A salary starts at $\$28000$ in year $1$ and increases by $\$1200$ at the start of each later year. Find the salary in year $6$, and the total earned over the first $6$ years.
Question HKC78-AISL-SEQ-01
[Maximum mark: 5]
A salary starts at $\$28000$ in year $1$ and increases by $\$1200$ at the start of each later year. Find the salary in year $6$, and the total earned over the first $6$ years.
Mark Scheme
This is arithmetic with $u_{1} = 28000$ and $d = 1200$.
Year $6$: $u_{6} = 28000 + 5 cdot 1200 = 34000$. [2]
Total: $S_{6} = dfrac{6}{2}(28000 + 34000) = 3 cdot 62000 = 186000$. [3]
Total: 5 marks
A population is modelled by $u_{n} = 500(1.04)^{n-1}$, where $u_{n}$ is the population at the start of year $n$.
(a) Find the population at the start of year $5$, to the nearest whole number.
(b) Find the smallest $n$ such that $u_{n} > 700$.
Question HKC78-AISL-SEQ-02
[Maximum mark: 6]
A population is modelled by $u_{n} = 500(1.04)^{n-1}$, where $u_{n}$ is the population at the start of year $n$.
(a) Find the population at the start of year $5$, to the nearest whole number.
(b) Find the smallest $n$ such that $u_{n} > 700$.
Mark Scheme
(a) $u_{5} = 500(1.04)^{4} = 584.929ldots$, which rounds to $585$. [2]
(b) $500(1.04)^{n-1} > 700$, so $(1.04)^{n-1} > 1.4$. [1]
$n – 1 > dfrac{ln 1.4}{ln 1.04} approx 8.579$, so $n – 1 ge 9$ and $n ge 10$. [2]
Check: $u_{9} = 500(1.04)^{8} approx 684.3$ and $u_{10} = 500(1.04)^{9} approx 711.7$. The smallest such $n$ is $10$. [1]
Total: 6 marks
