2022 HKDSE Mathematics Answers & Marking Scheme

(a) Since \(P(x)\) is divisible by \(3x - 1\), by the factor theorem, we have \(P(1/3) = 0\). So, \(3(1/3)^3 - k(1/3)^2 - 13(1/3) + 4 = 0\). This gives \(1/9 - k/9 - 13/3 + 4 = 0\). Multiplying by 9, we get \(1 - k - 39 + 36 = 0\), which simplifies to \(-k - 2 = 0\), hence \(k = -2\). (b) Now, \(P(x) = 3x^3 + 2x^2 - 13x + 4\). The remainder when \(P(x)\) is divided by \(x + 2\) is given by \(P(-2) = 3(-2)^3 + 2(-2)^2 - 13(-2) + 4 = -24 + 8 + 26 + 4 = 14\).

For (a): 1M for applying factor theorem \(P(1/3) = 0\), 1A for \(k = -2\). For (b): 1M for substituting \(x = -2\), 0.88A for remainder = 14.

(a) Let \(r\) be the radius of \(C\). Since \(C\) passes through \(A(2, 6)\) and has center at \(O(0,0)\), \(r^2 = (2 - 0)^2 + (6 - 0)^2 = 4 + 36 = 40\). Thus, the equation of \(C\) is \(x^2 + y^2 = 40\). (b) The slope of \(OA\) is \((6 - 0)/(2 - 0) = 3\). Since the tangent at \(A\) is perpendicular to \(OA\), the slope of the tangent is \(-1/3\). The equation of the tangent is \(y - 6 = -1/3(x - 2)\), which simplifies to \(3y - 18 = -x + 2\), and thus \(x + 3y - 20 = 0\).

For (a): 1M for finding radius squared, 0.88A for \(x^2 + y^2 = 40\). For (b): 1M for finding slope of tangent, 1A for \(x + 3y - 20 = 0\).

(a) Since the mean height is \(161\text{ cm}\), we have \((155 + 158 + 160 + 162 + 165 + x) / 6 = 161\). So, \(800 + x = 966\), which gives \(x = 166\). (b) The heights of the 6 students are \(155, 158, 160, 162, 165, 166\). The mean is \(161\). The standard deviation \(\sigma = \sqrt{\frac{(155-161)^2 + (158-161)^2 + (160-161)^2 + (162-161)^2 + (165-161)^2 + (166-161)^2}{6}} = \sqrt{\frac{36 + 9 + 1 + 1 + 16 + 25}{6}} = \sqrt{\frac{88}{6}} = \sqrt{\frac{44}{3}} \approx 3.83\text{ cm}\).

For (a): 1M for using mean equation, 0.88A for \(x = 166\). For (b): 1M for formula of standard deviation, 1A for \(3.83\text{ cm}\).

(a) By the cosine formula, \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos\angle ABC = 8^2 + 5^2 - 2(8)(5)\cos 120^\circ = 64 + 25 - 80(-0.5) = 89 + 40 = 129\). Thus, \(AC = \sqrt{129} \approx 11.4\text{ cm}\). (b) The area of \(\triangle ABC = \frac{1}{2}(AB)(BC)\sin\angle ABC = \frac{1}{2}(8)(5)\sin 120^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2\).

For (a): 1.38M for applying cosine formula, 1A for \(AC \approx 11.4\text{ cm}\). For (b): 1M for applying area formula, 0.5A for \(10\sqrt{3}\text{ cm}^2\).

(a) Let \(a\) be the first term and \(d\) be the common difference. We have \(a + 2d = 14\) and \(a + 6d = 30\). Subtracting the first equation from the second gives \(4d = 16\), so \(d = 4\). Substituting \(d = 4\) into the first equation, we get \(a + 8 = 14\), so \(a = 6\). (b) The sum of the first 20 terms is \(S_{20} = \frac{20}{2} [2a + (20-1)d] = 10 [2(6) + 19(4)] = 10 [12 + 76] = 10(88) = 880\).

For (a): 1M for setting up simultaneous equations, 1A for \(a = 6\) and \(d = 4\). For (b): 1M for applying arithmetic series sum formula, 0.88A for \(880\).

