Prove that if $a$ and $b$ are even integers, then $a + b$ is even.
Question HKC78-AASL-PRF-01
[Maximum mark: 4]
Prove that if $a$ and $b$ are even integers, then $a + b$ is even.
Mark Scheme
Let the even integers be $2a$ and $2b$, where $a$ and $b$ are integers. [1]
Their sum is $2a + 2b = 2(a + b)$. [2]
$a + b$ is an integer, so the sum is even. [1]
Total: 4 marks
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Question Q1
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Mark Scheme
Step 1: Identify given information.
$u_1 = 5$, $d = 3$, $n = 10$.
Step 2: Use formula $u_n = u_1 + (n-1)d$.
$u_{10} = 5 + (10-1)3$
Step 3: Calculate.
$u_{10} = 5 + (9)3 = 5 + 27 = 32$.
Answer: 32
A theatre has $18$ seats in the first row. Each later row has $3$ more seats than the row in front of it. Find the number of seats in row $11$, and the total number of seats in the first $11$ rows.
Question HKC78-AASL-SEQ-01
[Maximum mark: 5]
A theatre has $18$ seats in the first row. Each later row has $3$ more seats than the row in front of it. Find the number of seats in row $11$, and the total number of seats in the first $11$ rows.
Mark Scheme
This is arithmetic with $u_{1} = 18$ and $d = 3$.
Row $11$: $u_{11} = 18 + 10 cdot 3 = 48$. [2]
Total seats: $S_{11} = dfrac{11}{2}(18 + 48) = dfrac{11}{2} cdot 66 = 363$. [3]
Total: 5 marks
A geometric sequence has $u_{1} = 8$ and $u_{4} = 27$. The common ratio is positive. Find the common ratio and the sixth term.
Question HKC78-AASL-SEQ-02
[Maximum mark: 5]
A geometric sequence has $u_{1} = 8$ and $u_{4} = 27$. The common ratio is positive. Find the common ratio and the sixth term.
Mark Scheme
$u_{4} = 8r^{3} = 27$, so $r^{3} = dfrac{27}{8}$ and $r = dfrac{3}{2}$. [3]
$u_{6} = 8left(dfrac{3}{2}right)^{5} = 8 cdot dfrac{243}{32} = dfrac{243}{4}$. [2]
Total: 5 marks
Solve $\log_{3}(x – 1) + \log_{3}(x + 1) = 2$. Give the exact solution and the restriction that selects it.
Question HKC78-AASL-LOG-02
[Maximum mark: 6]
Solve $\log_{3}(x – 1) + \log_{3}(x + 1) = 2$. Give the exact solution and the restriction that selects it.
Mark Scheme
Both logarithms are defined only when $x – 1 > 0$ and $x + 1 > 0$, so $x > 1$. [1]
$log_{3}big((x-1)(x+1)big) = 2$, hence $log_{3}(x^{2} – 1) = 2$. [2]
$x^{2} – 1 = 3^{2} = 9$, so $x^{2} = 10$ and $x = pmsqrt{10}$. [2]
Only $x = sqrt{10}$ satisfies $x > 1$. [1]
Total: 6 marks
Prove that the square of an even integer is divisible by $4$.
Question HKC78-AASL-PRF-02
[Maximum mark: 4]
Prove that the square of an even integer is divisible by $4$.
Mark Scheme
Let the even integer be $n = 2k$, where $k$ is an integer. [1]
$n^{2} = 4k^{2}$. [2]
$k^{2}$ is an integer, so $4$ divides $n^{2}$. [1]
Total: 4 marks
For the sequence in Q1a, find the sum of the first 20 terms.
Question Q2
For the sequence in Q1a, find the sum of the first 20 terms.
Mark Scheme
Step 1: Identify given information.
$u_1 = 5$, $d = 3$, $n = 20$.
Step 2: Use formula $S_n = frac{n}{2}(2u_1 + (n-1)d)$.
$S_{20} = frac{20}{2}(2(5) + (20-1)3)$
Step 3: Calculate.
$S_{20} = 10(10 + (19)3) = 10(10 + 57) = 10(67) = 670$.
Answer: 670
The sum of the first $n$ terms of an arithmetic series is $S_{n} = 2n^{2} + n$. Find the first term and the common difference.
Question HKC78-AASL-SEQ-03
[Maximum mark: 5]
The sum of the first $n$ terms of an arithmetic series is $S_{n} = 2n^{2} + n$. Find the first term and the common difference.
Mark Scheme
$u_{1} = S_{1} = 2 + 1 = 3$. [1]
$S_{2} = 2(4) + 2 = 10$, so $u_{2} = S_{2} – S_{1} = 7$. [2]
Common difference $d = 7 – 3 = 4$. [1]
In general, $u_{n} = S_{n} – S_{n-1} = 4n – 1$ for $n ge 2$, which has first term $3$ and difference $4$. [1]
Total: 5 marks
Express $\log_{2} 20 – \log_{2} 5$ as an integer.
Question HKC78-AASL-LOG-03
[Maximum mark: 3]
Express $\log_{2} 20 – \log_{2} 5$ as an integer.
Mark Scheme
$log_{2} 20 – log_{2} 5 = log_{2}left(dfrac{20}{5}right) = log_{2} 4$. [2]
$log_{2} 4 = 2$. [1]
Total: 3 marks
