A-Level Mathematics October/November 2024 Q6: The first term of a convergent geometric progression is 10. The sum of the first 4 term…

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  The first term is $a = 10$. The sum of the first $n$ terms of a geometric progression is given by $S_n = \frac{a(1-r^n)}{1-r}$.

  The sum of the first 4 terms is $p = S_4 = \frac{10(1-r^4)}{1-r}$. The sum of the first 8 terms is $q = S_8 = \frac{10(1-r^8)}{1-r}$.

  We are given that $\frac{q}{p} = \frac{17}{16}$. Substituting the expressions for $p$ and $q$: $\frac{\frac{10(1-r^8)}{1-r}}{\frac{10(1-r^4)}{1-r}} = \frac{17}{16}$

  The terms $\frac{10}{1-r}$ cancel out, leaving: $\frac{1-r^8}{1-r^4} = \frac{17}{16}$

  Recognising $1-r^8$ as a difference of two squares, $(1-r^4)(1+r^4)$: $\frac{(1-r^4)(1+r^4)}{1-r^4} = \frac{17}{16}$ $1+r^4 = \frac{17}{16}$

  Now, we solve for $r$: $r^4 = \frac{17}{16} - 1$ $r^4 = \frac{1}{16}$ Taking the fourth root gives two possible real values for $r$: $r = \frac{1}{2} \quad \text{or} \quad r = -\frac{1}{2}$ Both values satisfy the condition for a convergent series, $|r|<1$.

  The sum to infinity is given by $S_\infty = \frac{a}{1-r}$. We find the two possible values.

  Case 1: $r = \frac{1}{2}$ $S_\infty = \frac{10}{1 - \frac{1}{2}} = \frac{10}{\frac{1}{2}} = 20$

  Case 2: $r = -\frac{1}{2}$ $S_\infty = \frac{10}{1 - (-\frac{1}{2})} = \frac{10}{1 + \frac{1}{2}} = \frac{10}{\frac{3}{2}} = \frac{20}{3}$

  The two possible values of the sum to infinity are 20 and $\frac{20}{3}$.

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  How the marks are awarded

  • M1 — For setting up the correct equation by substituting the formulas for the sum of a GP into the given ratio, i.e., $\frac{S_8}{S_4} = \frac{17}{16}$.
  • DM1 — For correctly simplifying the algebraic fraction by cancelling terms and using the difference of squares identity on $1-r^8$ to obtain the equation $1+r^4 = \frac{17}{16}$.
  • A1 — For correctly solving for $r$ to find $r^4 = \frac{1}{16}$ and obtaining the two real roots $r = \pm\frac{1}{2}$.
  • DM1 — For using the correct sum to infinity formula, $S_\infty = \frac{a}{1-r}$, with at least one of their found values of $r$ (where $|r|<1$).
  • A1 — For obtaining both correct final answers, 20 and $\frac{20}{3}$ (or 6.67 or better), with no extra incorrect answers.

  Common mistakes

  • After finding $r^4 = 1/16$, only considering the positive root $r = 1/2$ and therefore finding only one value for the sum to infinity.
  • Failing to simplify the fraction $\frac{1-r^8}{1-r^4}$ by not spotting the difference of squares, leading to a much more difficult or unsolvable equation.
  • Using an incorrect formula, such as the nth term formula $ar^{n-1}$ instead of the sum formula $S_n = \frac{a(1-r^n)}{1-r}$.
  • Making an algebraic slip when solving $1+r^4 = 17/16$, for example by forgetting to subtract the 1.

  Examiner tip: Practice recognising and applying algebraic identities, such as the difference of two squares, as they are often the key to simplifying complex expressions in series questions.

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