Standard +0.3 This is a straightforward geometric progression question requiring recall of standard formulas (second term = ar, sum to infinity = a/(1-r)) and solving simultaneous equations. The algebra is routine (substitution leading to a quadratic), making it slightly easier than average for A-level.
The second term of a geometric progression is 24. The sum to infinity of this progression is 150. Write down two equations in \(a\) and \(r\), where \(a\) is the first term and \(r\) is the common ratio. Solve your equations to find the possible values of \(a\) and \(r\). [5]
eg subst. of \(a = 150(1-r)\) or \(r = \frac{150-a}{150}\) in (i); alternatively, subst. of \(a = \frac{24}{r}\) or \(r = \frac{24}{a}\) in (ii)
\(r = 0.8\) or \(0.2\)
A1
if M0, B1 for both values of \(r\) and B1 for both values of \(a\), or B1 for each pair of correct values; NB \(150r^2 - 150r + 24 = [0]\) and \(a^2 - 150a +3600 = [0]\)
\(a = 30\) or \(a = 120\)
A1
A0 if wrongly attributed; A0 if wrongly attributed
Total: [5]
$ar = 24$ (i) | B1* | allow $ar^{-1} = 24$
$\frac{a}{1-r} = 150$ (ii) | B1* |
correct substitution to eliminate one unknown | M1dep* | eg subst. of $a = 150(1-r)$ or $r = \frac{150-a}{150}$ in (i); alternatively, subst. of $a = \frac{24}{r}$ or $r = \frac{24}{a}$ in (ii)
$r = 0.8$ or $0.2$ | A1 | if M0, B1 for both values of $r$ and B1 for both values of $a$, or B1 for each pair of correct values; NB $150r^2 - 150r + 24 = [0]$ and $a^2 - 150a +3600 = [0]$
$a = 30$ or $a = 120$ | A1 | A0 if wrongly attributed; A0 if wrongly attributed
**Total: [5]**
---
The second term of a geometric progression is 24. The sum to infinity of this progression is 150. Write down two equations in $a$ and $r$, where $a$ is the first term and $r$ is the common ratio. Solve your equations to find the possible values of $a$ and $r$. [5]
\hfill \mbox{\textit{OCR MEI C2 2014 Q7 [5]}}