| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.3 This is a straightforward geometric series question testing standard formulas (sum to infinity, nth term, sum of n terms). Part (a) is routine algebra using S∞ = a/(1-r). Parts (b-c) apply standard formulas. Part (d) requires minor algebraic manipulation but follows directly from the formula. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks |
|---|---|
| \(\frac{a}{1-r} = \frac{1200}{1-r} = 960\) | M1 A1 |
| \(960(1-r) = 1200\) resulting in \(r = -\frac{1}{4}\) | A1 |
| (3 marks) |
| Answer | Marks |
|---|---|
| \(T_9 = 1200 \times (-0.25)^8\) (or \(T_{10}\)) | M1 |
| Difference \(= T_9 - T_{10} = 0.0183105\ldots - (-0.0045776\ldots)\) | M1 |
| \(= 0.023\) (or \(-0.023\)) | A1 |
| (3 marks) |
| Answer | Marks |
|---|---|
| \(S_n = \frac{1200\left(1-(-0.25)^n\right)}{1-(-0.25)}\) | M1 A1 |
| (2 marks) |
| Answer | Marks |
|---|---|
| Since \(n\) is odd, \((-0.25)^n\) is negative, | M1 |
| so \(S_n = 960(1 + 0.25^n)\) | A1 |
| (2 marks) | |
| (10 marks total) |
## (a)
$\frac{a}{1-r} = \frac{1200}{1-r} = 960$ | M1 A1 |
$960(1-r) = 1200$ resulting in $r = -\frac{1}{4}$ | A1 |
| (3 marks) |
## (b)
$T_9 = 1200 \times (-0.25)^8$ (or $T_{10}$) | M1 |
Difference $= T_9 - T_{10} = 0.0183105\ldots - (-0.0045776\ldots)$ | M1 |
$= 0.023$ (or $-0.023$) | A1 |
| (3 marks) |
## (c)
$S_n = \frac{1200\left(1-(-0.25)^n\right)}{1-(-0.25)}$ | M1 A1 |
| (2 marks) |
## (d)
Since $n$ is odd, $(-0.25)^n$ is negative, | M1 |
so $S_n = 960(1 + 0.25^n)$ | A1 |
| (2 marks) |
| **(10 marks total)** |
---A geometric series has first term $1200$. Its sum to infinity is $960$.
\begin{enumerate}[label=(\alph*)]
\item Show that the common ratio of the series is $-\frac{1}{4}$. [3]
\item Find, to 3 decimal places, the difference between the ninth and tenth terms of the series. [3]
\item Write down an expression for the sum of the first $n$ terms of the series. [2]
\end{enumerate}
Given that $n$ is odd,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item prove that the sum of the first $n$ terms of the series is $960(1 + 0.25^n)$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [10]}}