(a) Let \(z = \frac{k x^2}{\sqrt{y}}\), where \(k \neq 0\) is a constant. Substituting \(x = 3, y = 16, z = 18\), we get \(18 = \frac{k(3^2)}{\sqrt{16}} \Rightarrow 18 = \frac{9k}{4} \Rightarrow k = 8\). Thus, \(z = \frac{8x^2}{\sqrt{y}}\). (b) Let \(x'\) and \(y'\) be the new values. \(x' = 2x\) and \(y' = (1 - 75\%)y = 0.25y\). The new value of \(z\) is \(z' = \frac{8(x')^2}{\sqrt{y'}} = \frac{8(2x)^2}{\sqrt{0.25y}} = \frac{32x^2}{0.5\sqrt{y}} = \frac{64x^2}{\sqrt{y}} = 8 \left(\frac{8x^2}{\sqrt{y}}\right) = 8z\). The percentage change in \(z = \frac{8z - z}{z} \times 100\% = 700\%\).

For (a): 1M for variation equation, 0.88A for \(z = \frac{8x^2}{\sqrt{y}}\). For (b): 1M for expressing new \(z\) in terms of old \(z\), 1A for \(700\%\).

(a) For \(3x - 5 < 7x + 11\), we have \(-16 < 4x \Rightarrow x > -4\). For \(\frac{5 - 2x}{3} \ge x - 5\), we have \(5 - 2x \ge 3(x - 5) \Rightarrow 5 - 2x \ge 3x - 15 \Rightarrow 20 \ge 5x \Rightarrow x \le 4\). Combining the two results, we get the solution: \(-4 < x \le 4\). (b) The integers satisfying the inequality are \(-3, -2, -1, 0, 1, 2, 3, 4\). Thus, there are 8 integers.

For (a): 1M for solving \(3x - 5 < 7x + 11\), 1M for solving \(\frac{5 - 2x}{3} \ge x - 5\), 0.88A for \(-4 < x \le 4\). For (b): 1A for 8.

(a) From \(\log_2(x + 5) - \log_2(x - 1) = 2\), we have \(\log_2\left(\frac{x+5}{x-1}\right) = 2\). This implies \(\frac{x+5}{x-1} = 2^2 = 4\). Then \(x + 5 = 4(x - 1) \Rightarrow x + 5 = 4x - 4 \Rightarrow 3x = 9 \Rightarrow x = 3\). Checking shows \(x+5 > 0\) and \(x-1 > 0\) are satisfied. (b) Comparing \(\log_2(2^y + 5) - \log_2(2^y - 1) = 2\) with the equation in (a), we have \(2^y = x\). Since \(x = 3\), \(2^y = 3\), which gives \(y = \log_2 3\).

For (a): 1M for using quotient law, 1M for converting to algebraic equation, 0.38A for \(x = 3\). For (b): 1M for substitution \(2^y = x\), 0.5A for \(y = \log_2 3\).

(a) For the quadratic equation \(x^2 + 2kx + (3k + 4) = 0\) to have equal real roots, the discriminant must be zero. So, \(\Delta = (2k)^2 - 4(1)(3k+4) = 0 \Rightarrow 4k^2 - 12k - 16 = 0 \Rightarrow k^2 - 3k - 4 = 0\). Factoring gives \((k-4)(k+1) = 0\), which yields \(k = 4\) or \(k = -1\). (b) For the positive value of \(k\), we have \(k = 4\). The equation becomes \(x^2 + 8x + 16 = 0\), which is \((x + 4)^2 = 0\). Thus, \(x = -4\).

For (a): 1M for discriminant \(\Delta = 0\), 1M for solving the quadratic equation in \(k\), 0.88A for \(k = 4\) or \(k = -1\). For (b): 1A for \(x = -4\).

(a) Since \(P(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 2a + b = -5\) -- (1)
Since \(P(-1) = -12\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12 \implies -2 + a - b - 6 = -12 \implies a - b = -4\) -- (2)
From (2), we have \(b = a + 4\). Substituting into (1):
\(2a + (a + 4) = -5 \implies 3a = -9 \implies a = -3\).
Then \(b = -3 + 4 = 1\).

