Edexcel C2 — Question 9 11 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeNew GP from transformation
DifficultyStandard +0.3 This is a straightforward application of geometric sequences with common ratio 1.5. Part (a) requires finding the 4th term using r³, part (b) involves summing diameters (routine GP sum formula), and part (c) combines GP sum with area formula. All steps are standard C2 techniques with no novel insight required, making it slightly easier than average.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{05006f1f-ebf0-4d70-9dbb-68221c09043e-4_325_662_1345_520} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows part of a design being produced by a computer program.
The program draws a series of circles with each one touching the previous one and such that their centres lie on a horizontal straight line. The radii of the circles form a geometric sequence with first term 1 mm and second term 1.5 mm . The width of the design is \(w\) as shown.
  1. Find the radius of the fourth circle to be drawn.
  2. Show that when eight circles have been drawn, \(w = 98.5 \mathrm {~mm}\) to 3 significant figures.
  3. Find the total area of the design in square centimetres when ten circles have been drawn.

AnswerMarks Guidance
(a) \(r = 1.5\)
\(u_4 = 1 \times (1.5)^3 = 3.375\) mmM1 A1
(b) \(w = 2 \times S_6; GP, a = 1, r = 1.5\)M1
\(= 2 \times \frac{1[(1.5)^6 - 1]}{1.5 - 1}\)M1 A1
\(= 98.516 = 98.5\) mm (3sf)A1
(c) areas form GP, \(a = \pi \times 1^2 = \pi, r = (1.5)^2 = 2.25\)B2
total area \(= \frac{\pi[(2.25)^{10} - 1]}{2.25 - 1} = 8354.8\) mm²M1 A1
\(= \frac{8354.8}{10^2}\) cm² \(= 83.5\) cm² (3sf)A1 (11)
Total: (75)
**(a)** $r = 1.5$ | |

$u_4 = 1 \times (1.5)^3 = 3.375$ mm | M1 A1 |

**(b)** $w = 2 \times S_6; GP, a = 1, r = 1.5$ | M1 |

$= 2 \times \frac{1[(1.5)^6 - 1]}{1.5 - 1}$ | M1 A1 |

$= 98.516 = 98.5$ mm (3sf) | A1 |

**(c)** areas form GP, $a = \pi \times 1^2 = \pi, r = (1.5)^2 = 2.25$ | B2 |

total area $= \frac{\pi[(2.25)^{10} - 1]}{2.25 - 1} = 8354.8$ mm² | M1 A1 |

$= \frac{8354.8}{10^2}$ cm² $= 83.5$ cm² (3sf) | A1 | **(11)**

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**Total: (75)**
9.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{05006f1f-ebf0-4d70-9dbb-68221c09043e-4_325_662_1345_520}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Figure 3 shows part of a design being produced by a computer program.\\
The program draws a series of circles with each one touching the previous one and such that their centres lie on a horizontal straight line.

The radii of the circles form a geometric sequence with first term 1 mm and second term 1.5 mm . The width of the design is $w$ as shown.
\begin{enumerate}[label=(\alph*)]
\item Find the radius of the fourth circle to be drawn.
\item Show that when eight circles have been drawn, $w = 98.5 \mathrm {~mm}$ to 3 significant figures.
\item Find the total area of the design in square centimetres when ten circles have been drawn.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2  Q9 [11]}}