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
Solve the equation: $2^{x} = 32$
Question Q3
Solve the equation: $2^{x} = 32$
Mark Scheme
We know that $32 = 2^5$.
So, $2^x = 2^5$.
Therefore, $x=5$.
Given that $\log_{a} 2 = x$ and $\log_{a} 3 = y$, express $\log_{a} 12$ in terms of $x$ and $y$.
Question Q4
Given that $\log_{a} 2 = x$ and $\log_{a} 3 = y$, express $\log_{a} 12$ in terms of $x$ and $y$.
Mark Scheme
Use log laws: $log_a(mn) = log_a m + log_a n$ and $log_a(m^k) = klog_a m$.
$log_a 12 = log_a (4 imes 3) = log_a (2^2 imes 3)$
$= log_a(2^2) + log_a 3$
$= 2log_a 2 + log_a 3$
Substitute given values:
$= 2x + y$.
Solve the inequality: $|2x – 5| < 7$
Question Q5
Solve the inequality: $|2x – 5| < 7$
Mark Scheme
For $|A| < k$, we have $-k < A < k$.
$-7 < 2x - 5 < 7$
Add 5 to all parts:
$-2 < 2x < 12$
Divide by 2:
$-1 < x < 6$
Expand $(2x + 3)^5$ using the binomial theorem.
Question Q6
Expand $(2x + 3)^5$ using the binomial theorem.
Mark Scheme
Using Binomial Theorem row 5: 1, 5, 10, 10, 5, 1.
$(2x+3)^5 = 1(2x)^5(3)^0 + 5(2x)^4(3)^1 + 10(2x)^3(3)^2 + 10(2x)^2(3)^3 + 5(2x)^1(3)^4 + 1(2x)^0(3)^5$
$= 32x^5 + 5(16x^4)(3) + 10(8x^3)(9) + 10(4x^2)(27) + 5(2x)(81) + 243$
$= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243$
Find the coefficient of the $x^3$ term in the expansion of $(x – 2)^{10}$.
Question Q7
Find the coefficient of the $x^3$ term in the expansion of $(x – 2)^{10}$.
Mark Scheme
General term is $inom{n}{r} a^{n-r} b^r$. Here $n=10, a=x, b=-2$.
We want $x^3$, so we need $n-r = 3$, which means $10-r=3$, so $r=7$.
Term = $inom{10}{7} (x)^3 (-2)^7$
$inom{10}{7} = frac{10 imes 9 imes 8}{3 imes 2 imes 1} = 120$.
$(-2)^7 = -128$.
Coefficient = $120 imes (-128) = -15360$.
A complex number is given by $z = 3 + 4i$. Find the modulus and argument of z.
Question Q8
A complex number is given by $z = 3 + 4i$. Find the modulus and argument of z.
Mark Scheme
$z = 3 + 4i$
Modulus $|z| = sqrt{3^2 + 4^2} = sqrt{9+16} = sqrt{25} = 5$.
Argument is in Quadrant 1. $ an heta = frac{4}{3}$.
$ heta = arctan(frac{4}{3}) approx 0.927$ radians (or $53.1^circ$).
Given $z_1 = 2 \operatorname{cis} \frac{\pi}{3}$ and $z_2 = 4 \operatorname{cis} \frac{\pi}{6}$, find $z_1 z_2$ in the form $a+bi$.
Question Q9
Given $z_1 = 2 \operatorname{cis} \frac{\pi}{3}$ and $z_2 = 4 \operatorname{cis} \frac{\pi}{6}$, find $z_1 z_2$ in the form $a+bi$.
Mark Scheme
Using polar form multiplication: $z_1 z_2 = r_1 r_2 ext{cis}( heta_1 + heta_2)$.
$z_1 z_2 = (2)(4) ext{cis}(frac{pi}{3} + frac{pi}{6})$
$= 8 ext{cis}(frac{2pi}{6} + frac{pi}{6}) = 8 ext{cis}(frac{pi}{2})$
$= 8(cos frac{pi}{2} + i sin frac{pi}{2})$
$= 8(0 + i(1)) = 8i$.
Form $a+bi$ is $0 + 8i$.
