Standard +0.3 This is a straightforward geometric progression problem requiring students to set up two equations (ar = 54 and a/(1-r) = 243) and solve for a and r, then calculate the tenth term. While it involves algebraic manipulation and the constraint r > 1/2, it's a standard textbook exercise with well-practiced techniques and no novel insight required, making it slightly easier than average.
6 The second term of a geometric progression is 54 and the sum to infinity of the progression is 243 . The common ratio is greater than \(\frac { 1 } { 2 }\).
Find the tenth term, giving your answer in exact form.
\(ar=54\) and \(\dfrac{a\text{ or }\textit{their}\,a}{1-r}=243\)
B1
SOI
\(\frac{54}{r}=243(1-r)\) leading to \(243r^2-243r+54[=0]\) \([9r^2-9r+2=0]\) OR \(a^2-243a+13122[=0]\)
*M1
Forming a 3-term quadratic expression in \(r\) or \(a\) using *their* 2nd term and \(S_\infty\). Allow \(\pm\) sign errors
\(k(3r-2)(3r-1)[=0]\) OR \((a-81)(a-162)[=0]\)
DM1
Solving *their* 3-term quadratic using factorisation, formula or completing the square. If factorising, factors must expand to give \(\pm\textit{their}\) coefficient of \(r^2\)
\(54\div\left(\textit{their}\,\frac{2}{3}\right)=a\) OR \(54\div(\textit{their}\,81)=r\)
DM1
May be implied by final answer
Tenth term \(=\dfrac{512}{243}\) \(\left[\text{OR }81\times\left(\frac{2}{3}\right)^9\text{ OR }54\times\left(\frac{2}{3}\right)^8\right]\)
A1
OE. Must be exact. Special case: If B1M1DM0DM1 scored then SC B1 can be awarded for the correct final answer
Total
5
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ar=54$ and $\dfrac{a\text{ or }\textit{their}\,a}{1-r}=243$ | B1 | SOI |
| $\frac{54}{r}=243(1-r)$ leading to $243r^2-243r+54[=0]$ $[9r^2-9r+2=0]$ OR $a^2-243a+13122[=0]$ | *M1 | Forming a 3-term quadratic expression in $r$ or $a$ using *their* 2nd term and $S_\infty$. Allow $\pm$ sign errors |
| $k(3r-2)(3r-1)[=0]$ OR $(a-81)(a-162)[=0]$ | DM1 | Solving *their* 3-term quadratic using factorisation, formula or completing the square. If factorising, factors must expand to give $\pm\textit{their}$ coefficient of $r^2$ |
| $54\div\left(\textit{their}\,\frac{2}{3}\right)=a$ OR $54\div(\textit{their}\,81)=r$ | DM1 | May be implied by final answer |
| Tenth term $=\dfrac{512}{243}$ $\left[\text{OR }81\times\left(\frac{2}{3}\right)^9\text{ OR }54\times\left(\frac{2}{3}\right)^8\right]$ | A1 | OE. Must be exact. **Special case:** If B1M1DM0DM1 scored then SC B1 can be awarded for the correct final answer |
| **Total** | **5** | |
6 The second term of a geometric progression is 54 and the sum to infinity of the progression is 243 . The common ratio is greater than $\frac { 1 } { 2 }$.
Find the tenth term, giving your answer in exact form.\\
\hfill \mbox{\textit{CAIE P1 2021 Q6 [5]}}