(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres. (i) How many...
Question 1 Report
(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres.
(i) How many days will it take the jogger to reach a distance of 15km in training?
(ii) Calculate the total distance he would have run in the training.
(b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.
Answer Details
(a) Daily distances form an AP: first term \(a=500\text{m}\), common difference \(d=250\text{m}\).
(i) The daily run reaches \(15\text{km}=15000\text{m}\) when the \(n\)th term equals \(15000\):
\[a+(n-1)d=15000\Rightarrow 500+250(n-1)=15000\]
\[250(n-1)=14500\Rightarrow n-1=58\Rightarrow n=59\text{ days}\]
(ii) Total distance is the sum of 59 terms:
\[S_{59}=\frac{n}{2}(a+l)=\frac{59}{2}(500+15000)=\frac{59}{2}(15500)=456250\text{ m}=456.25\text{ km}\]
(b) GP: second term \(ar=-3\); sum to infinity \(\dfrac{a}{1-r}=\dfrac{25}{2}\).
From \(ar=-3\), \(a=-\dfrac{3}{r}\). Substitute:
\[\frac{-3/r}{1-r}=\frac{25}{2}\Rightarrow -6=25r(1-r)\Rightarrow 25r^{2}-25r-6=0\]
\[r=\frac{25\pm\sqrt{625+600}}{50}=\frac{25\pm35}{50}\Rightarrow r=\tfrac65\ \text{or}\ r=-\tfrac15\]
A sum to infinity requires \(|r|<1\), so \(\boxed{r=-\tfrac15}\) (reject \(\tfrac65\)).
Download The App On Google Playstore
Everything you need to excel in your exams
