| Exam Board | SPS |
|---|---|
| Module | SPS SM Pure (SPS SM Pure) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Moderate -0.3 This is a straightforward geometric sequence question requiring standard formulas (nth term = ar^(n-1), sum formula). Part (a) involves dividing two terms to eliminate 'a' and finding r, part (b) is direct substitution, and part (c) requires solving an inequality using the sum formula. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
Solutions relying entirely on calculator technology or numerical methods are not acceptable.
A geometric sequence has first term $a$ and common ratio $r$, where $r > 0$\\
Given that
\begin{itemize}
\item the 3rd term is 20
\item the 5th term is 12.8\\
(a) show that $r = 0.8$\\
(b) Hence find the value of $a$.
\end{itemize}
Given that the sum of the first $n$ terms of this sequence is greater than 156\\
(c) find the smallest possible value of $n$.\\
\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q12 [7]}}