(b) If \(k = 12\), then \(Q(x) = 2x^3 - 3x^2 + x - 6 + 12 = 2x^3 - 3x^2 + x + 6\).
Note that \(Q(-1) = 2(-1)^3 - 3(-1)^2 + (-1) + 6 = -2 - 3 - 1 + 6 = 0\).
So \(x+1\) is a factor of \(Q(x)\).
By long division, \(Q(x) = (x+1)(2x^2 - 5x + 6)\).
For the equation \(2x^2 - 5x + 6 = 0\), the discriminant is:
\(\Delta = (-5)^2 - 4(2)(6) = 25 - 48 = -23 < 0\).
Thus, the quadratic equation \(2x^2 - 5x + 6 = 0\) has no real roots.
Therefore, \(Q(x) = 0\) has only one real root (which is \(x = -1\)).
Thus, the claim is disagreed.

(a) 1M for applying factor theorem to set up equation (1); 1M for applying remainder theorem to set up equation (2); 1A for \(a = -3\); 1A for \(b = 1\).
(b) 1M for factorizing \(Q(x)\) as \((x+1)(2x^2-5x+6)\); 1M for checking discriminant \(\Delta = -23 < 0\); 1A for correct explanation and conclusion.

(a) The equation of the circle \(C\) is \(x^2 + y^2 - 8x - 6y + 20 = 0\).
Center \(G = \left(-\frac{-8}{2}, -\frac{-6}{2}\right) = (4, 3)\).
Radius \(r = \sqrt{4^2 + 3^2 - 20} = \sqrt{16 + 9 - 20} = \sqrt{5}\).

(b)(i) Let the equation of \(L\) be \(y = mx \implies mx - y = 0\).
Since \(L\) is tangent to \(C\), the perpendicular distance from \(G(4, 3)\) to \(L\) is equal to \(r\):
\(\frac{|4m - 3|}{\sqrt{m^2 + 1}} = \sqrt{5}\)
\((4m - 3)^2 = 5(m^2 + 1)\)
\(16m^2 - 24m + 9 = 5m^2 + 5\)
\(11m^2 - 24m + 4 = 0\)
\((11m - 2)(m - 2) = 0\)
Since \(m > 1\), we have \(m = 2\).

(b)(ii) Substitute \(y = 2x\) into the circle equation:
\(x^2 + (2x)^2 - 8x - 6(2x) + 20 = 0\)
\(5x^2 - 20x + 20 = 0\)
\(x^2 - 4x + 4 = 0\)
\((x-2)^2 = 0 \implies x = 2\).
Substitute \(x = 2\) back into \(y = 2x\) to get \(y = 4\).
Thus, the coordinates of \(P\) are \((2, 4)\).

(a) 1A for center \((4,3)\); 1A for radius \(\sqrt{5}\).
(b)(i) 1M for setting up distance formula equal to radius; 1M for simplifying to quadratic in \(m\); 1A for \(m = 2\) (rejecting \(m = 2/11\) since \(m > 1\)).
(b)(ii) 1M for substituting \(y = 2x\) into circle equation; 1A for coordinates \((2, 4)\).

(a) Let \(u_i\) (\(i = 1, \dots, 8\)) be the 8 numbers.
Sum of the numbers \(= 8 \times 15 = 120\).
Since standard deviation \(\sigma = 4\), variance \(\sigma^2 = 16\).
Using the formula \(\sigma^2 = \frac{\sum u_i^2}{8} - \bar{u}^2\):
\(16 = \frac{\sum u_i^2}{8} - 15^2 \implies 16 = \frac{\sum u_i^2}{8} - 225\)
\(\frac{\sum u_i^2}{8} = 241 \implies \sum u_i^2 = 1928\).
So, the sum of squares is \(1928\).

(b) When \(x\) and \(y\) are added, the new mean \(\bar{u}_{\text{new}} = \frac{120 + (x + y)}{10} = \frac{120 + 30}{10} = 15\).
Since the new standard deviation is \(\sqrt{17.8}\), the new variance is \(17.8\).
Using the new variance formula:
\(17.8 = \frac{\sum u_i^2 + x^2 + y^2}{10} - \bar{u}_{\text{new}}^2\)
\(17.8 = \frac{1928 + x^2 + y^2}{10} - 225\)
\(242.8 = \frac{1928 + x^2 + y^2}{10}\)
\(2428 = 1928 + x^2 + y^2 \implies x^2 + y^2 = 500\).
Since \(x + y = 30 \implies y = 30 - x\):
\(x^2 + (30 - x)^2 = 500\)
\(x^2 + 900 - 60x + x^2 = 500\)
\(2x^2 - 60x + 400 = 0 \implies x^2 - 30x + 200 = 0\)
\((x-10)(x-20) = 0 \implies x = 10 \text{ or } x = 20\).
If \(x = 10\), then \(y = 20\); if \(x = 20\), then \(y = 10\).
Therefore, the values are \(10\) and \(20\).

