OCR MEI C2 — Question 7 10 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeProve term relationship
DifficultyStandard +0.3 This is a standard geometric progression problem requiring recall of formulas (second term = ar, sum to infinity = a/(1-r)) and solving simultaneous equations to find two solutions. Part (ii) involves straightforward algebraic manipulation of the ratio of nth terms. While it has multiple steps and 10 marks total, it follows predictable patterns 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
A geometric progression has first term \(a\) and common ratio \(r\). The second term is 6 and the sum to infinity is 25.
  1. Write down two equations in \(a\) and \(r\). Show that one possible value of \(a\) is 10 and find the other possible value of \(a\). Write down the corresponding values of \(r\). [7]
  2. Show that the ratio of the \(n\)th terms of the two geometric progressions found in part (i) can be written as \(2^{n-2} : 3^{n-2}\). [3]
Question 7:
AnswerMarks Guidance
7(i) ar = 6 oe
a
25 oe
1r
a
25
16
a
a2 – 25a + 150 [= 0]
a = 10 obtained from formula, factorising,
Factor theorem or completing the square
a = 15
AnswerMarks
r = 0.4 and 0.6B1
B1
M1
A1
A1
A1
A1
AnswerMarks
[7]must be in a and r
must be in a and r
6
or 25(1r)
r
or 25r2 – 25r + 6 [= 0]
r = 0.4 and r = 0.6
a = 15
6
a 10oe
AnswerMarks
0.6NB assuming a = 10 earns M0
All signs may be reversed
if M0, B1 for r = 0.4 and 0.6 and B1 for
a = 15 by trial and improvement
mark to benefit of candidate
AnswerMarks Guidance
7(ii) 10 × (3/5)n – 1 and 15 × (2/5)n – 1 seen
2n1 3n1
15 × 2n – 1 : 10 × 3n-1 or 3 :2
5n1 5n1
AnswerMarks
3 × 2n – 1 : 2 × 3n – 1M1
M1
A1
AnswerMarks
[3]may be implied by 3 × 2n – 1 : 2 × 3n – 1
and completion to given answer wwwcondone ratio reversed
condone ratio reversed
Question 7:
7 | (i) | ar = 6 oe
a
25 oe
1r
a
25
16
a
a2 – 25a + 150 [= 0]
a = 10 obtained from formula, factorising,
Factor theorem or completing the square
a = 15
r = 0.4 and 0.6 | B1
B1
M1
A1
A1
A1
A1
[7] | must be in a and r
must be in a and r
6
or 25(1r)
r
or 25r2 – 25r + 6 [= 0]
r = 0.4 and r = 0.6
a = 15
6
a 10oe
0.6 | NB assuming a = 10 earns M0
All signs may be reversed
if M0, B1 for r = 0.4 and 0.6 and B1 for
a = 15 by trial and improvement
mark to benefit of candidate
7 | (ii) | 10 × (3/5)n – 1 and 15 × (2/5)n – 1 seen
2n1 3n1
15 × 2n – 1 : 10 × 3n-1 or 3 :2
5n1 5n1
3 × 2n – 1 : 2 × 3n – 1 | M1
M1
A1
[3] | may be implied by 3 × 2n – 1 : 2 × 3n – 1
and completion to given answer www | condone ratio reversed
condone ratio reversed
A geometric progression has first term $a$ and common ratio $r$. The second term is 6 and the sum to infinity is 25.

\begin{enumerate}[label=(\roman*)]
\item Write down two equations in $a$ and $r$. Show that one possible value of $a$ is 10 and find the other possible value of $a$. Write down the corresponding values of $r$. [7]
\item Show that the ratio of the $n$th terms of the two geometric progressions found in part (i) can be written as $2^{n-2} : 3^{n-2}$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C2  Q7 [10]}}
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Q1 5 Q2 12 Q3 5 Q4 5 Q5 3 Q6 4 Q7 10