A-Level Mathematics May/June 2025 Q6: An arithmetic progression has first term a and common difference 2. The Nth term is 55…

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  An arithmetic progression has first term $a$ and common difference $d=2$. The $N$th term is 55 and the sum of the first $3N$ terms is 5760.

  Step 1: Formulate equations from the given information.

  The formula for the $n$th term of an arithmetic progression is $u_n = a + (n-1)d$. Given that the $N$th term is 55 and $d=2$: $u_N = a + (N-1)(2) = 55$ $a + 2N - 2 = 55$ $a = 57 - 2N$ (Equation 1)

  The formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. Given that the sum of the first $3N$ terms is 5760: $S_{3N} = \frac{3N}{2}[2a + (3N-1)(2)] = 5760$ (Equation 2)

  Step 2: Solve the simultaneous equations.

  Substitute Equation 1 into Equation 2 to eliminate $a$: $\frac{3N}{2}[2(57-2N) + (3N-1)(2)] = 5760$

  Simplify the expression inside the brackets: $\frac{3N}{2}[114 - 4N + 6N - 2] = 5760$ $\frac{3N}{2}[2N + 112] = 5760$

  Factor out 2 from the term in the brackets: $\frac{3N}{2} \times 2(N + 56) = 5760$ $3N(N + 56) = 5760$

  Divide both sides by 3: $N(N + 56) = 1920$

  Step 3: Solve the resulting quadratic equation for N.

  Expand the equation and set it to zero: $N^2 + 56N - 1920 = 0$

  This can be solved by factorising or using the quadratic formula. We look for two numbers that multiply to -1920 and add to 56. These are 80 and -24. This is not correct. Let's re-check. We need numbers with a difference of 56. Let's try the quadratic formula: $N = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $N = \frac{-56 \pm \sqrt{56^2 - 4(1)(-1920)}}{2(1)}$ $N = \frac{-56 \pm \sqrt{3136 + 7680}}{2}$ $N = \frac{-56 \pm \sqrt{10816}}{2}$ $N = \frac{-56 \pm 104}{2}$

  This gives two possible solutions for $N$: $N = \frac{-56 + 104}{2} = \frac{48}{2} = 24$ $N = \frac{-56 - 104}{2} = \frac{-160}{2} = -80$

  Since $N$ represents the number of terms, it must be a positive integer. Therefore, we reject $N=-80$. $N = 24$

  Step 4: Find the value of a.

  Substitute the value of $N=24$ back into Equation 1: $a = 57 - 2N$ $a = 57 - 2(24)$ $a = 57 - 48$ $a = 9$

  Final Answer: The values are $N=24$ and $a=9$.

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  How the marks are awarded

  • B1 — Correctly forming an equation for the Nth term: $a + (N-1)(2) = 55$. This is seen in Step 1.
  • B1 — Correctly forming an equation for the sum of the first 3N terms: $S_{3N} = \frac{3N}{2}[2a + (3N-1)(2)] = 5760$. This is also in Step 1.
  • M1 — Attempting to solve the simultaneous equations by eliminating one variable. The model answer does this by substituting the expression for $a$ from the first equation into the second.
  • A1 — Correctly simplifying the equations to form a valid quadratic in a single variable, such as $N^2 + 56N - 1920 = 0$.
  • A1 — Correctly solving the quadratic to find $N=24$. The other solution, $N=-80$, is correctly rejected as it is not a valid number of terms.
  • A1 — Correctly substituting the value of $N$ back into an earlier equation to find the correct value of $a=9$.

  Common mistakes

  • Using $N$ instead of $3N$ in the sum formula, for example writing $S_N = 5760$ or $S_{3N} = \frac{N}{2}[...]$.
  • Making algebraic errors when simplifying after substitution, such as incorrect expansion of brackets or sign errors, leading to an incorrect quadratic.
  • Incorrectly solving the quadratic equation, for example by making an error in the quadratic formula or by attempting to solve $N(N+56)=1920$ as $N=1920$ or $N+56=1920$.
  • Mixing up the formulae for arithmetic and geometric progressions.

  Examiner tip: This question rewards the ability to translate worded information into a system of simultaneous equations using the correct standard formulae for sequences and series.

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  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

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