Edexcel P2 2021 October — Question 7 10 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2021
SessionOctober
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeFind year when threshold exceeded
DifficultyModerate -0.3 This is a multi-part question testing standard geometric sequence formulas with straightforward application. Part (i) requires logarithms to solve an inequality (routine A-level technique), part (ii) involves algebraic manipulation of sum to infinity and term formulas leading to a given quadratic, then solving it and applying the convergence condition |r|<1. All steps are textbook exercises 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|<11.06g Equations with exponentials: solve a^x = b
7. (i) A geometric sequence has first term 4 and common ratio 6 Given that the \(n ^ { \text {th } }\) term is greater than \(10 ^ { 100 }\), find the minimum possible value of \(n\).
(ii) A different geometric sequence has first term \(a\) and common ratio \(r\). Given that
  • the second term of the sequence is - 6
  • the sum to infinity of the series is 25
    1. show that
$$25 r ^ { 2 } - 25 r - 6 = 0$$
  • Write down the solutions of $$25 r ^ { 2 } - 25 r - 6 = 0$$ Hence,
  • state the value of \(r\), giving a reason for your answer,
  • find the sum of the first 4 terms of the series. \includegraphics[max width=\textwidth, alt={}, center]{124ee19f-8a49-42df-9f4b-5a1cc2139be9-23_70_37_2617_1914}
    • Question 7:
    Part (i):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    States or implies \(4 \times 6^{n-1} > 10^{100}\)B1 Can use = or any inequality
    \(4 \times 6^{n-1} > 10^{100} \Rightarrow \log 4 + (n-1)\log 6 > 100\log 10\)M1 Must correctly take logs to remove powers; allow slips in rearranging
    \(n = 129\)A1
    Part (ii)(a):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    States \(ar = -6\) and \(\frac{a}{1-r} = 25\)B1 Both equations required
    Combines to form equation in \(r\): \(\frac{-6}{r(1-r)} = 25\)M1
    \(\Rightarrow -6 = 25r(1-r) \Rightarrow 25r^2 - 25r - 6 = 0\)A1* At least one correct simplified intermediate line, no errors; answer given so proof required
    Part (ii)(b):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    \(r = \frac{6}{5}, -\frac{1}{5}\)B1 Award when seen even if not in part (b)
    Part (ii)(c):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    \(r = -\frac{1}{5}\) as \(r < 1\) (for \(S_\infty\) to exist)
    Part (ii)(d):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Attempts \(S_4 = \frac{a(1-r^n)}{1-r}\) with \(n=4\), \(r = \text{their}(c)\), \(a = \frac{-6}{\text{their}(c)}\)M1 If no attempt at (c), accept either value from (b) for \(r\); if correct formula quoted allow slips in substitution
    \(S_4 = \frac{30\left(1-\left(-\frac{1}{5}\right)^4\right)}{1-\left(-\frac{1}{5}\right)} = 24.96\)A1 Accept \(\frac{624}{25}\) 
    ## Question 7:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States or implies $4 \times 6^{n-1} > 10^{100}$ | B1 | Can use = or any inequality |
| $4 \times 6^{n-1} > 10^{100} \Rightarrow \log 4 + (n-1)\log 6 > 100\log 10$ | M1 | Must correctly take logs to remove powers; allow slips in rearranging |
| $n = 129$ | A1 | |

### Part (ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States $ar = -6$ and $\frac{a}{1-r} = 25$ | B1 | Both equations required |
| Combines to form equation in $r$: $\frac{-6}{r(1-r)} = 25$ | M1 | |
| $\Rightarrow -6 = 25r(1-r) \Rightarrow 25r^2 - 25r - 6 = 0$ | A1* | At least one correct simplified intermediate line, no errors; answer given so proof required |

### Part (ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{6}{5}, -\frac{1}{5}$ | B1 | Award when seen even if not in part (b) |

### Part (ii)(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = -\frac{1}{5}$ as $|r| < 1$ (for $S_\infty$ to exist) | B1 | Requires minimal reason; reject $\frac{6}{5}$ since all terms negative so cannot give positive sum; do not accept just "GS is convergent" |

### Part (ii)(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $S_4 = \frac{a(1-r^n)}{1-r}$ with $n=4$, $r = \text{their}(c)$, $a = \frac{-6}{\text{their}(c)}$ | M1 | If no attempt at (c), accept either value from (b) for $r$; if correct formula quoted allow slips in substitution |
| $S_4 = \frac{30\left(1-\left(-\frac{1}{5}\right)^4\right)}{1-\left(-\frac{1}{5}\right)} = 24.96$ | A1 | Accept $\frac{624}{25}$ |

---
    7. (i) A geometric sequence has first term 4 and common ratio 6

Given that the $n ^ { \text {th } }$ term is greater than $10 ^ { 100 }$, find the minimum possible value of $n$.\\
(ii) A different geometric sequence has first term $a$ and common ratio $r$.

Given that

\begin{itemize}
  \item the second term of the sequence is - 6
  \item the sum to infinity of the series is 25
\begin{enumerate}[label=(\alph*)]
\item show that
\end{itemize}

$$25 r ^ { 2 } - 25 r - 6 = 0$$
\item Write down the solutions of

$$25 r ^ { 2 } - 25 r - 6 = 0$$

Hence,
\item state the value of $r$, giving a reason for your answer,
\item find the sum of the first 4 terms of the series.

\includegraphics[max width=\textwidth, alt={}, center]{124ee19f-8a49-42df-9f4b-5a1cc2139be9-23_70_37_2617_1914}
\end{enumerate}

\hfill \mbox{\textit{Edexcel P2 2021 Q7 [10]}}
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