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A-LevelMathematicsSeries9709/12 · Feb/Mar 2025 · Q87 Marks

A geometric progression is such that its second term is – 120 and its sum to infinity is 160.

  1. (a)Find the common ratio.[4]
  2. (b)The first nine terms of the progression are now removed. Find the sum to infinity of the remaining terms of the progression.[3]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

  1. [B1] ar = -120 and a/(1-r) = 160
  2. [M1] (-120/r) / (1-r) = 160 or r / (1-r) * (-120/r) = 160 Elimination of either a or r. Condone ± errors for this mark.
  3. [A1] 4r² - 4r - 3[=0] or a² - 160a - 19200[=0] OE Rearrange to arrive at a three-term quadratic.
  4. [A1] r = -1/2 only
  5. [B1] [a =] 240 FT -120÷(their r), where -1<r<1, r≠0.
  6. [M1] 160 - 240(1-(-1/2)⁹) / (1-(-1/2)) With (their 240) and (their r) as long as -1<r<1, r≠0. Condone reversed subtraction.
  7. [A1] 5/16 or -0.3125 A0 for -0.313 without sight of -5/16 or -0.3125. Condone use of r = 3/2 to provide a second solution.

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About this question

Topic
Series, A-Level Mathematics (syllabus 9709). Structured question, 7 marks.
Source
A-Level Mathematics, Feb/Mar 2025 Paper 1 Variant 2, question 8.

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