A-Level Mathematics May/June 2024 Q8(b): The first three terms of a geometric progression are 25, 4q-1 and 13-q, where q is a po…

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  For a geometric progression, the ratio of consecutive terms is constant. Let the terms be $u_1, u_2, u_3$.

  $u_1 = 25$ $u_2 = 4q-1$ $u_3 = 13-q$

  The common ratio $r$ is given by $r = \frac{u_2}{u_1} = \frac{u_3}{u_2}$.

  Therefore, $\frac{4q-1}{25} = \frac{13-q}{4q-1}$.

  Cross-multiplying gives: $(4q-1)^2 = 25(13-q)$ $16q^2 - 8q + 1 = 325 - 25q$ $16q^2 - 8q + 25q + 1 - 325 = 0$ $16q^2 + 17q - 324 = 0$

  To solve the quadratic equation, we can factorise it: $(16q+81)(q-4) = 0$

  This gives two possible solutions for $q$: $q = -\frac{81}{16}$ or $q = 4$

  Since the question states that $q$ is a positive constant, we must use $q=4$.

  Now we find the common ratio, $r$, using $q=4$. $u_2 = 4(4) - 1 = 15$ $r = \frac{u_2}{u_1} = \frac{15}{25} = \frac{3}{5}$

  For the sum to infinity to exist, we must have $|r| < 1$. Since $|\frac{3}{5}| < 1$, the sum to infinity exists.

  The formula for the sum to infinity is $S_\infty = \frac{a}{1-r}$, where $a$ is the first term.

  $a = 25$ and $r = \frac{3}{5}$.

  $S_\infty = \frac{25}{1 - \frac{3}{5}} = \frac{25}{\frac{2}{5}}$

  $S_\infty = 25 \times \frac{5}{2} = \frac{125}{2}$

  So, the sum to infinity is $\frac{125}{2}$ or $62.5$.

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  How the marks are awarded

  • M1 — The M1 is awarded for correctly using the property of a geometric progression, $u_2/u_1 = u_3/u_2$, to form the equation $(4q-1)^2 = 25(13-q)$ and expanding it into the correct 3-term quadratic equation $16q^2 + 17q - 324 = 0$.
  • M1 — This M1 is for a correct method to solve the 3-term quadratic equation, such as factorisation into $(q-4)(16q+81)=0$, and identifying the positive solution $q=4$ as required by the question.
  • A1 — The A1 mark is earned for correctly calculating the common ratio, $r$. After finding $q=4$, the terms are 25, 15, 9, so $r = 15/25 = 3/5$.
  • A1 — The final A1 mark is for correctly substituting the values of $a$ and $r$ into the sum to infinity formula, $S_\infty = a/(1-r)$, and calculating the final answer as $125/2$.

  Common mistakes

  • Confusing a geometric progression with an arithmetic progression and setting up the incorrect initial equation, e.g., $(4q-1) - 25 = (13-q) - (4q-1)$.
  • Making an algebraic error when expanding the brackets, e.g., $(4q-1)^2 = 16q^2 - 1$ or $16q^2+1$, leading to an incorrect quadratic equation.
  • Correctly solving the quadratic to find two values for $q$ but using the negative value ($q=-81/16$), instead of rejecting it as the question states $q$ is a positive constant.
  • Using an incorrect formula for the sum to infinity, such as $S_\infty = a/(r-1)$, or making a calculation error with the fractions, e.g., $1 - 3/5 = -2/5$.

  Examiner tip: Recognise that for any three consecutive terms $u_1, u_2, u_3$ of a geometric progression, the relationship $u_2^2 = u_1 u_3$ holds, which quickly leads to a solvable equation.

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