Step-by-Step Solution
Key Concept: Case study on arithmetic progressions.
(a) How many rows are there in the stack? [1 Mark]
$a=30,d=-1,l=1$. $1=30+(n-1)(-1)\Rightarrow n-1=29\Rightarrow n=30$. There are $30$ rows. [1.0 Mark]
(b) How many pipes are in the 12th row from the bottom? [1 Mark]
$a_{12}=30+11(-1)=30-11=19$ pipes. [1.0 Mark]
(c) Find the total number of pipes used in the stack. [1 Mark]
$S_{30}=\dfrac{30}{2}(30+1)=15\times31=465$ pipes. [1.0 Mark]
(d) If the company only had 300 pipes available and stacked as many complete rows as possible from the bottom (starting with 30 pipes and decreasing by 1 each row), how many complete rows could they build? [1 Mark]
$S_n=\dfrac n2[2(30)+(n-1)(-1)]\leq300\Rightarrow\dfrac n2[60-n+1]\leq300\Rightarrow n(61-n)\leq600$. [0.5 Mark]
$61n-n^2\leq600\Rightarrow n^2-61n+600\geq0$. Solving $n^2-61n+600=0$: $n=\dfrac{61\pm\sqrt{3721-2400}}{2}=\dfrac{61\pm\sqrt{1321}}{2}\approx\dfrac{61\pm36.3}{2}$, giving $n\approx12.3$ or $n\approx48.7$. [0.5 Mark]
Correct Answer: $a=30,d=-1,l=1$. $1=30+(n-1)(-1)\Rightarrow n-1=29\Rightarrow n=30$. There are $30$ rows. [1.0 Mark] | $a_{12}=30+11(-1)=30-11=19$ pipes. [1.0 Mark] | $S_{30}=\dfrac{30}{2}(30+1)=15\times31=465$ pipes. [1.0 Mark] | $S_n=\dfrac n2[2(30)+(n-1)(-1)]\leq300\Rightarrow\dfrac n2[60-n+1]\leq300\Rightarrow n(61-n)\leq600$. [0.5 Mark]