Arithmetic & Geometric Progressions: Mixed AP/GP: find terms & convergence — EJC 2025 H2 Math Prelim Paper 1
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An arithmetic sequence \(u_1, u_2, u_3, \ldots\) has \(u_1 = 9\) and \(u_3 = b\), where \(b\) is a constant.
- Find \(u_2\) in terms of \(b\).
A geometric sequence \(v_1, v_2, v_3, \ldots\) has \(v_1 = 9\) and \(v_3 = b\).
- Find the possible values of \(v_2\) in terms of \(b\).
It is now given that \(u_2 - v_2 = 8\), and the geometric sequence \(v_1, v_2, v_3, \ldots\) is convergent.
- Find the value of \(b\). Hence find \(u_2\) and \(v_2\).
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In an arithmetic sequence, the middle term equals the average of its neighbours: \[ u_2 = \frac{u_1 + u_3}{2} = \frac{9 + b}{2} \]
\(u_2 = \dfrac{9+b}{2}\)In a geometric sequence, \(v_2^2 = v_1 v_3 = 9b\), so \(v_2 = \pm\sqrt{9b} = \pm 3\sqrt{b}\).
\(v_2 = \pm 3\sqrt{b}\)From \(u_2 - v_2 = 8\): \(v_2 = u_2 - 8 = \dfrac{9+b}{2} - 8 = \dfrac{b-7}{2}\).
Also \(v_2^2 = 9b\): \[\begin{aligned} \left(\frac{b-7}{2}\right)^2 &= 9b\\ b^2 - 14b + 49 &= 36b\\ b^2 - 50b + 49 &= 0\\ (b-1)(b-49) &= 0 \end{aligned}\]
So \(b=1\) or \(b=49\).
If \(b=49\): common ratio \(|r| = \sqrt{49/9} = \tfrac{7}{3} > 1\), so the sequence is not convergent. Reject.
Hence \(b = 1\): \[ u_2 = \frac{9+1}{2} = 5, \qquad v_2 = 5 - 8 = -3 \]