CAIE P1 2011 June — Question 6 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2011
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeFind first term from conditions
DifficultyStandard +0.3 Part (a) requires setting up two equations from given conditions (a·r² = 20 and a/(1-r) = 3a) and solving simultaneously, which is a standard GP problem with one moderately tricky algebraic step. Part (b) is a straightforward proof using standard AP formulas with simple algebraic manipulation. Both parts are routine applications of series formulas with no novel insight required, making this slightly easier than average.
Spec1.04h Arithmetic sequences: nth term and sum formulae1.04j Sum to infinity: convergent geometric series |r|<1

6
  1. A geometric progression has a third term of 20 and a sum to infinity which is three times the first term. Find the first term.
  2. An arithmetic progression is such that the eighth term is three times the third term. Show that the sum of the first eight terms is four times the sum of the first four terms.

AnswerMarks Guidance
(a) \(ar^2 = 20\)B1 co
\(\frac{a}{1-r} = 3a\)B1 co
Soln of equations \(\rightarrow (r = \frac{2}{3})\) \(a = 45\)M1 A1 Complete method to find \(a\). co
[4]
(b) \(a + 7d = 3(a + 2d)\)M1 Use of \(a + (n-1)d\)
\(\rightarrow 2a = d\)A1 co
\(S_8 = 4(2a + 7d) = 32d\) or \(64a\)M1 correct use of \(S_n\) formula once.
\(S_4 = 2(2a + 3d) = 8d\) or \(16a\)A1 ag
[4]
(a) $ar^2 = 20$ | B1 | co
$\frac{a}{1-r} = 3a$ | B1 | co
Soln of equations $\rightarrow (r = \frac{2}{3})$ $a = 45$ | M1 A1 | Complete method to find $a$. co
| [4] |

(b) $a + 7d = 3(a + 2d)$ | M1 | Use of $a + (n-1)d$
$\rightarrow 2a = d$ | A1 | co
$S_8 = 4(2a + 7d) = 32d$ or $64a$ | M1 | correct use of $S_n$ formula once.
$S_4 = 2(2a + 3d) = 8d$ or $16a$ | A1 | ag
| [4] |
6
\begin{enumerate}[label=(\alph*)]
\item A geometric progression has a third term of 20 and a sum to infinity which is three times the first term. Find the first term.
\item An arithmetic progression is such that the eighth term is three times the third term. Show that the sum of the first eight terms is four times the sum of the first four terms.
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2011 Q6 [8]}}