Moderate -0.3 This is a straightforward multi-part geometric sequence question requiring standard formulas (ar^n for terms, sum formulas). Part (a) is guided ('show that'), making it easier. All parts involve direct application of memorized formulas with no conceptual challenges or novel problem-solving, making it slightly easier than average but not trivial due to the multi-step nature and algebraic manipulation required.
M1 for forming an equation for \(r^3\) based on 96 and 324 (e.g. \(96r^3 = 324\) scores M1). The equation must involve multiplication/division rather than addition/subtraction.
Do not penalise solutions with working in decimals, providing these are correctly rounded or truncated to at least 2dp and the final answer 2/3 is seen.
Alternative (verification):
M1 Using \(r^3 = \frac{8}{27}\) and multiplying 324 by this (or multiplying by \(r = \frac{2}{3}\) three times).
A1 Obtaining 96 (cso). (A conclusion is not required.)
\(324 \times \left(\frac{2}{3}\right)^3 = 96\) (no real evidence of calculation) is not quite enough and scores M1 A0.
Part (b):
M1 for the use of a correct formula or for 'working back' by dividing by \(\frac{2}{3}\) (or by their \(r\)) twice from 324 (or 5 times from 96).
Exceptionally, allow M1 also for using \(ar^3 = 324\) or \(ar^6 = 96\) instead of \(ar^2 = 324\) or \(ar^3 = 96\), or for dividing by \(r\) three times from 324 (or 6 times from 96)... but no other exceptions are allowed.
Part (c):
M1 for use of sum to 15 terms formula with values of \(a\) and \(r\). If the wrong power is used, e.g. 14, the M mark is scored only if the correct sum formula is stated.
\(1^{\text{st}}\) A1ft for a correct expression or correct ft their \(a\) with \(r = \frac{2}{3}\).
\(2^{\text{nd}}\) A1 for awrt 2180, even following 'minor inaccuracies'.
Condone missing brackets round the \(\frac{2}{3}\) for the marks in part (c).
Alternative:
M1 for adding 15 terms and \(1^{\text{st}}\) A1ft for adding the 15 terms that ft from their \(a\) and \(r = \frac{2}{3}\).
Part (d):
Answer
Marks
Guidance
M1 for use of correct sum to infinity formula with their \(a\). For this mark, if a value of \(r\) different from the given value is being used, M1 can still be allowed providing \(
r
< 1\).
**Part (a)**
$324r^3 = 96$ or $r^3 = \frac{96}{324}$ or $r^3 = \frac{8}{27}$ | M1 |
$r = \frac{2}{3}$ | A1cso |
**Part (b)**
$a\left(\frac{2}{3}\right)^3 = 324$ or $a\left(\frac{2}{3}\right)^3 = 96$ | M1, A1 | $a = ...$, $729$ |
**Part (c)**
$S_{15} = \frac{729\left[1 - \left(\frac{2}{3}\right)^{15}\right]}{1 - \frac{2}{3}}$, $= 2182.00...$ | M1A1ft, (3) | (AWRT 2180) |
**Part (d)**
$S_\infty = \frac{729}{1 - \frac{2}{3}} = 2187$ | M1, A1 |
**Guidance:**
**Part (a):**
M1 for forming an equation for $r^3$ based on 96 and 324 (e.g. $96r^3 = 324$ scores M1). The equation must involve multiplication/division rather than addition/subtraction.
Do not penalise solutions with working in decimals, providing these are correctly rounded or truncated to at least 2dp and the final answer 2/3 is seen.
**Alternative (verification):**
M1 Using $r^3 = \frac{8}{27}$ and multiplying 324 by this (or multiplying by $r = \frac{2}{3}$ three times).
A1 Obtaining 96 (cso). (A conclusion is not required.)
$324 \times \left(\frac{2}{3}\right)^3 = 96$ (no real evidence of calculation) is not quite enough and scores M1 A0.
**Part (b):**
M1 for the use of a correct formula or for 'working back' by dividing by $\frac{2}{3}$ (or by their $r$) twice from 324 (or 5 times from 96).
Exceptionally, allow M1 also for using $ar^3 = 324$ or $ar^6 = 96$ instead of $ar^2 = 324$ or $ar^3 = 96$, or for dividing by $r$ three times from 324 (or 6 times from 96)... but no other exceptions are allowed.
**Part (c):**
M1 for use of sum to 15 terms formula with values of $a$ and $r$. If the wrong power is used, e.g. 14, the M mark is scored only if the correct sum formula is stated.
$1^{\text{st}}$ A1ft for a correct expression or correct ft their $a$ with $r = \frac{2}{3}$.
$2^{\text{nd}}$ A1 for awrt 2180, even following 'minor inaccuracies'.
Condone missing brackets round the $\frac{2}{3}$ for the marks in part (c).
**Alternative:**
M1 for adding 15 terms and $1^{\text{st}}$ A1ft for adding the 15 terms that ft from their $a$ and $r = \frac{2}{3}$.
**Part (d):**
M1 for use of correct sum to infinity formula with their $a$. For this mark, if a value of $r$ different from the given value is being used, M1 can still be allowed providing $|r| < 1$.
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\begin{enumerate}
\item The third term of a geometric sequence is 324 and the sixth term is 96\\
(a) Show that the common ratio of the sequence is $\frac { 2 } { 3 }$\\
(b) Find the first term of the sequence.\\
(c) Find the sum of the first 15 terms of the sequence.\\
(d) Find the sum to infinity of the sequence.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2009 Q5 [9]}}