| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| 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 | Moderate -0.3 This is a straightforward geometric series question requiring standard formula applications (S_∞ = a/(1-r), finding terms, S_n formula). Part (a) is given as 'show that', making it easier. Part (d) requires solving an inequality but with direct substitution into the S_n formula. All techniques are routine C2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{a}{1-r} = 480\) | M1 | |
| \(\frac{120}{1-r} = 480 \Rightarrow 120 = 480(1-r)\) | M1 | |
| \(1-r = \frac{1}{4} \Rightarrow r = \frac{3}{4}\) * | A1 cso | (3 marks) |
| (b) \(u_5 = 120 \times \left(\frac{3}{4}\right)^4 [= 37.96875]\) either | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Difference \(= 9.49\) | A1 (allow \(\pm\)) | (2 marks) |
| (c) \(S_7 = \frac{120(1-(0.75)^7)}{1-0.75}\) | M1 | |
| \(= 415.9277\ldots\) | (AWRT) 416 | A1 |
| (d) \(\frac{120(1-(0.75)^n)}{1-0.75} > 300\) | M1 | |
| \(1-(0.75)^n > \frac{300}{480}\) | A1 (or better) | |
| \(n > \frac{\log(0.375)}{\log(0.75)}\) | M1 (=\(3.409\ldots\)) | |
| \(n = 4\) | A1 cso | (4 marks) |
(a) $\frac{a}{1-r} = 480$ | M1
$\frac{120}{1-r} = 480 \Rightarrow 120 = 480(1-r)$ | M1
$1-r = \frac{1}{4} \Rightarrow r = \frac{3}{4}$ * | A1 cso | (3 marks)
(b) $u_5 = 120 \times \left(\frac{3}{4}\right)^4 [= 37.96875]$ either | M1
$u_6 = 120 \times \left(\frac{3}{4}\right)^5 [= 28.4765625]$
Difference $= 9.49$ | A1 (allow $\pm$) | (2 marks)
(c) $S_7 = \frac{120(1-(0.75)^7)}{1-0.75}$ | M1
$= 415.9277\ldots$ | (AWRT) 416 | A1 | (2 marks)
(d) $\frac{120(1-(0.75)^n)}{1-0.75} > 300$ | M1
$1-(0.75)^n > \frac{300}{480}$ | A1 (or better)
$n > \frac{\log(0.375)}{\log(0.75)}$ | M1 (=$3.409\ldots$)
$n = 4$ | A1 cso | (4 marks)
**Guidance:**
- (a) 1st M1 for use of $S_\infty$; 2nd M1 for substituting for $a$ and moving $(1-r)$ to form linear equation in $r$.
- (b) M1 for some correct use of $ar^{n-1}$ [$120\left(\frac{3}{4}\right)^2 - 120\left(\frac{3}{4}\right)^0$ is M0]
- (c) M1 for a correct expression (need use of $a$ and $r$)
- (d) 1st M1 for attempting $S_n > 300$ [or $=300$] (need use of $a$ and some use of $r$); 2nd M1 for valid attempt to solve $r^n = p(r, p<1)$, must give linear eqn in $n$. Any correct log form will do.
**Trial & Imp.:**
- 1st M1 for attempting at least 2 values of $S_n$, one $n<4$ and one $n \geq 4$.
- 2nd M1 for attempting $S_3$ and $S_4$.
- 1st A1 for both values correct to 2 s.f. or better.
- 2nd A1 for $n=4$.
**For Information:** $u_1=120$, $u_3=90$, $u_4=67.5$, $u_4=50.625$, $S_2=210$, $S_3=277.5$, $S_4=328.125$, $S_5=366.09\ldots$
**Total: 11 marks**
---\begin{enumerate}
\item The first term of a geometric series is 120 . The sum to infinity of the series is 480 .\\
(a) Show that the common ratio, $r$, is $\frac { 3 } { 4 }$.\\
(b) Find, to 2 decimal places, the difference between the 5th and 6th term.\\
(c) Calculate the sum of the first 7 terms.
\end{enumerate}
The sum of the first $n$ terms of the series is greater than 300 .\\
(d) Calculate the smallest possible value of $n$.\\
\hfill \mbox{\textit{Edexcel C2 2006 Q4 [11]}}