Edexcel C2 2005 January — Question 6 8 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2005
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeFind sum to infinity
DifficultyModerate -0.3 This is a straightforward multi-part geometric series question requiring standard formulas and calculator work. Parts (a)-(c) involve routine algebraic manipulation (finding r from ar³/ar = ratio, then finding a, then applying sum formula), while part (d) tests understanding that S∞ - S₅₀ exists when |r| < 1. All techniques are standard C2 content with no novel problem-solving 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

  1. The second and fourth terms of a geometric series are 7.2 and 5.832 respectively.
The common ratio of the series is positive.
For this series, find
  1. the common ratio,
  2. the first term,
  3. the sum of the first 50 terms, giving your answer to 3 decimal places,
  4. the difference between the sum to infinity and the sum of the first 50 terms, giving your answer to 3 decimal places.

(a) \(ar = 7.2, ar^3 = 5.832 \Rightarrow r^2 = \frac{5.832}{7.2} (= 0.81)\)
AnswerMarks Guidance
\(r = 0.9\)M1, A1 (2 marks)
(b) \(a = \frac{7.2}{(a)} = 8\)M1, A1 (2 marks)
(c) \(s_{50} = \frac{8(1-(0.9)^{50})}{1-0.9} = 79.588\)(3dp) M1, A1 c.a.o
(d) \(s_\infty = \frac{8}{1-0.9} (= 80)\)
AnswerMarks Guidance
\(s_\infty - s_{50} = 80 - (c) = 0.412\)(Awrt 3 dp) M1, A1
(Subtotal: 8 marks)
Guidance: (a) M1: for full method \(\to r^2\) or \(r\). N.B. \(ar^2 = 7.2, ar^4 = 5.832 \to r = 0.9\) scores M1A1 in part (a) but probably M0A0 in (b).
(c) M1: their "\(a\)", "\(r\)" in \(s_{50}\) formula. (d) M1: their "\(a\)", "\(r\)" in \(s_\infty\). A1: for \(80 -\) their (c), i.e., their (c) only.
**(a)** $ar = 7.2, ar^3 = 5.832 \Rightarrow r^2 = \frac{5.832}{7.2} (= 0.81)$
$r = 0.9$ | M1, A1 | (2 marks)

**(b)** $a = \frac{7.2}{(a)} = 8$ | M1, A1 | (2 marks)

**(c)** $s_{50} = \frac{8(1-(0.9)^{50})}{1-0.9} = 79.588$ | (3dp) | M1, A1 c.a.o | (2 marks)

**(d)** $s_\infty = \frac{8}{1-0.9} (= 80)$
$s_\infty - s_{50} = 80 - (c) = 0.412$ | (Awrt 3 dp) | M1, A1 | (2 marks)

**(Subtotal: 8 marks)**

Guidance: (a) M1: for full method $\to r^2$ or $r$. N.B. $ar^2 = 7.2, ar^4 = 5.832 \to r = 0.9$ scores M1A1 in part (a) but probably M0A0 in (b).
(c) M1: their "$a$", "$r$" in $s_{50}$ formula. (d) M1: their "$a$", "$r$" in $s_\infty$. A1: for $80 -$ their (c), i.e., their (c) only.
\begin{enumerate}
  \item The second and fourth terms of a geometric series are 7.2 and 5.832 respectively.
\end{enumerate}

The common ratio of the series is positive.\\
For this series, find\\
(a) the common ratio,\\
(b) the first term,\\
(c) the sum of the first 50 terms, giving your answer to 3 decimal places,\\
(d) the difference between the sum to infinity and the sum of the first 50 terms, giving your answer to 3 decimal places.\\

\hfill \mbox{\textit{Edexcel C2 2005 Q6 [8]}}