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Find the general terms of both sequences.

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TOPIC 01

2025 JM01

The negative square root is excluded by the given positive ratio.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Find the general terms of both sequences.

Sn=n2+2n,b1=2,b3=2a4,bn geometric with r>0S_n=n^2+2n,\quad b_1=2,\quad b_3=2a_4,\quad b_n\text{ geometric with }r>0

Official paper · jm01-2025 · II.1(a) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use aₙ=Sₙ−Sₙ₋₁ and separately check n=1.
Hint 2
Use b₃=b₁r² and the positive-ratio condition.
Worked solution
  1. Subtract consecutive partial sums; the same formula also gives a₁=3.

    an=(n2+2n)−[(n−1)2+2(n−1)]=2n+1a_n=(n^2+2n)-[(n-1)^2+2(n-1)]=2n+1
  2. Use a₄=9 to determine the ratio.

    2r2=18, r>0  ⟹  r=3,bn=2⋅3n−12r^2=18,\ r>0\implies r=3,\quad b_n=2\cdot3^{n-1}

aₙ=2n+1; bₙ=2·3ⁿ⁻¹.

Checks and common pitfalls: The negative square root is excluded by the given positive ratio.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Subtract consecutive partial sums; the same formula also gives a₁=3.
    an=(n2+2n)−[(n−1)2+2(n−1)]=2n+1a_n=(n^2+2n)-[(n-1)^2+2(n-1)]=2n+1
  • Use a₄=9 to determine the ratio.
    2r2=18, r>0  ⟹  r=3,bn=2⋅3n−12r^2=18,\ r>0\implies r=3,\quad b_n=2\cdot3^{n-1}

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