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
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
Find the coefficient of $x^{2}$ in the expansion of $(3 – x)^{5}$.
Question HKC78-AASL-BIN-01
[Maximum mark: 3]
Find the coefficient of $x^{2}$ in the expansion of $(3 – x)^{5}$.
Mark Scheme
The $x^{2}$ term is $dbinom{5}{2}(3)^{3}(-x)^{2} = 10 cdot 27 cdot x^{2} = 270x^{2}$. [2]
The coefficient is $270$. [1]
Total: 3 marks
The third term in the expansion of $(1 + kx)^{6}$, in ascending powers of $x$, is $60x^{2}$. Given that $k > 0$, find $k$.
Question HKC78-AASL-BIN-02
[Maximum mark: 4]
The third term in the expansion of $(1 + kx)^{6}$, in ascending powers of $x$, is $60x^{2}$. Given that $k > 0$, find $k$.
Mark Scheme
In ascending powers, the third term is the $x^{2}$ term: $dbinom{6}{2}(kx)^{2} = 15k^{2}x^{2}$. [2]
$15k^{2} = 60$, so $k^{2} = 4$. [1]
$k > 0$, hence $k = 2$. [1]
Total: 4 marks
