2025 EJC P1 Q7

2025 EJC P1 Q7

9 marks
Prelims

Do not use a calculator in answering this question.

An arithmetic sequence \({{u}_{1}}\), \({{u}_{2}}\), \({{u}_{3}}\), \(\ldots \) has \({{u}_{1}}=9\) and \({{u}_{3}}=b\), where \(b\) is a constant.

  1. Find \({{u}_{2}}\) in terms of \(b\).[2]

A geometric sequence \({{v}_{1}}\), \({{v}_{2}}\), \({{v}_{3}}\), \(\ldots \) has \({{v}_{1}}=9\) and \({{v}_{3}}=b\).

  1. Find the possible values of \({{v}_{2}}\) in terms of \(b\).[2]

It is now given that \({{u}_{2}}-{{v}_{2}}=8\), and the geometric sequence \({{v}_{1}}\), \({{v}_{2}}\), \({{v}_{3}}\), \(\ldots \) is convergent.

  1. Find the value of \(b\).
    Hence find \({{u}_{2}}\) and \({{v}_{2}}\).[5]

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Answer:(a) \(u_2=\frac{b+9}{2}\) (b) \(v_2=\pm3\sqrt b\) (c) \(b=1,\ u_2=5,\ v_2=-3\)

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