| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Topic | Geometric Sequences and Series |
| Type | Shared terms between AP and GP |
| Difficulty | Challenging +1.2 This question requires setting up equations from given conditions about AP and GP terms, algebraic manipulation to find the common ratio (part i), and proving a divisibility/membership result (part iii(b)). While it involves multiple steps and some algebraic insight, the techniques are standard for A-level: solving simultaneous equations, using nth term formulas, and showing one sequence is contained in another. The proof in (iii)(b) requires recognizing that g_n = 5·3^(n-1) and a_n = 5 + 10(n-1), then showing 3^(n-1) ≡ 1 (mod 2) makes g_n always odd and thus expressible as a_k for some k. This is moderately challenging but within reach of strong A-level students, placing it above average difficulty. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
**Question 10: Arithmetic and geometric sequences**
**(i)**
- $a_1 = g_1 = a$ **M1** Attempt at least one equation linking $a$, $d$, $r$
- $a + d = ar$, $a + 4d = ar^2$ **A1** Obtain two correct equations
- $a + 4(ar - a) = ar^2$, so $ar^2 - 4ar + a = 0$ **M1** Eliminate $d$ from equations
- $a(r^2 - 4r + 3) = 0$, $a(r-1)(r-3) = 0$ **M1** Attempt value for $r$
- $a = 0,\ r = 1,\ r = 3$ **A1** Obtain $r = 3$ (ignore second solution if given)
- $a_1 \neq a_2$ so $r = 3$ **B1** Justify $r = 3$ as only valid solution; could be stating that $r = 1$ or $d = 0$ does not give a valid solution
**(ii)**
- $d = 2a_1$ **B1** Correct expression for $d$ (B0 if $a$ not $a_1$)
**(iii)(a)**
- Geometric: $5, 15, 45$ **B1** Correct three terms for geometric sequence
- Arithmetic: $5, 15, 25$ **B1** Correct three terms for arithmetic sequence
**(iii)(b)**
- e.g. The product of 3 and an integer ending $\ldots 5$ ends with $\ldots 5$, so the terms of the geometric sequence all end $\ldots 5$ **M1** Consider terms of geometric sequence. Could be worded argument, or could justify with $5\times 3^{n-1}$ (allow $5\times 3^n$ as general term)
- The arithmetic sequence covers *all* odd multiples of 5 **M1** Consider terms of arithmetic sequence. No further evidence required, but must make it clear that *all* terms are contained in AP
- So the terms of the geometric sequence are all in the arithmetic sequence. **A1** Conclude appropriately
Alternative approach: setting up GP as $5\times 3^{n-1}$ and AP as $5(2n-1)$, and then comparing $3^{n-1}$ and $2n-1$:
- **M1** justify $3^{n-1}$ as always odd
- **M1** $2n-1$ as *all* odd numbers
- **A1** hence terms of GP are all in the AP
**Total: 11 marks**10 An arithmetic sequence and a geometric sequence have $n$th terms $a _ { n }$ and $g _ { n }$ respectively, where $n = 1,2,3 , \ldots$. It is given that $a _ { 1 } = g _ { 1 } , a _ { 2 } = g _ { 2 } , a _ { 5 } = g _ { 3 } , a _ { 1 } \neq a _ { 2 }$ and $a _ { 1 } \neq 0$.\\
(i) Show that the common ratio of the geometric sequence is 3 .\\
(ii) Find the common difference of the arithmetic sequence in terms of $a _ { 1 }$.\\
(iii) Let $a _ { 1 } = g _ { 1 } = 5$.
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms of both sequences.
\item Show that every term of the geometric sequence is also a term of the arithmetic sequence.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2017 Q10 [11]}}