| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.8 This is a straightforward geometric series question requiring only standard formulas and routine calculations. Finding the common ratio from consecutive terms (r = 144/192 = 3/4), then working backwards to find the first term, applying the sum to infinity formula, and solving an inequality are all textbook procedures with no conceptual challenges or novel problem-solving required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\{ar = 192\) and \(ar^2 = 144\}\) | ||
| \(r = \frac{144}{192}\) | M1 | Attempt to eliminate \(a\) |
| \(r = \frac{3}{4}\) or 0.75 | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(a(0.75) = 192\) | M1 | Insert their \(r\) into correct equation |
| \(a = \frac{192}{0.75} = 256\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(S_\infty = \frac{256}{1-0.75}\) | M1 | Applies \(\frac{a}{1-r}\) correctly using both their \(a\) and \( |
| \(S_\infty = 1024\) | A1 cao | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{256(1-(0.75)^n)}{1-0.75} > 1000\) | M1 | Applies \(S_n\) with their \(a\), \(r\) and "uses" 1000 |
| \((0.75)^n < 1 - \frac{1000(0.25)}{256} = \frac{6}{256}\) | M1 | Attempt to isolate \(+(r)^n\) from \(S_n\) formula |
| \(n\log(0.75) < \log\left(\frac{6}{256}\right)\) | M1 | Uses power law of logarithms correctly |
| \(n > \frac{\log\left(\frac{6}{256}\right)}{\log(0.75)} = 13.047... \Rightarrow n = 14\) | A1 cso | Note direction of inequality reversal [4] |
## Question 6:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\{ar = 192$ and $ar^2 = 144\}$ | | |
| $r = \frac{144}{192}$ | M1 | Attempt to eliminate $a$ |
| $r = \frac{3}{4}$ or 0.75 | A1 | **[2]** |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $a(0.75) = 192$ | M1 | Insert their $r$ into correct equation |
| $a = \frac{192}{0.75} = 256$ | A1 | **[2]** |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $S_\infty = \frac{256}{1-0.75}$ | M1 | Applies $\frac{a}{1-r}$ correctly using both their $a$ and $|r|<1$ |
| $S_\infty = 1024$ | A1 cao | **[2]** |
### Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{256(1-(0.75)^n)}{1-0.75} > 1000$ | M1 | Applies $S_n$ with their $a$, $r$ and "uses" 1000 |
| $(0.75)^n < 1 - \frac{1000(0.25)}{256} = \frac{6}{256}$ | M1 | Attempt to isolate $+(r)^n$ from $S_n$ formula |
| $n\log(0.75) < \log\left(\frac{6}{256}\right)$ | M1 | Uses power law of logarithms correctly |
| $n > \frac{\log\left(\frac{6}{256}\right)}{\log(0.75)} = 13.047... \Rightarrow n = 14$ | A1 cso | Note direction of inequality reversal **[4]** |
---\begin{enumerate}
\item The second and third terms of a geometric series are 192 and 144 respectively.
\end{enumerate}
For this series, find\\
(a) the common ratio,\\
(b) the first term,\\
(c) the sum to infinity,\\
(d) the smallest value of $n$ for which the sum of the first $n$ terms of the series exceeds 1000.\\
\hfill \mbox{\textit{Edexcel C2 2011 Q6 [10]}}