(a) 1A for sum = 120; 1M for using the variance formula; 1A for sum of squares = 1928.
(b) 1M for finding the new mean is 15; 1M for setting up the new variance equation to obtain \(x^2 + y^2 = 500\); 1M for substituting \(y = 30 - x\) and forming a quadratic; 1A for \(x = 10, y = 20\) (or vice versa).

(a) In \(\triangle ABC\), by the Cosine Formula:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos\angle ABC\)
\(AC^2 = 10^2 + 6^2 - 2(10)(6)\cos 120^\circ\)
\(AC^2 = 100 + 36 - 120(-0.5)\)
\(AC^2 = 136 + 60 = 196\)
\(AC = 14\text{ cm}\).

(b) In \(\triangle ACD\), by the Sine Formula:
\(\frac{\sin\angle ADC}{AC} = \frac{\sin\angle CAD}{CD}\)
\(\frac{\sin\angle ADC}{14} = \frac{\sin 30^\circ}{8}\)
\(\sin\angle ADC = \frac{14 \sin 30^\circ}{8} = \frac{7}{8} = 0.875\).
Thus, the acute angle is \(\angle ADC = \sin^{-1}(0.875) \approx 61.045^\circ \approx 61.0^\circ\).
Or the obtuse angle is \(\angle ADC = 180^\circ - 61.045^\circ \approx 118.955^\circ \approx 119.0^\circ\).
Since both values yield valid triangles where \(AD > CD\) (or equivalently, \(\angle ACD > 30^\circ\)), both are acceptable.
Therefore, the two possible values of \(\angle ADC\) are \(61.0^\circ\) and \(119.0^\circ\).

(a) 1M for using cosine formula; 1M for substituting \(\cos 120^\circ = -0.5\); 1A for \(AC = 14\text{ cm}\) (must have units, or ignore if general standard).
(b) 1M for using sine formula; 1A for \(\sin\angle ADC = 0.875\); 1A for \(61.0^\circ\); 1A for \(119.0^\circ\).

(a) Let the first term be \(a\) and the common difference be \(d\).
\(a_3 = a + 2d = 15\) -- (1)
\(a_7 = a + 6d = 39\) -- (2)
Subtracting (1) from (2):
\(4d = 24 \implies d = 6\).
Substituting \(d = 6\) back into (1):
\(a + 12 = 15 \implies a = 3\).
So the first term is \(3\) and the common difference is \(6\).

(b)(i) The sum of the first \(n\) terms is \(T_n = \frac{n}{2}[2a + (n-1)d]\).
\(T_2 = a_1 + a_2 = 3 + 9 = 12\).
\(T_4 = T_2 + a_3 + a_4 = 12 + 15 + 21 = 48\).
For the geometric sequence:
\(g_1 = b = T_2 \implies b = 12\).
\(g_3 = b r^2 = T_4 \implies 12 r^2 = 48 \implies r^2 = 4\).
Since \(r > 0\), we have \(r = 2\).

(b)(ii) The sum of the first \(n\) terms of \(g_n\) is:
\(S_n = \frac{b(r^n - 1)}{r - 1} = \frac{12(2^n - 1)}{2 - 1} = 12(2^n - 1)\).
We require \(S_n > 10^6\):
\(12(2^n - 1) > 10^6 \implies 2^n - 1 > 83333.33\)
\(2^n > 83334.33\)
\(n \log 2 > \log 83334.33\)
\(n > \frac{\log 83334.33}{\log 2} \approx 16.35\).
Since \(n\) must be an integer, the least value of \(n\) is \(17\).

(a) 1M for setting up equations (1) and (2); 1A for \(a=3\) and \(d=6\).
(b)(i) 1M for calculating \(T_2\) and \(T_4\) correctly; 1A for \(b=12\) and \(r=2\).
(b)(ii) 1M for formulating the inequality \(12(2^n-1) > 10^6\); 1M for solving the inequality using logarithms; 1A for \(n = 17\).

