(n = 29, 4, 29\) or \(30\) | A1 | For correct solution for their index – allow integer values either side |
| i.e. 30th year / 30 years / year is 2030 | A1 | For correctly linking their index to date or |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) Total is \(\frac{100000(1 - 0.9^n)}{1 - 0.9} = 957609\) | M1 | For relevant use of \(\frac{a(1-r^n)}{1-r}\) |
| A1∨ | For correct (unsimplified) statement for their integer \(n\) (if no \(n\) stated then use their year – 2000) | |
| A1 | For answer 958000 or better, including decimal |
**(i)** $100 \, 000 \times 0.9^3 = 72900$ | M1 | For relevant use of $ar^n$ or equiv
**(ii)** $100 \, 000 \times 0.9^5 = 5000$ | A1, B1 | For the correct answer 72900; For a correct equation or inequality
Hence $\log 0.9 = \log 0.05$ | M1 | For complete solution method by logs or trial
So $x = 28.4, 28$ or $29$; or $n = 29, 4, 29$ or $30$ | A1 | For correct solution for their index – allow integer values either side
i.e. 30th year / 30 years / year is 2030 | A1 | For correctly linking their index to date or
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## Question 8 (continued):
**(iii)** Total is $\frac{100000(1 - 0.9^n)}{1 - 0.9} = 957609$ | M1 | For relevant use of $\frac{a(1-r^n)}{1-r}$
| A1∨ | For correct (unsimplified) statement for their integer $n$ (if no $n$ stated then use their year – 2000)
| A1 | For answer 958000 or better, including decimal
---8 The amounts of oil pumped from an oil well in each of the years 2001 to 2004 formed a geometric progression with common ratio 0.9 . The amount pumped in 2001 was 100000 barrels.\\
(i) Calculate the amount pumped in 2004.
It is assumed that the amounts of oil pumped in future years will continue to follow the same geometric progression. Production from the well will stop at the end of the first year in which the amount pumped is less than 5000 barrels.\\
(ii) Calculate in which year the amount pumped will fall below 5000 barrels.\\
(iii) Calculate the total amount of oil pumped from the well from the year 2001 up to and including the final year of production.
\hfill \mbox{\textit{OCR C2 2005 Q8 [9]}}