A-Level Mathematics May/June 2025 Q2: The first two terms of a geometric progression are 4 sin²θ, 8 sin³θ, where θ is an angl…

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  The first term is $a = 4 \sin^2\theta$. The second term is $ar = 8 \sin^3\theta$.

  First, find the common ratio, $r$: $r = \frac{\text{second term}}{\text{first term}} = \frac{8 \sin^3\theta}{4 \sin^2\theta} = 2\sin\theta$

  The sum to infinity of a geometric progression is given by the formula $S_\infty = \frac{a}{1-r}$, for $|r| < 1$. We are given that $S_\infty = \frac{1}{2}$.

  Substitute the expressions for $a$, $r$, and the value of $S_\infty$ into the formula: $\frac{1}{2} = \frac{4 \sin^2\theta}{1 - 2\sin\theta}$

  Now, rearrange this equation to solve for $\theta$. Cross-multiply to get: $1(1 - 2\sin\theta) = 2(4 \sin^2\theta)$ $1 - 2\sin\theta = 8 \sin^2\theta$

  This can be written as a quadratic equation in terms of $\sin\theta$: $8 \sin^2\theta + 2\sin\theta - 1 = 0$

  Let $x = \sin\theta$. The equation is $8x^2 + 2x - 1 = 0$. Factorise the quadratic: $(4x - 1)(2x + 1) = 0$

  This gives two possible solutions for $x$: $4x - 1 = 0 \implies x = \frac{1}{4}$ $2x + 1 = 0 \implies x = -\frac{1}{2}$

  So, we have two possible values for $\sin\theta$: $\sin\theta = \frac{1}{4}$ or $\sin\theta = -\frac{1}{2}$

  The question states that $0 < \theta < \frac{\pi}{2}$. In this interval (the first quadrant), $\sin\theta$ must be positive. Therefore, we must reject the solution $\sin\theta = -\frac{1}{2}$.

  The only valid solution is $\sin\theta = \frac{1}{4}$.

  The question asks for the answer in the form $\sin^{-1}k$. $\theta = \sin^{-1}\left(\frac{1}{4}\right)$

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  How the marks are awarded

  • B1 — The common ratio, $r$, is correctly found by dividing the second term by the first term, resulting in the unsimplified or simplified expression $r = \frac{8 \sin^3\theta}{4 \sin^2\theta}$ or $r = 2\sin\theta$.
  • M1 — The formula for the sum to infinity, $S_\infty = \frac{a}{1-r}$, is used correctly by substituting the given value of $S_\infty = \frac{1}{2}$ and the expressions for $a$ and $r$ to form the equation $\frac{1}{2} = \frac{4 \sin^2\theta}{1 - 2\sin\theta}$.
  • DM1 — The equation is correctly rearranged into the three-term quadratic $8 \sin^2\theta + 2\sin\theta - 1 = 0$ and a valid method (factorisation or quadratic formula) is used to solve for at least one value of $\sin\theta$.
  • A1 — The correct value of $\sin\theta = \frac{1}{4}$ is selected, the other solution is correctly rejected based on the given domain, and the final answer is stated in the required form $\theta = \sin^{-1}\left(\frac{1}{4}\right)$.

  Common mistakes

  • Calculating the common ratio incorrectly, for example as $r = \frac{1}{2\sin\theta}$ by inverting the division.
  • Making a sign error when rearranging the equation to form the quadratic, such as getting $8 \sin^2\theta - 2\sin\theta + 1 = 0$, which leads to no real solutions.
  • Failing to reject the negative solution for $\sin\theta$, and either giving two answers for $\theta$ or giving only the incorrect one.
  • Forgetting the condition for the sum to infinity to exist, $|r|<1$, which in this case means $|2\sin\theta|<1$. While not essential for this specific solution path, ignoring it leads to the failure to reject the invalid solution.

  Examiner tip: Recognise when a problem combines topics, and be prepared to treat trigonometric functions as algebraic variables to simplify and solve equations.

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