(a) Since \(x-2\) is a factor, \(P(2) = 0\), which gives \(16 + 4p + 2q - 10 = 0 \implies 2p + q = -3\). Since the remainder of dividing by \(x+1\) is \(-9\), \(P(-1) = -9\), which gives \(-2 + p - q - 10 = -9 \implies p - q = 3\). Solving these equations simultaneously, we get \(p = 0\) and \(q = -3\). (b) Substituting these values, we have \(P(x) = 2x^3 - 3x - 10 = 0\). Since \(x=2\) is a root, by polynomial division, \(2x^3 - 3x - 10 = (x-2)(2x^2 + 4x + 5) = 0\). For the quadratic equation \(2x^2 + 4x + 5 = 0\), the discriminant is \(\Delta = 4^2 - 4(2)(5) = 16 - 40 = -24 < 0\). Since \(\Delta < 0\), the quadratic equation has no real roots. Thus, \(x = 2\) is the only real root of \(P(x) = 0\).

(a) 1M for using Factor Theorem P(2)=0, 1M for using Remainder Theorem P(-1)=-9, 1A for both p=0 and q=-3. (b) 1M for polynomial division to get 2x^2 + 4x + 5, 1A for getting x=2 or the factored form, 1M for calculating the discriminant, 1A for concluding that there is only one real root since discriminant is negative.

(a) The equation of the line \(L\) is \(y = mx \implies mx - y = 0\). The circle \(C\) can be written as \((x-5)^2 + y^2 = 9\), with center \(G(5, 0)\) and radius \(R = 3\). Since \(L\) is tangent to \(C\), the perpendicular distance from \(G\) to \(L\) is equal to the radius \(R\). Therefore, \(\frac{|5m - 0|}{\sqrt{m^2 + (-1)^2}} = 3 \implies |5m| = 3\sqrt{m^2 + 1} \implies 25m^2 = 9(m^2 + 1) \implies 16m^2 = 9\). Hence, \(m = \pm \frac{3}{4}\). (b) Since \(OA\) and \(OB\) are tangents to the circle \(C\) at \(A\) and \(B\) respectively, we have \(\angle OAG = \angle OBG = 90^\circ\). Thus, \(OAGB\) is a cyclic quadrilateral, and \(OG\) must be the diameter of the circle passing through \(O\), \(A\), and \(B\). The center of the new circle is the midpoint of \(OG\), which is \((2.5, 0)\), and its radius is \(\frac{5}{2} = 2.5\). The equation of the circle is \((x-2.5)^2 + y^2 = 2.5^2 \implies x^2 - 5x + 6.25 + y^2 = 6.25 \implies x^2 + y^2 - 5x = 0\).

(a) 1M for setting the line equation, 1M for using the distance from center to line equals radius, 1A for 16m^2 = 9, 1A for m = 3/4 or -3/4. (b) 1M for recognizing OG as the diameter of the circle, 1M for finding center and radius of the new circle, 1A for the final equation x^2 + y^2 - 5x = 0.

(a) Let \(x_1, x_2, \dots, x_8\) be the 8 numbers. Mean \(\bar{x} = \frac{\sum x_i}{8} = 15 \implies \sum x_i = 120\). The standard deviation \(\sigma = 4\), so variance \(\sigma^2 = 16\). Using \(\sigma^2 = \frac{\sum x_i^2}{8} - \bar{x}^2\), we have \(16 = \frac{\sum x_i^2}{8} - 15^2 \implies \frac{\sum x_i^2}{8} = 16 + 225 = 241 \implies \sum x_i^2 = 1928\). (b) When 9 and 21 are removed, the new sum of numbers is \(120 - 9 - 21 = 90\). The new mean is \(\bar{x}' = \frac{90}{6} = 15\). The new sum of squares is \(1928 - 9^2 - 21^2 = 1928 - 81 - 441 = 1406\). The new variance is \((\sigma')^2 = \frac{1406}{6} - (15)^2 = \frac{703}{3} - 225 = \frac{28}{3}\). The new standard deviation is \(\sigma' = \sqrt{\frac{28}{3}} \approx 3.06\).

(a) 1M for finding the sum = 120, 1M for setting up the variance formula, 1A for sum of squares = 1928. (b) 1A for new mean = 15, 1M for finding the new sum of squares = 1406, 1M for calculating the new variance, 1A for new standard deviation = 3.06 (accept 3.055 to 3.060).

