ESAT Maths 1 · free worked solution

ESAT Maths 1 quadratic sequence: find u₁₂

Quadratic-sequence solution: use three terms to find the coefficients and calculate u₁₂ = 279.

Quadratic sequencesSpecification M4.19Reviewed 9 August 2026

Answer first

Option D

The correct answer is

279279

Substitute the known term numbers to create three linear equations for a, b and c.

This is original UniGenius practice material from the reviewed Maths 1 bank. It is not an official UAT-UK question, past paper, or recalled question.

The question

Read the setup, then choose the shortest valid route

The nth term of a quadratic sequence is un=an2+bn+cu_n=a n^2+b n+c. Given u2=9u_2=9, u4=31u_4=31 and u9=156u_9=156, what is u12u_{12}?

  • A259259
  • B299299
  • C319319
  • D279279Answer
  • E339339

Worked solution

The route in 5 steps

  1. Substitution gives 4a+2b+c=94a+2b+c=9, 16a+4b+c=3116a+4b+c=31 and 81a+9b+c=15681a+9b+c=156.

  2. Subtracting the first equation from the second gives 6a+b=116a+b=11.

  3. Subtracting the second from the third gives 13a+b=2513a+b=25, so 7a=147a=14 and a=2a=2.

  4. Then b=−1b=-1 and substitution into the first equation gives c=3c=3.

  5. Therefore u12=2(12)2−12+3=279u_{12}=2(12)^2-12+3=279.

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