(a) Let \(O\) be the centroid of the equilateral triangle \(ABC\). The height of \(\triangle ABC\) is \(12 \sin 60^\circ = 6\sqrt{3}\text{ cm}\). Since \(O\) is the centroid, \(AO = \frac{2}{3} \times 6\sqrt{3} = 4\sqrt{3}\text{ cm}\). In the right-angled triangle \(VOA\), \(VA = \sqrt{VO^2 + AO^2} = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{112} = 4\sqrt{7}\text{ cm} \approx 10.6\text{ cm}\). (b) Since \(M\) is the midpoint of \(BC\) and \(\triangle ABC\) is equilateral, \(AM \perp BC\). By symmetry, \(VM \perp BC\). The angle between face \(VBC\) and the base \(ABC\) is \(\angle VMO\). Since \(O\) is the centroid, \(OM = \frac{1}{3} \times 6\sqrt{3} = 2\sqrt{3}\text{ cm}\). In the right-angled triangle \(VOM\), \(\tan \angle VMO = \frac{VO}{OM} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}}\). Thus, \(\angle VMO = \tan^{-1}\left(\frac{4}{\sqrt{3}}\right) \approx 66.6^\circ\). (c) The line of greatest slope of face \(VBC\) is \(VM\), and the projection of \(VA\) onto plane \(VBC\) lies on \(VM\). Thus, the angle between the line \(VA\) and face \(VBC\) is \(\angle AVM\). In \(\triangle VAM\), we have \(VA = 4\sqrt{7}\text{ cm}\), \(AM = 6\sqrt{3}\text{ cm}\), and \(VM = \sqrt{VO^2 + OM^2} = \sqrt{8^2 + (2\sqrt{3})^2} = \sqrt{76} = 2\sqrt{19}\text{ cm}\). By the cosine rule in \(\triangle VAM\): \(\cos \angle AVM = \frac{VA^2 + VM^2 - AM^2}{2 \cdot VA \cdot VM} = \frac{112 + 76 - 108}{2 \cdot (4\sqrt{7}) \cdot (2\sqrt{19})} = \frac{80}{16\sqrt{133}} = \frac{5}{\sqrt{133}}\). Therefore, \(\angle AVM = \cos^{-1}\left(\frac{5}{\sqrt{133}}\right) \approx 64.3^\circ\).

(a) 1M for finding AO, 1A for VA = 4\sqrt{7} cm (or 10.6 cm). (b) 1M for identifying the angle is \angle VMO, 1A for 66.6 degrees. (c) 1M for identifying the required angle is \angle AVM, 1M for applying the cosine rule in \triangle VAM, 1A for 64.3 degrees.

(a) Let the arithmetic sequence have first term \(a\) and common difference \(d\). The first, second, and fifth terms are \(a\), \(a+d\), and \(a+4d\) respectively. Since they form a geometric sequence, we have: \((a+d)^2 = a(a+4d) \implies a^2 + 2ad + d^2 = a^2 + 4ad \implies d^2 = 2ad\). Since \(d \neq 0\), we can divide by \(d\) to get \(d = 2a\). The common ratio is \(r = \frac{a+d}{a} = \frac{a+2a}{a} = 3\). (b) (i) For the second geometric sequence, the first term is \(a\) and the common ratio is \(\frac{1}{r} = \frac{1}{3}\). Since the sum to infinity is 12: \(\frac{a}{1 - 1/3} = 12 \implies \frac{a}{2/3} = 12 \implies a = 8\). (ii) Since \(a = 8\), we have \(d = 2(8) = 16\). The sum of the first \(n\) terms of the arithmetic sequence is: \(S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(8) + (n-1)16] = \frac{n}{2}[16n] = 8n^2\). We require \(S_n > 2024 \implies 8n^2 > 2024 \implies n^2 > 253 \implies n > \sqrt{253} \approx 15.9\). Since \(n\) must be an integer, the least value of \(n\) is 16.

(a) 1M for setting up the relation (a+d)^2 = a(a+4d), 1M for simplifying to d = 2a, 1A for finding r = 3. (b) 1A for finding a = 8, 1M for establishing S_n = 8n^2, 1M for setting up the inequality 8n^2 > 2024, 1A for concluding that the least value of n is 16.